QUANTITATIVE TRAIT LOCI (QTL) MAPPING
1 MATHEMATICAL PROBLEMS IN QTL MAPPING
1.1 PROBLEM 1: LOD SCORE CALCULATION
Calculate the LOD score for a putative QTL given the following information:
• Likelihood under the alternative hypothesis (QTL present): 1.5 × 10−12
• Likelihood under the null hypothesis (no QTL): 2.7 × 10−15
Solution: The LOD score is calculated as:
𝐿𝑂𝐷 = log10 (Likelihood (QTL present)
Likelihood (no QTL) )
=log10 (1.5×10−12
2.7×10−15)
=log10(555.56)
= 2.74
Therefore, the
LOD score for this putative QTL is 2.74.
1.2 PROBLEM 2: HERITABILITY ESTIMATION
In a QTL mapping study for plant height in rice, you’ve identified 5 QTLs with the following
effect sizes:
• QTL 1: 15% of phenotypic variance
• QTL 2: 10% of phenotypic variance
• QTL 3: 8% of phenotypic variance
• QTL 4: 5% of phenotypic variance
• QTL 5: 3% of phenotypic variance
Calculate the total heritability explained by these QTLs. If the total heritability of plant height is
estimated to be 0.65, what percentage of the genetic variance remains unexplained?
Solution:
Step 1: Calculate total heritability explained by QTLs
ℎ𝑄𝑇𝐿
2= 0.15 + 0.10 + 0.08 + 0.05 + 0.03
= 0.41
Step 2: Calculate unexplained heritability
ℎ𝑢𝑛𝑒𝑥𝑝𝑙𝑎𝑖𝑛𝑒𝑑
2= ℎ𝑡𝑜𝑡𝑎𝑙
2− ℎ𝑄𝑇𝐿
2
= 0.65 − 0.41
= 0.24
Step 3: Calculate percentage of unexplained genetic variance
Percentage unexplained =ℎ𝑢𝑛𝑒𝑥𝑝𝑙𝑎𝑖𝑛𝑒𝑑
2
ℎ𝑡𝑜𝑡𝑎𝑙
2×100%
=0.24
0.65 ×100%
=36.92%
Therefore, 36.92% of the genetic variance remains unexplained.
1.3 PROBLEM 3: RECOMBINATION FRACTION ESTIMATION
In a QTL mapping study, you observe the following genotype frequencies in an F2 population of
1000 individuals:
• AA BB: 240
• AA Bb: 260
• AA bb: 10
• Aa BB: 255
• Aa Bb: 220
• Aa bb: 5
• aa BB: 8
• aa Bb: 2
• aa bb: 0
Estimate the recombination fraction between loci A and B.
Solution:
Step 1: Calculate the number of recombinant gametes Recombinant gametes are AB and ab.
𝑁𝐴𝐵 = 2(240)+255 +260 + 8 = 1003
𝑁𝑎𝑏 = 2(0)+5+2+10 =17
Total recombinant =1003 +17 =1020
Step 2: Calculate the total number of gametes
Total gametes = 2 × 1000 =2000
Step 3: Calculate the recombination fraction
𝑟 = Number of recombinant gametes
Total number of gametes
=1020
2000
= 0.51
The estimated recombination fraction between loci A and B is 0.51, suggesting that these loci
are unlinked (as r ≈ 0.5).
1.4 PROBLEM 4: QTL EFFECT SIZE CALCULATION
In a QTL mapping study for seed weight in soybeans, you’ve identified a QTL with the following
genotypic values:
• AA: 25g
• Aa: 22g
• aa: 18g
Calculate the additive (a) and dominance (d) effects of this QTL.
Solution:
Step 1: Calculate the additive effect (a) The additive effect is half the difference between the
homozygous genotypes:
𝑎 = 1
2(𝐴𝐴 −𝑎𝑎)
=1
2(25𝑔 − 18𝑔)
= 3.5𝑔
Step 2: Calculate the dominance effect (d) The dominance effect is the deviation of the
heterozygote from the average of the homozygotes:
𝑑 = 𝐴𝑎 −1
2(𝐴𝐴 +𝑎𝑎)
=22𝑔 − 1
2(25𝑔 + 18𝑔)
=22𝑔 − 21.5𝑔
= 0.5𝑔
Therefore, the additive effect (a) is 3.5g and the dominance effect (d) is 0.5g.
1.5 PROBLEM 5: POWER CALCULATION
You’re planning a QTL mapping study and want to determine the sample size needed to detect
a QTL that explains 5% of the phenotypic variance with 80% power at a significance level of
0.05. Assume you’re using an F2 population and a single marker analysis. Use the
approximation formula:
𝑛 = 2(𝑧𝛼/2 + 𝑧𝛽)2
ℎ𝑞
2
where 𝑛 is the required sample size, 𝑧𝛼/2 is the critical value of the standard normal distribution
for 𝛼/2, 𝑧𝛽 is the critical value for 𝛽 (1 - power), and ℎ𝑞
2 is the heritability explained by the QTL.
Solution:
Given:
• ℎ𝑞
2= 0.05 (5% of phenotypic variance)
• Power = 80%, so 𝛽 = 0.2
• Significance level 𝛼 = 0.05
Step 1: Find critical values
𝑧𝛼/2 = 𝑧0.025 = 1.96 (from standard normal table)
𝑧𝛽= 𝑧0.2 = 0.84 (from standard normal table)
Step 2: Apply the formula
𝑛 = 2(𝑧𝛼/2 + 𝑧𝛽)2
ℎ𝑞
2
=2(1.96 + 0.84)2
0.05
=2(2.8)2
0.05
=15.68
0.05
=313.6
Rounding up, you need a sample size of at least 314 F2 individuals to detect this QTL with 80%
power at a significance level of 0.05.
1.6 PROBLEM 1: LOD SCORE CALCULATION
Calculate the LOD score for a putative QTL given the following information:
• Likelihood under the alternative hypothesis (QTL present): 1.5 × 10−12
• Likelihood under the null hypothesis (no QTL): 2.7 × 10−15
Solution: The LOD score is calculated as:
𝐿𝑂𝐷 = log10 (Likelihood (QTL present)
Likelihood (no QTL) )
=log10 (1.5×10−12
2.7×10−15)
=log10(555.56)
= 2.74
Therefore, the
LOD score for this putative QTL is 2.74.
1.7 PROBLEM 2: HERITABILITY ESTIMATION
In a QTL mapping study for plant height in rice, you’ve identified 5 QTLs with the following
effect sizes:
• QTL 1: 15% of phenotypic variance
• QTL 2: 10% of phenotypic variance
• QTL 3: 8% of phenotypic variance
• QTL 4: 5% of phenotypic variance
• QTL 5: 3% of phenotypic variance
Calculate the total heritability explained by these QTLs. If the total heritability of plant height is
estimated to be 0.65, what percentage of the genetic variance remains unexplained?
Solution:
Step 1: Calculate total heritability explained by QTLs
ℎ𝑄𝑇𝐿
2= 0.15 + 0.10 + 0.08 + 0.05 + 0.03
= 0.41
Step 2: Calculate unexplained heritability
ℎ𝑢𝑛𝑒𝑥𝑝𝑙𝑎𝑖𝑛𝑒𝑑
2= ℎ𝑡𝑜𝑡𝑎𝑙
2− ℎ𝑄𝑇𝐿
2
= 0.65 − 0.41
= 0.24
Step 3: Calculate percentage of unexplained genetic variance
Percentage unexplained =ℎ𝑢𝑛𝑒𝑥𝑝𝑙𝑎𝑖𝑛𝑒𝑑
2
ℎ𝑡𝑜𝑡𝑎𝑙
2×100%
=0.24
0.65 ×100%
=36.92%
Therefore, 36.92% of the genetic variance remains unexplained.
1.8 PROBLEM 3: RECOMBINATION FRACTION ESTIMATION
In a QTL mapping study, you observe the following genotype frequencies in an F2 population of
1000 individuals:
• AA BB: 240
• AA Bb: 260
• AA bb: 10
• Aa BB: 255
• Aa Bb: 220
• Aa bb: 5
• aa BB: 8
• aa Bb: 2
• aa bb: 0
Estimate the recombination fraction between loci A and B.
Solution:
Step 1: Calculate the number of recombinant gametes Recombinant gametes are AB and ab.
𝑁𝐴𝐵 = 2(240)+255 +260 + 8 = 1003
𝑁𝑎𝑏 = 2(0)+5+2+10 =17
Total recombinant =1003 +17 =1020
Step 2: Calculate the total number of gametes
Total gametes = 2 × 1000 =2000
Step 3: Calculate the recombination fraction
𝑟 = Number of recombinant gametes
Total number of gametes
=1020
2000
= 0.51
The estimated recombination fraction between loci A and B is 0.51, suggesting that these loci
are unlinked (as r ≈ 0.5).
1.9 PROBLEM 4: QTL EFFECT SIZE CALCULATION
In a QTL mapping study for seed weight in soybeans, you’ve identified a QTL with the following
genotypic values:
• AA: 25g
• Aa: 22g
• aa: 18g
Calculate the additive (a) and dominance (d) effects of this QTL.
Solution:
Step 1: Calculate the additive effect (a) The additive effect is half the difference between the
homozygous genotypes:
𝑎 = 1
2(𝐴𝐴 −𝑎𝑎)
=1
2(25𝑔 − 18𝑔)
= 3.5𝑔
Step 2: Calculate the dominance effect (d) The dominance effect is the deviation of the
heterozygote from the average of the homozygotes:
𝑑 = 𝐴𝑎 −1
2(𝐴𝐴 +𝑎𝑎)
=22𝑔 − 1
2(25𝑔 + 18𝑔)
=22𝑔 − 21.5𝑔
= 0.5𝑔
Therefore, the additive effect (a) is 3.5g and the dominance effect (d) is 0.5g.
1.10 PROBLEM 5: POWER CALCULATION
You’re planning a QTL mapping study and want to determine the sample size needed to detect
a QTL that explains 5% of the phenotypic variance with 80% power at a significance level of
0.05. Assume you’re using an F2 population and a single marker analysis. Use the
approximation formula:
𝑛 = 2(𝑧𝛼/2 + 𝑧𝛽)2
ℎ𝑞
2
where 𝑛 is the required sample size, 𝑧𝛼/2 is the critical value of the standard normal distribution
for 𝛼/2, 𝑧𝛽 is the critical value for 𝛽 (1 - power), and ℎ𝑞
2 is the heritability explained by the QTL.
Solution:
Given:
• ℎ𝑞
2= 0.05 (5% of phenotypic variance)
• Power = 80%, so 𝛽 = 0.2
• Significance level 𝛼 = 0.05
Step 1: Find critical values
𝑧𝛼/2 = 𝑧0.025 = 1.96 (from standard normal table)
𝑧𝛽= 𝑧0.2 = 0.84 (from standard normal table)
Step 2: Apply the formula
𝑛 = 2(𝑧𝛼/2 + 𝑧𝛽)2
ℎ𝑞
2
=2(1.96 + 0.84)2
0.05
=2(2.8)2
0.05
=15.68
0.05
=313.6
Rounding up, you need a sample size of at least 314 F2 individuals to detect this QTL with 80%
power at a significance level of 0.05.
1.11 PROBLEM 1: LOD SCORE CALCULATION
Calculate the LOD score for a putative QTL given the following information:
• Likelihood under the alternative hypothesis (QTL present): 1.5 × 10−12
• Likelihood under the null hypothesis (no QTL): 2.7 × 10−15
Solution: The LOD score is calculated as:
𝐿𝑂𝐷 = log10 (Likelihood (QTL present)
Likelihood (no QTL) )
=log10 (1.5×10−12
2.7×10−15)
=log10(555.56)
= 2.74
Therefore, the
LOD score for this putative QTL is 2.74.
1.12 PROBLEM 2: HERITABILITY ESTIMATION
In a QTL mapping study for plant height in rice, you’ve identified 5 QTLs with the following
effect sizes:
• QTL 1: 15% of phenotypic variance
• QTL 2: 10% of phenotypic variance
• QTL 3: 8% of phenotypic variance
• QTL 4: 5% of phenotypic variance
• QTL 5: 3% of phenotypic variance
Calculate the total heritability explained by these QTLs. If the total heritability of plant height is
estimated to be 0.65, what percentage of the genetic variance remains unexplained?
Solution:
Step 1: Calculate total heritability explained by QTLs
ℎ𝑄𝑇𝐿
2= 0.15 + 0.10 + 0.08 + 0.05 + 0.03
= 0.41
Step 2: Calculate unexplained heritability
ℎ𝑢𝑛𝑒𝑥𝑝𝑙𝑎𝑖𝑛𝑒𝑑
2= ℎ𝑡𝑜𝑡𝑎𝑙
2− ℎ𝑄𝑇𝐿
2
= 0.65 − 0.41
= 0.24
Step 3: Calculate percentage of unexplained genetic variance
Percentage unexplained =ℎ𝑢𝑛𝑒𝑥𝑝𝑙𝑎𝑖𝑛𝑒𝑑
2
ℎ𝑡𝑜𝑡𝑎𝑙
2×100%
=0.24
0.65 ×100%
=36.92%
Therefore, 36.92% of the genetic variance remains unexplained.
1.13 PROBLEM 3: RECOMBINATION FRACTION ESTIMATION
In a QTL mapping study, you observe the following genotype frequencies in an F2 population of
1000 individuals:
• AA BB: 240
• AA Bb: 260
• AA bb: 10
• Aa BB: 255
• Aa Bb: 220
• Aa bb: 5
• aa BB: 8
• aa Bb: 2
• aa bb: 0
Estimate the recombination fraction between loci A and B.
Solution:
Step 1: Calculate the number of recombinant gametes Recombinant gametes are AB and ab.
𝑁𝐴𝐵 = 2(240)+255 +260 + 8 = 1003
𝑁𝑎𝑏 = 2(0)+5+2+10 =17
Total recombinant =1003 +17 =1020
Step 2: Calculate the total number of gametes
Total gametes = 2 × 1000 =2000
Step 3: Calculate the recombination fraction
𝑟 = Number of recombinant gametes
Total number of gametes
=1020
2000
= 0.51
The estimated recombination fraction between loci A and B is 0.51, suggesting that these loci
are unlinked (as r ≈ 0.5).
1.14 PROBLEM 4: QTL EFFECT SIZE CALCULATION
In a QTL mapping study for seed weight in soybeans, you’ve identified a QTL with the following
genotypic values:
• AA: 25g
• Aa: 22g
• aa: 18g
Calculate the additive (a) and dominance (d) effects of this QTL.
Solution:
Step 1: Calculate the additive effect (a) The additive effect is half the difference between the
homozygous genotypes:
𝑎 = 1
2(𝐴𝐴 −𝑎𝑎)
=1
2(25𝑔 − 18𝑔)
= 3.5𝑔
Step 2: Calculate the dominance effect (d) The dominance effect is the deviation of the
heterozygote from the average of the homozygotes:
𝑑 = 𝐴𝑎 −1
2(𝐴𝐴 +𝑎𝑎)
=22𝑔 − 1
2(25𝑔 + 18𝑔)
=22𝑔 − 21.5𝑔
= 0.5𝑔
Therefore, the additive effect (a) is 3.5g and the dominance effect (d) is 0.5g.
1.15 PROBLEM 5: POWER CALCULATION
You’re planning a QTL mapping study and want to determine the sample size needed to detect
a QTL that explains 5% of the phenotypic variance with 80% power at a significance level of
0.05. Assume you’re using an F2 population and a single marker analysis. Use the
approximation formula:
𝑛 = 2(𝑧𝛼/2 + 𝑧𝛽)2
ℎ𝑞
2
where 𝑛 is the required sample size, 𝑧𝛼/2 is the critical value of the standard normal distribution
for 𝛼/2, 𝑧𝛽 is the critical value for 𝛽 (1 - power), and ℎ𝑞
2 is the heritability explained by the QTL.
Solution:
Given:
• ℎ𝑞
2= 0.05 (5% of phenotypic variance)
• Power = 80%, so 𝛽 = 0.2
• Significance level 𝛼 = 0.05
Step 1: Find critical values
𝑧𝛼/2 = 𝑧0.025 = 1.96 (from standard normal table)
𝑧𝛽= 𝑧0.2 = 0.84 (from standard normal table)
Step 2: Apply the formula
𝑛 = 2(𝑧𝛼/2 + 𝑧𝛽)2
ℎ𝑞
2
=2(1.96 + 0.84)2
0.05
=2(2.8)2
0.05
=15.68
0.05
=313.6
Rounding up, you need a sample size of at least 314 F2 individuals to detect this QTL with 80%
power at a significance level of 0.05.
1.16 PROBLEM 1: LOD SCORE CALCULATION
Calculate the LOD score for a putative QTL given the following information:
• Likelihood under the alternative hypothesis (QTL present): 1.5 × 10−12
• Likelihood under the null hypothesis (no QTL): 2.7 × 10−15
Solution: The LOD score is calculated as:
𝐿𝑂𝐷 = log10 (Likelihood (QTL present)
Likelihood (no QTL) )
=log10 (1.5×10−12
2.7×10−15)
=log10(555.56)
= 2.74
Therefore, the
LOD score for this putative QTL is 2.74.
1.17 PROBLEM 2: HERITABILITY ESTIMATION
In a QTL mapping study for plant height in rice, you’ve identified 5 QTLs with the following
effect sizes:
• QTL 1: 15% of phenotypic variance
• QTL 2: 10% of phenotypic variance
• QTL 3: 8% of phenotypic variance
• QTL 4: 5% of phenotypic variance
• QTL 5: 3% of phenotypic variance
Calculate the total heritability explained by these QTLs. If the total heritability of plant height is
estimated to be 0.65, what percentage of the genetic variance remains unexplained?
Solution:
Step 1: Calculate total heritability explained by QTLs
ℎ𝑄𝑇𝐿
2= 0.15 + 0.10 + 0.08 + 0.05 + 0.03
= 0.41
Step 2: Calculate unexplained heritability
ℎ𝑢𝑛𝑒𝑥𝑝𝑙𝑎𝑖𝑛𝑒𝑑
2= ℎ𝑡𝑜𝑡𝑎𝑙
2− ℎ𝑄𝑇𝐿
2
= 0.65 − 0.41
= 0.24
Step 3: Calculate percentage of unexplained genetic variance
Percentage unexplained =ℎ𝑢𝑛𝑒𝑥𝑝𝑙𝑎𝑖𝑛𝑒𝑑
2
ℎ𝑡𝑜𝑡𝑎𝑙
2×100%
=0.24
0.65 ×100%
=36.92%
Therefore, 36.92% of the genetic variance remains unexplained.
1.18 PROBLEM 3: RECOMBINATION FRACTION ESTIMATION
In a QTL mapping study, you observe the following genotype frequencies in an F2 population of
1000 individuals:
• AA BB: 240
• AA Bb: 260
• AA bb: 10
• Aa BB: 255
• Aa Bb: 220
• Aa bb: 5
• aa BB: 8
• aa Bb: 2
• aa bb: 0
Estimate the recombination fraction between loci A and B.
Solution:
Step 1: Calculate the number of recombinant gametes Recombinant gametes are AB and ab.
𝑁𝐴𝐵 = 2(240)+255 +260 + 8 = 1003
𝑁𝑎𝑏 = 2(0)+5+2+10 =17
Total recombinant =1003 +17 =1020
Step 2: Calculate the total number of gametes
Total gametes = 2 × 1000 =2000
Step 3: Calculate the recombination fraction
𝑟 = Number of recombinant gametes
Total number of gametes
=1020
2000
= 0.51
The estimated recombination fraction between loci A and B is 0.51, suggesting that these loci
are unlinked (as r ≈ 0.5).
1.19 PROBLEM 4: QTL EFFECT SIZE CALCULATION
In a QTL mapping study for seed weight in soybeans, you’ve identified a QTL with the following
genotypic values:
• AA: 25g
• Aa: 22g
• aa: 18g
Calculate the additive (a) and dominance (d) effects of this QTL.
Solution:
Step 1: Calculate the additive effect (a) The additive effect is half the difference between the
homozygous genotypes:
𝑎 = 1
2(𝐴𝐴 −𝑎𝑎)
=1
2(25𝑔 − 18𝑔)
= 3.5𝑔
Step 2: Calculate the dominance effect (d) The dominance effect is the deviation of the
heterozygote from the average of the homozygotes:
𝑑 = 𝐴𝑎 −1
2(𝐴𝐴 +𝑎𝑎)
=22𝑔 − 1
2(25𝑔 + 18𝑔)
=22𝑔 − 21.5𝑔
= 0.5𝑔
Therefore, the additive effect (a) is 3.5g and the dominance effect (d) is 0.5g.
1.20 PROBLEM 5: POWER CALCULATION
You’re planning a QTL mapping study and want to determine the sample size needed to detect
a QTL that explains 5% of the phenotypic variance with 80% power at a significance level of
0.05. Assume you’re using an F2 population and a single marker analysis. Use the
approximation formula:
𝑛 = 2(𝑧𝛼/2 + 𝑧𝛽)2
ℎ𝑞
2
where 𝑛 is the required sample size, 𝑧𝛼/2 is the critical value of the standard normal distribution
for 𝛼/2, 𝑧𝛽 is the critical value for 𝛽 (1 - power), and ℎ𝑞
2 is the heritability explained by the QTL.
Solution:
Given:
• ℎ𝑞
2= 0.05 (5% of phenotypic variance)
• Power = 80%, so 𝛽 = 0.2
• Significance level 𝛼 = 0.05
Step 1: Find critical values
𝑧𝛼/2 = 𝑧0.025 = 1.96 (from standard normal table)
𝑧𝛽= 𝑧0.2 = 0.84 (from standard normal table)
Step 2: Apply the formula
𝑛 = 2(𝑧𝛼/2 + 𝑧𝛽)2
ℎ𝑞
2
=2(1.96 + 0.84)2
0.05
=2(2.8)2
0.05
=15.68
0.05
=313.6
Rounding up, you need a sample size of at least 314 F2 individuals to detect this QTL with 80%
power at a significance level of 0.05.
1.21 PROBLEM 1: LOD SCORE CALCULATION
Calculate the LOD score for a putative QTL given the following information:
• Likelihood under the alternative hypothesis (QTL present): 1.5 × 10−12
• Likelihood under the null hypothesis (no QTL): 2.7 × 10−15
Solution: The LOD score is calculated as:
𝐿𝑂𝐷 = log10 (Likelihood (QTL present)
Likelihood (no QTL) )
=log10 (1.5×10−12
2.7×10−15)
=log10(555.56)
= 2.74
Therefore, the
LOD score for this putative QTL is 2.74.
1.22 PROBLEM 2: HERITABILITY ESTIMATION
In a QTL mapping study for plant height in rice, you’ve identified 5 QTLs with the following
effect sizes:
• QTL 1: 15% of phenotypic variance
• QTL 2: 10% of phenotypic variance
• QTL 3: 8% of phenotypic variance
• QTL 4: 5% of phenotypic variance
• QTL 5: 3% of phenotypic variance
Calculate the total heritability explained by these QTLs. If the total heritability of plant height is
estimated to be 0.65, what percentage of the genetic variance remains unexplained?
Solution:
Step 1: Calculate total heritability explained by QTLs
ℎ𝑄𝑇𝐿
2= 0.15 + 0.10 + 0.08 + 0.05 + 0.03
= 0.41
Step 2: Calculate unexplained heritability
ℎ𝑢𝑛𝑒𝑥𝑝𝑙𝑎𝑖𝑛𝑒𝑑
2= ℎ𝑡𝑜𝑡𝑎𝑙
2− ℎ𝑄𝑇𝐿
2
= 0.65 − 0.41
= 0.24
Step 3: Calculate percentage of unexplained genetic variance
Percentage unexplained =ℎ𝑢𝑛𝑒𝑥𝑝𝑙𝑎𝑖𝑛𝑒𝑑
2
ℎ𝑡𝑜𝑡𝑎𝑙
2×100%
=0.24
0.65 ×100%
=36.92%
Therefore, 36.92% of the genetic variance remains unexplained.
1.23 PROBLEM 3: RECOMBINATION FRACTION ESTIMATION
In a QTL mapping study, you observe the following genotype frequencies in an F2 population of
1000 individuals:
• AA BB: 240
• AA Bb: 260
• AA bb: 10
• Aa BB: 255
• Aa Bb: 220
• Aa bb: 5
• aa BB: 8
• aa Bb: 2
• aa bb: 0
Estimate the recombination fraction between loci A and B.
Solution:
Step 1: Calculate the number of recombinant gametes Recombinant gametes are AB and ab.
𝑁𝐴𝐵 = 2(240)+255 +260 + 8 = 1003
𝑁𝑎𝑏 = 2(0)+5+2+10 =17
Total recombinant =1003 +17 =1020
Step 2: Calculate the total number of gametes
Total gametes = 2 × 1000 =2000
Step 3: Calculate the recombination fraction
𝑟 = Number of recombinant gametes
Total number of gametes
=1020
2000
= 0.51
The estimated recombination fraction between loci A and B is 0.51, suggesting that these loci
are unlinked (as r ≈ 0.5).
1.24 PROBLEM 4: QTL EFFECT SIZE CALCULATION
In a QTL mapping study for seed weight in soybeans, you’ve identified a QTL with the following
genotypic values:
• AA: 25g
• Aa: 22g
• aa: 18g
Calculate the additive (a) and dominance (d) effects of this QTL.
Solution:
Step 1: Calculate the additive effect (a) The additive effect is half the difference between the
homozygous genotypes:
𝑎 = 1
2(𝐴𝐴 −𝑎𝑎)
=1
2(25𝑔 − 18𝑔)
= 3.5𝑔
Step 2: Calculate the dominance effect (d) The dominance effect is the deviation of the
heterozygote from the average of the homozygotes:
𝑑 = 𝐴𝑎 −1
2(𝐴𝐴 +𝑎𝑎)
=22𝑔 − 1
2(25𝑔 + 18𝑔)
=22𝑔 − 21.5𝑔
= 0.5𝑔
Therefore, the additive effect (a) is 3.5g and the dominance effect (d) is 0.5g.
1.25 PROBLEM 5: POWER CALCULATION
You’re planning a QTL mapping study and want to determine the sample size needed to detect
a QTL that explains 5% of the phenotypic variance with 80% power at a significance level of
0.05. Assume you’re using an F2 population and a single marker analysis. Use the
approximation formula:
𝑛 = 2(𝑧𝛼/2 + 𝑧𝛽)2
ℎ𝑞
2
where 𝑛 is the required sample size, 𝑧𝛼/2 is the critical value of the standard normal distribution
for 𝛼/2, 𝑧𝛽 is the critical value for 𝛽 (1 - power), and ℎ𝑞
2 is the heritability explained by the QTL.
Solution:
Given:
• ℎ𝑞
2= 0.05 (5% of phenotypic variance)
• Power = 80%, so 𝛽 = 0.2
• Significance level 𝛼 = 0.05
Step 1: Find critical values
𝑧𝛼/2 = 𝑧0.025 = 1.96 (from standard normal table)
𝑧𝛽= 𝑧0.2 = 0.84 (from standard normal table)
Step 2: Apply the formula
𝑛 = 2(𝑧𝛼/2 + 𝑧𝛽)2
ℎ𝑞
2
=2(1.96 + 0.84)2
0.05
=2(2.8)2
0.05
=15.68
0.05
=313.6
Rounding up, you need a sample size of at least 314 F2 individuals to detect this QTL with 80%
power at a significance level of 0.05.
1.26 PROBLEM 1: LOD SCORE CALCULATION
Calculate the LOD score for a putative QTL given the following information:
• Likelihood under the alternative hypothesis (QTL present): 1.5 × 10−12
• Likelihood under the null hypothesis (no QTL): 2.7 × 10−15
Solution: The LOD score is calculated as:
𝐿𝑂𝐷 = log10 (Likelihood (QTL present)
Likelihood (no QTL) )
=log10 (1.5×10−12
2.7×10−15)
=log10(555.56)
= 2.74
Therefore, the
LOD score for this putative QTL is 2.74.
1.27 PROBLEM 2: HERITABILITY ESTIMATION
In a QTL mapping study for plant height in rice, you’ve identified 5 QTLs with the following
effect sizes:
• QTL 1: 15% of phenotypic variance
• QTL 2: 10% of phenotypic variance
• QTL 3: 8% of phenotypic variance
• QTL 4: 5% of phenotypic variance
• QTL 5: 3% of phenotypic variance
Calculate the total heritability explained by these QTLs. If the total heritability of plant height is
estimated to be 0.65, what percentage of the genetic variance remains unexplained?
Solution:
Step 1: Calculate total heritability explained by QTLs
ℎ𝑄𝑇𝐿
2= 0.15 + 0.10 + 0.08 + 0.05 + 0.03
= 0.41
Step 2: Calculate unexplained heritability
ℎ𝑢𝑛𝑒𝑥𝑝𝑙𝑎𝑖𝑛𝑒𝑑
2= ℎ𝑡𝑜𝑡𝑎𝑙
2− ℎ𝑄𝑇𝐿
2
= 0.65 − 0.41
= 0.24
Step 3: Calculate percentage of unexplained genetic variance
Percentage unexplained =ℎ𝑢𝑛𝑒𝑥𝑝𝑙𝑎𝑖𝑛𝑒𝑑
2
ℎ𝑡𝑜𝑡𝑎𝑙
2×100%
=0.24
0.65 ×100%
=36.92%
Therefore, 36.92% of the genetic variance remains unexplained.
1.28 PROBLEM 3: RECOMBINATION FRACTION ESTIMATION
In a QTL mapping study, you observe the following genotype frequencies in an F2 population of
1000 individuals:
• AA BB: 240
• AA Bb: 260
• AA bb: 10
• Aa BB: 255
• Aa Bb: 220
• Aa bb: 5
• aa BB: 8
• aa Bb: 2
• aa bb: 0
Estimate the recombination fraction between loci A and B.
Solution:
Step 1: Calculate the number of recombinant gametes Recombinant gametes are AB and ab.
𝑁𝐴𝐵 = 2(240)+255 +260 + 8 = 1003
𝑁𝑎𝑏 = 2(0)+5+2+10 =17
Total recombinant =1003 +17 =1020
Step 2: Calculate the total number of gametes
Total gametes = 2 × 1000 =2000
Step 3: Calculate the recombination fraction
𝑟 = Number of recombinant gametes
Total number of gametes
=1020
2000
= 0.51
The estimated recombination fraction between loci A and B is 0.51, suggesting that these loci
are unlinked (as r ≈ 0.5).
1.29 PROBLEM 4: QTL EFFECT SIZE CALCULATION
In a QTL mapping study for seed weight in soybeans, you’ve identified a QTL with the following
genotypic values:
• AA: 25g
• Aa: 22g
• aa: 18g
Calculate the additive (a) and dominance (d) effects of this QTL.
Solution:
Step 1: Calculate the additive effect (a) The additive effect is half the difference between the
homozygous genotypes:
𝑎 = 1
2(𝐴𝐴 −𝑎𝑎)
=1
2(25𝑔 − 18𝑔)
= 3.5𝑔
Step 2: Calculate the dominance effect (d) The dominance effect is the deviation of the
heterozygote from the average of the homozygotes:
𝑑 = 𝐴𝑎 −1
2(𝐴𝐴 +𝑎𝑎)
=22𝑔 − 1
2(25𝑔 + 18𝑔)
=22𝑔 − 21.5𝑔
= 0.5𝑔
Therefore, the additive effect (a) is 3.5g and the dominance effect (d) is 0.5g.
1.30 PROBLEM 5: POWER CALCULATION
You’re planning a QTL mapping study and want to determine the sample size needed to detect
a QTL that explains 5% of the phenotypic variance with 80% power at a significance level of
0.05. Assume you’re using an F2 population and a single marker analysis. Use the
approximation formula:
𝑛 = 2(𝑧𝛼/2 + 𝑧𝛽)2
ℎ𝑞
2
where 𝑛 is the required sample size, 𝑧𝛼/2 is the critical value of the standard normal distribution
for 𝛼/2, 𝑧𝛽 is the critical value for 𝛽 (1 - power), and ℎ𝑞
2 is the heritability explained by the QTL.
Solution:
Given:
• ℎ𝑞
2= 0.05 (5% of phenotypic variance)
• Power = 80%, so 𝛽 = 0.2
• Significance level 𝛼 = 0.05
Step 1: Find critical values
𝑧𝛼/2 = 𝑧0.025 = 1.96 (from standard normal table)
𝑧𝛽= 𝑧0.2 = 0.84 (from standard normal table)
Step 2: Apply the formula
𝑛 = 2(𝑧𝛼/2 + 𝑧𝛽)2
ℎ𝑞
2
=2(1.96 + 0.84)2
0.05
=2(2.8)2
0.05
=15.68
0.05
=313.6
Rounding up, you need a sample size of at least 314 F2 individuals to detect this QTL with 80%
power at a significance level of 0.05.
1.31 PROBLEM 1: LOD SCORE CALCULATION
Calculate the LOD score for a putative QTL given the following information:
• Likelihood under the alternative hypothesis (QTL present): 1.5 × 10−12
• Likelihood under the null hypothesis (no QTL): 2.7 × 10−15
Solution: The LOD score is calculated as:
𝐿𝑂𝐷 = log10 (Likelihood (QTL present)
Likelihood (no QTL) )
=log10 (1.5×10−12
2.7×10−15)
=log10(555.56)
= 2.74
Therefore, the
LOD score for this putative QTL is 2.74.
1.32 PROBLEM 2: HERITABILITY ESTIMATION
In a QTL mapping study for plant height in rice, you’ve identified 5 QTLs with the following
effect sizes:
• QTL 1: 15% of phenotypic variance
• QTL 2: 10% of phenotypic variance
• QTL 3: 8% of phenotypic variance
• QTL 4: 5% of phenotypic variance
• QTL 5: 3% of phenotypic variance
Calculate the total heritability explained by these QTLs. If the total heritability of plant height is
estimated to be 0.65, what percentage of the genetic variance remains unexplained?
Solution:
Step 1: Calculate total heritability explained by QTLs
ℎ𝑄𝑇𝐿
2= 0.15 + 0.10 + 0.08 + 0.05 + 0.03
= 0.41
Step 2: Calculate unexplained heritability
ℎ𝑢𝑛𝑒𝑥𝑝𝑙𝑎𝑖𝑛𝑒𝑑
2= ℎ𝑡𝑜𝑡𝑎𝑙
2− ℎ𝑄𝑇𝐿
2
= 0.65 − 0.41
= 0.24
Step 3: Calculate percentage of unexplained genetic variance
Percentage unexplained =ℎ𝑢𝑛𝑒𝑥𝑝𝑙𝑎𝑖𝑛𝑒𝑑
2
ℎ𝑡𝑜𝑡𝑎𝑙
2×100%
=0.24
0.65 ×100%
=36.92%
Therefore, 36.92% of the genetic variance remains unexplained.
1.33 PROBLEM 3: RECOMBINATION FRACTION ESTIMATION
In a QTL mapping study, you observe the following genotype frequencies in an F2 population of
1000 individuals:
• AA BB: 240
• AA Bb: 260
• AA bb: 10
• Aa BB: 255
• Aa Bb: 220
• Aa bb: 5
• aa BB: 8
• aa Bb: 2
• aa bb: 0
Estimate the recombination fraction between loci A and B.
Solution:
Step 1: Calculate the number of recombinant gametes Recombinant gametes are AB and ab.
𝑁𝐴𝐵 = 2(240)+255 +260 + 8 = 1003
𝑁𝑎𝑏 = 2(0)+5+2+10 =17
Total recombinant =1003 +17 =1020
Step 2: Calculate the total number of gametes
Total gametes = 2 × 1000 =2000
Step 3: Calculate the recombination fraction
𝑟 = Number of recombinant gametes
Total number of gametes
=1020
2000
= 0.51
The estimated recombination fraction between loci A and B is 0.51, suggesting that these loci
are unlinked (as r ≈ 0.5).
1.34 PROBLEM 4: QTL EFFECT SIZE CALCULATION
In a QTL mapping study for seed weight in soybeans, you’ve identified a QTL with the following
genotypic values:
• AA: 25g
• Aa: 22g
• aa: 18g
Calculate the additive (a) and dominance (d) effects of this QTL.
Solution:
Step 1: Calculate the additive effect (a) The additive effect is half the difference between the
homozygous genotypes:
𝑎 = 1
2(𝐴𝐴 −𝑎𝑎)
=1
2(25𝑔 − 18𝑔)
= 3.5𝑔
Step 2: Calculate the dominance effect (d) The dominance effect is the deviation of the
heterozygote from the average of the homozygotes:
𝑑 = 𝐴𝑎 −1
2(𝐴𝐴 +𝑎𝑎)
=22𝑔 − 1
2(25𝑔 + 18𝑔)
=22𝑔 − 21.5𝑔
= 0.5𝑔
Therefore, the additive effect (a) is 3.5g and the dominance effect (d) is 0.5g.
1.35 PROBLEM 5: POWER CALCULATION
You’re planning a QTL mapping study and want to determine the sample size needed to detect
a QTL that explains 5% of the phenotypic variance with 80% power at a significance level of
0.05. Assume you’re using an F2 population and a single marker analysis. Use the
approximation formula:
𝑛 = 2(𝑧𝛼/2 + 𝑧𝛽)2
ℎ𝑞
2
where 𝑛 is the required sample size, 𝑧𝛼/2 is the critical value of the standard normal distribution
for 𝛼/2, 𝑧𝛽 is the critical value for 𝛽 (1 - power), and ℎ𝑞
2 is the heritability explained by the QTL.
Solution:
Given:
• ℎ𝑞
2= 0.05 (5% of phenotypic variance)
• Power = 80%, so 𝛽 = 0.2
• Significance level 𝛼 = 0.05
Step 1: Find critical values
𝑧𝛼/2 = 𝑧0.025 = 1.96 (from standard normal table)
𝑧𝛽= 𝑧0.2 = 0.84 (from standard normal table)
Step 2: Apply the formula
𝑛 = 2(𝑧𝛼/2 + 𝑧𝛽)2
ℎ𝑞
2
=2(1.96 + 0.84)2
0.05
=2(2.8)2
0.05
=15.68
0.05
=313.6
Rounding up, you need a sample size of at least 314 F2 individuals to detect this QTL with 80%
power at a significance level of 0.05.
1.36 PROBLEM 1: LOD SCORE CALCULATION
Calculate the LOD score for a putative QTL given the following information:
• Likelihood under the alternative hypothesis (QTL present): 1.5 × 10−12
• Likelihood under the null hypothesis (no QTL): 2.7 × 10−15
Solution: The LOD score is calculated as:
𝐿𝑂𝐷 = log10 (Likelihood (QTL present)
Likelihood (no QTL) )
=log10 (1.5×10−12
2.7×10−15)
=log10(555.56)
= 2.74
Therefore, the
LOD score for this putative QTL is 2.74.
1.37 PROBLEM 2: HERITABILITY ESTIMATION
In a QTL mapping study for plant height in rice, you’ve identified 5 QTLs with the following
effect sizes:
• QTL 1: 15% of phenotypic variance
• QTL 2: 10% of phenotypic variance
• QTL 3: 8% of phenotypic variance
• QTL 4: 5% of phenotypic variance
• QTL 5: 3% of phenotypic variance
Calculate the total heritability explained by these QTLs. If the total heritability of plant height is
estimated to be 0.65, what percentage of the genetic variance remains unexplained?
Solution:
Step 1: Calculate total heritability explained by QTLs
ℎ𝑄𝑇𝐿
2= 0.15 + 0.10 + 0.08 + 0.05 + 0.03
= 0.41
Step 2: Calculate unexplained heritability
ℎ𝑢𝑛𝑒𝑥𝑝𝑙𝑎𝑖𝑛𝑒𝑑
2= ℎ𝑡𝑜𝑡𝑎𝑙
2− ℎ𝑄𝑇𝐿
2
= 0.65 − 0.41
= 0.24
Step 3: Calculate percentage of unexplained genetic variance
Percentage unexplained =ℎ𝑢𝑛𝑒𝑥𝑝𝑙𝑎𝑖𝑛𝑒𝑑
2
ℎ𝑡𝑜𝑡𝑎𝑙
2×100%
=0.24
0.65 ×100%
=36.92%
Therefore, 36.92% of the genetic variance remains unexplained.
1.38 PROBLEM 3: RECOMBINATION FRACTION ESTIMATION
In a QTL mapping study, you observe the following genotype frequencies in an F2 population of
1000 individuals:
• AA BB: 240
• AA Bb: 260
• AA bb: 10
• Aa BB: 255
• Aa Bb: 220
• Aa bb: 5
• aa BB: 8
• aa Bb: 2
• aa bb: 0
Estimate the recombination fraction between loci A and B.
Solution:
Step 1: Calculate the number of recombinant gametes Recombinant gametes are AB and ab.
𝑁𝐴𝐵 = 2(240)+255 +260 + 8 = 1003
𝑁𝑎𝑏 = 2(0)+5+2+10 =17
Total recombinant =1003 +17 =1020
Step 2: Calculate the total number of gametes
Total gametes = 2 × 1000 =2000
Step 3: Calculate the recombination fraction
𝑟 = Number of recombinant gametes
Total number of gametes
=1020
2000
= 0.51
The estimated recombination fraction between loci A and B is 0.51, suggesting that these loci
are unlinked (as r ≈ 0.5).
1.39 PROBLEM 4: QTL EFFECT SIZE CALCULATION
In a QTL mapping study for seed weight in soybeans, you’ve identified a QTL with the following
genotypic values:
• AA: 25g
• Aa: 22g
• aa: 18g
Calculate the additive (a) and dominance (d) effects of this QTL.
Solution:
Step 1: Calculate the additive effect (a) The additive effect is half the difference between the
homozygous genotypes:
𝑎 = 1
2(𝐴𝐴 −𝑎𝑎)
=1
2(25𝑔 − 18𝑔)
= 3.5𝑔
Step 2: Calculate the dominance effect (d) The dominance effect is the deviation of the
heterozygote from the average of the homozygotes:
𝑑 = 𝐴𝑎 −1
2(𝐴𝐴 +𝑎𝑎)
=22𝑔 − 1
2(25𝑔 + 18𝑔)
=22𝑔 − 21.5𝑔
= 0.5𝑔
Therefore, the additive effect (a) is 3.5g and the dominance effect (d) is 0.5g.
1.40 PROBLEM 5: POWER CALCULATION
You’re planning a QTL mapping study and want to determine the sample size needed to detect
a QTL that explains 5% of the phenotypic variance with 80% power at a significance level of
0.05. Assume you’re using an F2 population and a single marker analysis. Use the
approximation formula:
𝑛 = 2(𝑧𝛼/2 + 𝑧𝛽)2
ℎ𝑞
2
where 𝑛 is the required sample size, 𝑧𝛼/2 is the critical value of the standard normal distribution
for 𝛼/2, 𝑧𝛽 is the critical value for 𝛽 (1 - power), and ℎ𝑞
2 is the heritability explained by the QTL.
Solution:
Given:
• ℎ𝑞
2= 0.05 (5% of phenotypic variance)
• Power = 80%, so 𝛽 = 0.2
• Significance level 𝛼 = 0.05
Step 1: Find critical values
𝑧𝛼/2 = 𝑧0.025 = 1.96 (from standard normal table)
𝑧𝛽= 𝑧0.2 = 0.84 (from standard normal table)
Step 2: Apply the formula
𝑛 = 2(𝑧𝛼/2 + 𝑧𝛽)2
ℎ𝑞
2
=2(1.96 + 0.84)2
0.05
=2(2.8)2
0.05
=15.68
0.05
=313.6
Rounding up, you need a sample size of at least 314 F2 individuals to detect this QTL with 80%
power at a significance level of 0.05.
1.41 PROBLEM 1: LOD SCORE CALCULATION
Calculate the LOD score for a putative QTL given the following information:
• Likelihood under the alternative hypothesis (QTL present): 1.5 × 10−12
• Likelihood under the null hypothesis (no QTL): 2.7 × 10−15
Solution: The LOD score is calculated as:
𝐿𝑂𝐷 = log10 (Likelihood (QTL present)
Likelihood (no QTL) )
=log10 (1.5×10−12
2.7×10−15)
=log10(555.56)
= 2.74
Therefore, the
LOD score for this putative QTL is 2.74.
1.42 PROBLEM 2: HERITABILITY ESTIMATION
In a QTL mapping study for plant height in rice, you’ve identified 5 QTLs with the following
effect sizes:
• QTL 1: 15% of phenotypic variance
• QTL 2: 10% of phenotypic variance
• QTL 3: 8% of phenotypic variance
• QTL 4: 5% of phenotypic variance
• QTL 5: 3% of phenotypic variance
Calculate the total heritability explained by these QTLs. If the total heritability of plant height is
estimated to be 0.65, what percentage of the genetic variance remains unexplained?
Solution:
Step 1: Calculate total heritability explained by QTLs
ℎ𝑄𝑇𝐿
2= 0.15 + 0.10 + 0.08 + 0.05 + 0.03
= 0.41
Step 2: Calculate unexplained heritability
ℎ𝑢𝑛𝑒𝑥𝑝𝑙𝑎𝑖𝑛𝑒𝑑
2= ℎ𝑡𝑜𝑡𝑎𝑙
2− ℎ𝑄𝑇𝐿
2
= 0.65 − 0.41
= 0.24
Step 3: Calculate percentage of unexplained genetic variance
Percentage unexplained =ℎ𝑢𝑛𝑒𝑥𝑝𝑙𝑎𝑖𝑛𝑒𝑑
2
ℎ𝑡𝑜𝑡𝑎𝑙
2×100%
=0.24
0.65 ×100%
=36.92%
Therefore, 36.92% of the genetic variance remains unexplained.
1.43 PROBLEM 3: RECOMBINATION FRACTION ESTIMATION
In a QTL mapping study, you observe the following genotype frequencies in an F2 population of
1000 individuals:
• AA BB: 240
• AA Bb: 260
• AA bb: 10
• Aa BB: 255
• Aa Bb: 220
• Aa bb: 5
• aa BB: 8
• aa Bb: 2
• aa bb: 0
Estimate the recombination fraction between loci A and B.
Solution:
Step 1: Calculate the number of recombinant gametes Recombinant gametes are AB and ab.
𝑁𝐴𝐵 = 2(240)+255 +260 + 8 = 1003
𝑁𝑎𝑏 = 2(0)+5+2+10 =17
Total recombinant =1003 +17 =1020
Step 2: Calculate the total number of gametes
Total gametes = 2 × 1000 =2000
Step 3: Calculate the recombination fraction
𝑟 = Number of recombinant gametes
Total number of gametes
=1020
2000
= 0.51
The estimated recombination fraction between loci A and B is 0.51, suggesting that these loci
are unlinked (as r ≈ 0.5).
1.44 PROBLEM 4: QTL EFFECT SIZE CALCULATION
In a QTL mapping study for seed weight in soybeans, you’ve identified a QTL with the following
genotypic values:
• AA: 25g
• Aa: 22g
• aa: 18g
Calculate the additive (a) and dominance (d) effects of this QTL.
Solution:
Step 1: Calculate the additive effect (a) The additive effect is half the difference between the
homozygous genotypes:
𝑎 = 1
2(𝐴𝐴 −𝑎𝑎)
=1
2(25𝑔 − 18𝑔)
= 3.5𝑔
Step 2: Calculate the dominance effect (d) The dominance effect is the deviation of the
heterozygote from the average of the homozygotes:
𝑑 = 𝐴𝑎 −1
2(𝐴𝐴 +𝑎𝑎)
=22𝑔 − 1
2(25𝑔 + 18𝑔)
=22𝑔 − 21.5𝑔
= 0.5𝑔
Therefore, the additive effect (a) is 3.5g and the dominance effect (d) is 0.5g.
1.45 PROBLEM 5: POWER CALCULATION
You’re planning a QTL mapping study and want to determine the sample size needed to detect
a QTL that explains 5% of the phenotypic variance with 80% power at a significance level of
0.05. Assume you’re using an F2 population and a single marker analysis. Use the
approximation formula:
𝑛 = 2(𝑧𝛼/2 + 𝑧𝛽)2
ℎ𝑞
2
where 𝑛 is the required sample size, 𝑧𝛼/2 is the critical value of the standard normal distribution
for 𝛼/2, 𝑧𝛽 is the critical value for 𝛽 (1 - power), and ℎ𝑞
2 is the heritability explained by the QTL.
Solution:
Given:
• ℎ𝑞
2= 0.05 (5% of phenotypic variance)
• Power = 80%, so 𝛽 = 0.2
• Significance level 𝛼 = 0.05
Step 1: Find critical values
𝑧𝛼/2 = 𝑧0.025 = 1.96 (from standard normal table)
𝑧𝛽= 𝑧0.2 = 0.84 (from standard normal table)
Step 2: Apply the formula
𝑛 = 2(𝑧𝛼/2 + 𝑧𝛽)2
ℎ𝑞
2
=2(1.96 + 0.84)2
0.05
=2(2.8)2
0.05
=15.68
0.05
=313.6
Rounding up, you need a sample size of at least 314 F2 individuals to detect this QTL with 80%
power at a significance level of 0.05.
1.46 PROBLEM 1: LOD SCORE CALCULATION
Calculate the LOD score for a putative QTL given the following information:
• Likelihood under the alternative hypothesis (QTL present): 1.5 × 10−12
• Likelihood under the null hypothesis (no QTL): 2.7 × 10−15
Solution: The LOD score is calculated as:
𝐿𝑂𝐷 = log10 (Likelihood (QTL present)
Likelihood (no QTL) )
=log10 (1.5×10−12
2.7×10−15)
=log10(555.56)
= 2.74
Therefore, the
LOD score for this putative QTL is 2.74.
1.47 PROBLEM 2: HERITABILITY ESTIMATION
In a QTL mapping study for plant height in rice, you’ve identified 5 QTLs with the following
effect sizes:
• QTL 1: 15% of phenotypic variance
• QTL 2: 10% of phenotypic variance
• QTL 3: 8% of phenotypic variance
• QTL 4: 5% of phenotypic variance
• QTL 5: 3% of phenotypic variance
Calculate the total heritability explained by these QTLs. If the total heritability of plant height is
estimated to be 0.65, what percentage of the genetic variance remains unexplained?
Solution:
Step 1: Calculate total heritability explained by QTLs
ℎ𝑄𝑇𝐿
2= 0.15 + 0.10 + 0.08 + 0.05 + 0.03
= 0.41
Step 2: Calculate unexplained heritability
ℎ𝑢𝑛𝑒𝑥𝑝𝑙𝑎𝑖𝑛𝑒𝑑
2= ℎ𝑡𝑜𝑡𝑎𝑙
2− ℎ𝑄𝑇𝐿
2
= 0.65 − 0.41
= 0.24
Step 3: Calculate percentage of unexplained genetic variance
Percentage unexplained =ℎ𝑢𝑛𝑒𝑥𝑝𝑙𝑎𝑖𝑛𝑒𝑑
2
ℎ𝑡𝑜𝑡𝑎𝑙
2×100%
=0.24
0.65 ×100%
=36.92%
Therefore, 36.92% of the genetic variance remains unexplained.
1.48 PROBLEM 3: RECOMBINATION FRACTION ESTIMATION
In a QTL mapping study, you observe the following genotype frequencies in an F2 population of
1000 individuals:
• AA BB: 240
• AA Bb: 260
• AA bb: 10
• Aa BB: 255
• Aa Bb: 220
• Aa bb: 5
• aa BB: 8
• aa Bb: 2
• aa bb: 0
Estimate the recombination fraction between loci A and B.
Solution:
Step 1: Calculate the number of recombinant gametes Recombinant gametes are AB and ab.
𝑁𝐴𝐵 = 2(240)+255 +260 + 8 = 1003
𝑁𝑎𝑏 = 2(0)+5+2+10 =17
Total recombinant =1003 +17 =1020
Step 2: Calculate the total number of gametes
Total gametes = 2 × 1000 =2000
Step 3: Calculate the recombination fraction
𝑟 = Number of recombinant gametes
Total number of gametes
=1020
2000
= 0.51
The estimated recombination fraction between loci A and B is 0.51, suggesting that these loci
are unlinked (as r ≈ 0.5).
1.49 PROBLEM 4: QTL EFFECT SIZE CALCULATION
In a QTL mapping study for seed weight in soybeans, you’ve identified a QTL with the following
genotypic values:
• AA: 25g
• Aa: 22g
• aa: 18g
Calculate the additive (a) and dominance (d) effects of this QTL.
Solution:
Step 1: Calculate the additive effect (a) The additive effect is half the difference between the
homozygous genotypes:
𝑎 = 1
2(𝐴𝐴 −𝑎𝑎)
=1
2(25𝑔 − 18𝑔)
= 3.5𝑔
Step 2: Calculate the dominance effect (d) The dominance effect is the deviation of the
heterozygote from the average of the homozygotes:
𝑑 = 𝐴𝑎 −1
2(𝐴𝐴 +𝑎𝑎)
=22𝑔 − 1
2(25𝑔 + 18𝑔)
=22𝑔 − 21.5𝑔
= 0.5𝑔
Therefore, the additive effect (a) is 3.5g and the dominance effect (d) is 0.5g.
1.50 PROBLEM 5: POWER CALCULATION
You’re planning a QTL mapping study and want to determine the sample size needed to detect
a QTL that explains 5% of the phenotypic variance with 80% power at a significance level of
0.05. Assume you’re using an F2 population and a single marker analysis. Use the
approximation formula:
𝑛 = 2(𝑧𝛼/2 + 𝑧𝛽)2
ℎ𝑞
2
where 𝑛 is the required sample size, 𝑧𝛼/2 is the critical value of the standard normal distribution
for 𝛼/2, 𝑧𝛽 is the critical value for 𝛽 (1 - power), and ℎ𝑞
2 is the heritability explained by the QTL.
Solution:
Given:
• ℎ𝑞
2= 0.05 (5% of phenotypic variance)
• Power = 80%, so 𝛽 = 0.2
• Significance level 𝛼 = 0.05
Step 1: Find critical values
𝑧𝛼/2 = 𝑧0.025 = 1.96 (from standard normal table)
𝑧𝛽= 𝑧0.2 = 0.84 (from standard normal table)
Step 2: Apply the formula
𝑛 = 2(𝑧𝛼/2 + 𝑧𝛽)2
ℎ𝑞
2
=2(1.96 + 0.84)2
0.05
=2(2.8)2
0.05
=15.68
0.05
=313.6
Rounding up, you need a sample size of at least 314 F2 individuals to detect this QTL with 80%
power at a significance level of 0.05.
1.51 PROBLEM 1: LOD SCORE CALCULATION
Calculate the LOD score for a putative QTL given the following information:
• Likelihood under the alternative hypothesis (QTL present): 1.5 × 10−12
• Likelihood under the null hypothesis (no QTL): 2.7 × 10−15
Solution: The LOD score is calculated as:
𝐿𝑂𝐷 = log10 (Likelihood (QTL present)
Likelihood (no QTL) )
=log10 (1.5×10−12
2.7×10−15)
=log10(555.56)
= 2.74
Therefore, the
LOD score for this putative QTL is 2.74.
1.52 PROBLEM 2: HERITABILITY ESTIMATION
In a QTL mapping study for plant height in rice, you’ve identified 5 QTLs with the following
effect sizes:
• QTL 1: 15% of phenotypic variance
• QTL 2: 10% of phenotypic variance
• QTL 3: 8% of phenotypic variance
• QTL 4: 5% of phenotypic variance
• QTL 5: 3% of phenotypic variance
Calculate the total heritability explained by these QTLs. If the total heritability of plant height is
estimated to be 0.65, what percentage of the genetic variance remains unexplained?
Solution:
Step 1: Calculate total heritability explained by QTLs
ℎ𝑄𝑇𝐿
2= 0.15 + 0.10 + 0.08 + 0.05 + 0.03
= 0.41
Step 2: Calculate unexplained heritability
ℎ𝑢𝑛𝑒𝑥𝑝𝑙𝑎𝑖𝑛𝑒𝑑
2= ℎ𝑡𝑜𝑡𝑎𝑙
2− ℎ𝑄𝑇𝐿
2
= 0.65 − 0.41
= 0.24
Step 3: Calculate percentage of unexplained genetic variance
Percentage unexplained =ℎ𝑢𝑛𝑒𝑥𝑝𝑙𝑎𝑖𝑛𝑒𝑑
2
ℎ𝑡𝑜𝑡𝑎𝑙
2×100%
=0.24
0.65 ×100%
=36.92%
Therefore, 36.92% of the genetic variance remains unexplained.
1.53 PROBLEM 3: RECOMBINATION FRACTION ESTIMATION
In a QTL mapping study, you observe the following genotype frequencies in an F2 population of
1000 individuals:
• AA BB: 240
• AA Bb: 260
• AA bb: 10
• Aa BB: 255
• Aa Bb: 220
• Aa bb: 5
• aa BB: 8
• aa Bb: 2
• aa bb: 0
Estimate the recombination fraction between loci A and B.
Solution:
Step 1: Calculate the number of recombinant gametes Recombinant gametes are AB and ab.
𝑁𝐴𝐵 = 2(240)+255 +260 + 8 = 1003
𝑁𝑎𝑏 = 2(0)+5+2+10 =17
Total recombinant =1003 +17 =1020
Step 2: Calculate the total number of gametes
Total gametes = 2 × 1000 =2000
Step 3: Calculate the recombination fraction
𝑟 = Number of recombinant gametes
Total number of gametes
=1020
2000
= 0.51
The estimated recombination fraction between loci A and B is 0.51, suggesting that these loci
are unlinked (as r ≈ 0.5).
1.54 PROBLEM 4: QTL EFFECT SIZE CALCULATION
In a QTL mapping study for seed weight in soybeans, you’ve identified a QTL with the following
genotypic values:
• AA: 25g
• Aa: 22g
• aa: 18g
Calculate the additive (a) and dominance (d) effects of this QTL.
Solution:
Step 1: Calculate the additive effect (a) The additive effect is half the difference between the
homozygous genotypes:
𝑎 = 1
2(𝐴𝐴 −𝑎𝑎)
=1
2(25𝑔 − 18𝑔)
= 3.5𝑔
Step 2: Calculate the dominance effect (d) The dominance effect is the deviation of the
heterozygote from the average of the homozygotes:
𝑑 = 𝐴𝑎 −1
2(𝐴𝐴 +𝑎𝑎)
=22𝑔 − 1
2(25𝑔 + 18𝑔)
=22𝑔 − 21.5𝑔
= 0.5𝑔
Therefore, the additive effect (a) is 3.5g and the dominance effect (d) is 0.5g.
1.55 PROBLEM 5: POWER CALCULATION
You’re planning a QTL mapping study and want to determine the sample size needed to detect
a QTL that explains 5% of the phenotypic variance with 80% power at a significance level of
0.05. Assume you’re using an F2 population and a single marker analysis. Use the
approximation formula:
𝑛 = 2(𝑧𝛼/2 + 𝑧𝛽)2
ℎ𝑞
2
where 𝑛 is the required sample size, 𝑧𝛼/2 is the critical value of the standard normal distribution
for 𝛼/2, 𝑧𝛽 is the critical value for 𝛽 (1 - power), and ℎ𝑞
2 is the heritability explained by the QTL.
Solution:
Given:
• ℎ𝑞
2= 0.05 (5% of phenotypic variance)
• Power = 80%, so 𝛽 = 0.2
• Significance level 𝛼 = 0.05
Step 1: Find critical values
𝑧𝛼/2 = 𝑧0.025 = 1.96 (from standard normal table)
𝑧𝛽= 𝑧0.2 = 0.84 (from standard normal table)
Step 2: Apply the formula
𝑛 = 2(𝑧𝛼/2 + 𝑧𝛽)2
ℎ𝑞
2
=2(1.96 + 0.84)2
0.05
=2(2.8)2
0.05
=15.68
0.05
=313.6
Rounding up, you need a sample size of at least 314 F2 individuals to detect this QTL with 80%
power at a significance level of 0.05.
1.56 PROBLEM 1: LOD SCORE CALCULATION
Calculate the LOD score for a putative QTL given the following information:
• Likelihood under the alternative hypothesis (QTL present): 1.5 × 10−12
• Likelihood under the null hypothesis (no QTL): 2.7 × 10−15
Solution: The LOD score is calculated as:
𝐿𝑂𝐷 = log10 (Likelihood (QTL present)
Likelihood (no QTL) )
=log10 (1.5×10−12
2.7×10−15)
=log10(555.56)
= 2.74
Therefore, the
LOD score for this putative QTL is 2.74.
1.57 PROBLEM 2: HERITABILITY ESTIMATION
In a QTL mapping study for plant height in rice, you’ve identified 5 QTLs with the following
effect sizes:
• QTL 1: 15% of phenotypic variance
• QTL 2: 10% of phenotypic variance
• QTL 3: 8% of phenotypic variance
• QTL 4: 5% of phenotypic variance
• QTL 5: 3% of phenotypic variance
Calculate the total heritability explained by these QTLs. If the total heritability of plant height is
estimated to be 0.65, what percentage of the genetic variance remains unexplained?
Solution:
Step 1: Calculate total heritability explained by QTLs
ℎ𝑄𝑇𝐿
2= 0.15 + 0.10 + 0.08 + 0.05 + 0.03
= 0.41
Step 2: Calculate unexplained heritability
ℎ𝑢𝑛𝑒𝑥𝑝𝑙𝑎𝑖𝑛𝑒𝑑
2= ℎ𝑡𝑜𝑡𝑎𝑙
2− ℎ𝑄𝑇𝐿
2
= 0.65 − 0.41
= 0.24
Step 3: Calculate percentage of unexplained genetic variance
Percentage unexplained =ℎ𝑢𝑛𝑒𝑥𝑝𝑙𝑎𝑖𝑛𝑒𝑑
2
ℎ𝑡𝑜𝑡𝑎𝑙
2×100%
=0.24
0.65 ×100%
=36.92%
Therefore, 36.92% of the genetic variance remains unexplained.
1.58 PROBLEM 3: RECOMBINATION FRACTION ESTIMATION
In a QTL mapping study, you observe the following genotype frequencies in an F2 population of
1000 individuals:
• AA BB: 240
• AA Bb: 260
• AA bb: 10
• Aa BB: 255
• Aa Bb: 220
• Aa bb: 5
• aa BB: 8
• aa Bb: 2
• aa bb: 0
Estimate the recombination fraction between loci A and B.
Solution:
Step 1: Calculate the number of recombinant gametes Recombinant gametes are AB and ab.
𝑁𝐴𝐵 = 2(240)+255 +260 + 8 = 1003
𝑁𝑎𝑏 = 2(0)+5+2+10 =17
Total recombinant =1003 +17 =1020
Step 2: Calculate the total number of gametes
Total gametes = 2 × 1000 =2000
Step 3: Calculate the recombination fraction
𝑟 = Number of recombinant gametes
Total number of gametes
=1020
2000
= 0.51
The estimated recombination fraction between loci A and B is 0.51, suggesting that these loci
are unlinked (as r ≈ 0.5).
1.59 PROBLEM 4: QTL EFFECT SIZE CALCULATION
In a QTL mapping study for seed weight in soybeans, you’ve identified a QTL with the following
genotypic values:
• AA: 25g
• Aa: 22g
• aa: 18g
Calculate the additive (a) and dominance (d) effects of this QTL.
Solution:
Step 1: Calculate the additive effect (a) The additive effect is half the difference between the
homozygous genotypes:
𝑎 = 1
2(𝐴𝐴 −𝑎𝑎)
=1
2(25𝑔 − 18𝑔)
= 3.5𝑔
Step 2: Calculate the dominance effect (d) The dominance effect is the deviation of the
heterozygote from the average of the homozygotes:
𝑑 = 𝐴𝑎 −1
2(𝐴𝐴 +𝑎𝑎)
=22𝑔 − 1
2(25𝑔 + 18𝑔)
=22𝑔 − 21.5𝑔
= 0.5𝑔
Therefore, the additive effect (a) is 3.5g and the dominance effect (d) is 0.5g.
1.60 PROBLEM 5: POWER CALCULATION
You’re planning a QTL mapping study and want to determine the sample size needed to detect
a QTL that explains 5% of the phenotypic variance with 80% power at a significance level of
0.05. Assume you’re using an F2 population and a single marker analysis. Use the
approximation formula:
𝑛 = 2(𝑧𝛼/2 + 𝑧𝛽)2
ℎ𝑞
2
where 𝑛 is the required sample size, 𝑧𝛼/2 is the critical value of the standard normal distribution
for 𝛼/2, 𝑧𝛽 is the critical value for 𝛽 (1 - power), and ℎ𝑞
2 is the heritability explained by the QTL.
Solution:
Given:
• ℎ𝑞
2= 0.05 (5% of phenotypic variance)
• Power = 80%, so 𝛽 = 0.2
• Significance level 𝛼 = 0.05
Step 1: Find critical values
𝑧𝛼/2 = 𝑧0.025 = 1.96 (from standard normal table)
𝑧𝛽= 𝑧0.2 = 0.84 (from standard normal table)
Step 2: Apply the formula
𝑛 = 2(𝑧𝛼/2 + 𝑧𝛽)2
ℎ𝑞
2
=2(1.96 + 0.84)2
0.05
=2(2.8)2
0.05
=15.68
0.05
=313.6
Rounding up, you need a sample size of at least 314 F2 individuals to detect this QTL with 80%
power at a significance level of 0.05.
1.61 PROBLEM 1: LOD SCORE CALCULATION
Calculate the LOD score for a putative QTL given the following information:
• Likelihood under the alternative hypothesis (QTL present): 1.5 × 10−12
• Likelihood under the null hypothesis (no QTL): 2.7 × 10−15
Solution: The LOD score is calculated as:
𝐿𝑂𝐷 = log10 (Likelihood (QTL present)
Likelihood (no QTL) )
=log10 (1.5×10−12
2.7×10−15)
=log10(555.56)
= 2.74
Therefore, the
LOD score for this putative QTL is 2.74.
1.62 PROBLEM 2: HERITABILITY ESTIMATION
In a QTL mapping study for plant height in rice, you’ve identified 5 QTLs with the following
effect sizes:
• QTL 1: 15% of phenotypic variance
• QTL 2: 10% of phenotypic variance
• QTL 3: 8% of phenotypic variance
• QTL 4: 5% of phenotypic variance
• QTL 5: 3% of phenotypic variance
Calculate the total heritability explained by these QTLs. If the total heritability of plant height is
estimated to be 0.65, what percentage of the genetic variance remains unexplained?
Solution:
Step 1: Calculate total heritability explained by QTLs
ℎ𝑄𝑇𝐿
2= 0.15 + 0.10 + 0.08 + 0.05 + 0.03
= 0.41
Step 2: Calculate unexplained heritability
ℎ𝑢𝑛𝑒𝑥𝑝𝑙𝑎𝑖𝑛𝑒𝑑
2= ℎ𝑡𝑜𝑡𝑎𝑙
2− ℎ𝑄𝑇𝐿
2
= 0.65 − 0.41
= 0.24
Step 3: Calculate percentage of unexplained genetic variance
Percentage unexplained =ℎ𝑢𝑛𝑒𝑥𝑝𝑙𝑎𝑖𝑛𝑒𝑑
2
ℎ𝑡𝑜𝑡𝑎𝑙
2×100%
=0.24
0.65 ×100%
=36.92%
Therefore, 36.92% of the genetic variance remains unexplained.
1.63 PROBLEM 3: RECOMBINATION FRACTION ESTIMATION
In a QTL mapping study, you observe the following genotype frequencies in an F2 population of
1000 individuals:
• AA BB: 240
• AA Bb: 260
• AA bb: 10
• Aa BB: 255
• Aa Bb: 220
• Aa bb: 5
• aa BB: 8
• aa Bb: 2
• aa bb: 0
Estimate the recombination fraction between loci A and B.
Solution:
Step 1: Calculate the number of recombinant gametes Recombinant gametes are AB and ab.
𝑁𝐴𝐵 = 2(240)+255 +260 + 8 = 1003
𝑁𝑎𝑏 = 2(0)+5+2+10 =17
Total recombinant =1003 +17 =1020
Step 2: Calculate the total number of gametes
Total gametes = 2 × 1000 =2000
Step 3: Calculate the recombination fraction
𝑟 = Number of recombinant gametes
Total number of gametes
=1020
2000
= 0.51
The estimated recombination fraction between loci A and B is 0.51, suggesting that these loci
are unlinked (as r ≈ 0.5).
1.64 PROBLEM 4: QTL EFFECT SIZE CALCULATION
In a QTL mapping study for seed weight in soybeans, you’ve identified a QTL with the following
genotypic values:
• AA: 25g
• Aa: 22g
• aa: 18g
Calculate the additive (a) and dominance (d) effects of this QTL.
Solution:
Step 1: Calculate the additive effect (a) The additive effect is half the difference between the
homozygous genotypes:
𝑎 = 1
2(𝐴𝐴 −𝑎𝑎)
=1
2(25𝑔 − 18𝑔)
= 3.5𝑔
Step 2: Calculate the dominance effect (d) The dominance effect is the deviation of the
heterozygote from the average of the homozygotes:
𝑑 = 𝐴𝑎 −1
2(𝐴𝐴 +𝑎𝑎)
=22𝑔 − 1
2(25𝑔 + 18𝑔)
=22𝑔 − 21.5𝑔
= 0.5𝑔
Therefore, the additive effect (a) is 3.5g and the dominance effect (d) is 0.5g.
1.65 PROBLEM 5: POWER CALCULATION
You’re planning a QTL mapping study and want to determine the sample size needed to detect
a QTL that explains 5% of the phenotypic variance with 80% power at a significance level of
0.05. Assume you’re using an F2 population and a single marker analysis. Use the
approximation formula:
𝑛 = 2(𝑧𝛼/2 + 𝑧𝛽)2
ℎ𝑞
2
where 𝑛 is the required sample size, 𝑧𝛼/2 is the critical value of the standard normal distribution
for 𝛼/2, 𝑧𝛽 is the critical value for 𝛽 (1 - power), and ℎ𝑞
2 is the heritability explained by the QTL.
Solution:
Given:
• ℎ𝑞
2= 0.05 (5% of phenotypic variance)
• Power = 80%, so 𝛽 = 0.2
• Significance level 𝛼 = 0.05
Step 1: Find critical values
𝑧𝛼/2 = 𝑧0.025 = 1.96 (from standard normal table)
𝑧𝛽= 𝑧0.2 = 0.84 (from standard normal table)
Step 2: Apply the formula
𝑛 = 2(𝑧𝛼/2 + 𝑧𝛽)2
ℎ𝑞
2
=2(1.96 + 0.84)2
0.05
=2(2.8)2
0.05
=15.68
0.05
=313.6
Rounding up, you need a sample size of at least 314 F2 individuals to detect this QTL with 80%
power at a significance level of 0.05.
1.66 PROBLEM 1: LOD SCORE CALCULATION
Calculate the LOD score for a putative QTL given the following information:
• Likelihood under the alternative hypothesis (QTL present): 1.5 × 10−12
• Likelihood under the null hypothesis (no QTL): 2.7 × 10−15
Solution: The LOD score is calculated as:
𝐿𝑂𝐷 = log10 (Likelihood (QTL present)
Likelihood (no QTL) )
=log10 (1.5×10−12
2.7×10−15)
=log10(555.56)
= 2.74
Therefore, the
LOD score for this putative QTL is 2.74.
1.67 PROBLEM 2: HERITABILITY ESTIMATION
In a QTL mapping study for plant height in rice, you’ve identified 5 QTLs with the following
effect sizes:
• QTL 1: 15% of phenotypic variance
• QTL 2: 10% of phenotypic variance
• QTL 3: 8% of phenotypic variance
• QTL 4: 5% of phenotypic variance
• QTL 5: 3% of phenotypic variance
Calculate the total heritability explained by these QTLs. If the total heritability of plant height is
estimated to be 0.65, what percentage of the genetic variance remains unexplained?
Solution:
Step 1: Calculate total heritability explained by QTLs
ℎ𝑄𝑇𝐿
2= 0.15 + 0.10 + 0.08 + 0.05 + 0.03
= 0.41
Step 2: Calculate unexplained heritability
ℎ𝑢𝑛𝑒𝑥𝑝𝑙𝑎𝑖𝑛𝑒𝑑
2= ℎ𝑡𝑜𝑡𝑎𝑙
2− ℎ𝑄𝑇𝐿
2
= 0.65 − 0.41
= 0.24
Step 3: Calculate percentage of unexplained genetic variance
Percentage unexplained =ℎ𝑢𝑛𝑒𝑥𝑝𝑙𝑎𝑖𝑛𝑒𝑑
2
ℎ𝑡𝑜𝑡𝑎𝑙
2×100%
=0.24
0.65 ×100%
=36.92%
Therefore, 36.92% of the genetic variance remains unexplained.
1.68 PROBLEM 3: RECOMBINATION FRACTION ESTIMATION
In a QTL mapping study, you observe the following genotype frequencies in an F2 population of
1000 individuals:
• AA BB: 240
• AA Bb: 260
• AA bb: 10
• Aa BB: 255
• Aa Bb: 220
• Aa bb: 5
• aa BB: 8
• aa Bb: 2
• aa bb: 0
Estimate the recombination fraction between loci A and B.
Solution:
Step 1: Calculate the number of recombinant gametes Recombinant gametes are AB and ab.
𝑁𝐴𝐵 = 2(240)+255 +260 + 8 = 1003
𝑁𝑎𝑏 = 2(0)+5+2+10 =17
Total recombinant =1003 +17 =1020
Step 2: Calculate the total number of gametes
Total gametes = 2 × 1000 =2000
Step 3: Calculate the recombination fraction
𝑟 = Number of recombinant gametes
Total number of gametes
=1020
2000
= 0.51
The estimated recombination fraction between loci A and B is 0.51, suggesting that these loci
are unlinked (as r ≈ 0.5).
1.69 PROBLEM 4: QTL EFFECT SIZE CALCULATION
In a QTL mapping study for seed weight in soybeans, you’ve identified a QTL with the following
genotypic values:
• AA: 25g
• Aa: 22g
• aa: 18g
Calculate the additive (a) and dominance (d) effects of this QTL.
Solution:
Step 1: Calculate the additive effect (a) The additive effect is half the difference between the
homozygous genotypes:
𝑎 = 1
2(𝐴𝐴 −𝑎𝑎)
=1
2(25𝑔 − 18𝑔)
= 3.5𝑔
Step 2: Calculate the dominance effect (d) The dominance effect is the deviation of the
heterozygote from the average of the homozygotes:
𝑑 = 𝐴𝑎 −1
2(𝐴𝐴 +𝑎𝑎)
=22𝑔 − 1
2(25𝑔 + 18𝑔)
=22𝑔 − 21.5𝑔
= 0.5𝑔
Therefore, the additive effect (a) is 3.5g and the dominance effect (d) is 0.5g.
1.70 PROBLEM 5: POWER CALCULATION
You’re planning a QTL mapping study and want to determine the sample size needed to detect
a QTL that explains 5% of the phenotypic variance with 80% power at a significance level of
0.05. Assume you’re using an F2 population and a single marker analysis. Use the
approximation formula:
𝑛 = 2(𝑧𝛼/2 + 𝑧𝛽)2
ℎ𝑞
2
where 𝑛 is the required sample size, 𝑧𝛼/2 is the critical value of the standard normal distribution
for 𝛼/2, 𝑧𝛽 is the critical value for 𝛽 (1 - power), and ℎ𝑞
2 is the heritability explained by the QTL.
Solution:
Given:
• ℎ𝑞
2= 0.05 (5% of phenotypic variance)
• Power = 80%, so 𝛽 = 0.2
• Significance level 𝛼 = 0.05
Step 1: Find critical values
𝑧𝛼/2 = 𝑧0.025 = 1.96 (from standard normal table)
𝑧𝛽= 𝑧0.2 = 0.84 (from standard normal table)
Step 2: Apply the formula
𝑛 = 2(𝑧𝛼/2 + 𝑧𝛽)2
ℎ𝑞
2
=2(1.96 + 0.84)2
0.05
=2(2.8)2
0.05
=15.68
0.05
=313.6
Rounding up, you need a sample size of at least 314 F2 individuals to detect this QTL with 80%
power at a significance level of 0.05.
1.71 PROBLEM 1: LOD SCORE CALCULATION
Calculate the LOD score for a putative QTL given the following information:
• Likelihood under the alternative hypothesis (QTL present): 1.5 × 10−12
• Likelihood under the null hypothesis (no QTL): 2.7 × 10−15
Solution: The LOD score is calculated as:
𝐿𝑂𝐷 = log10 (Likelihood (QTL present)
Likelihood (no QTL) )
=log10 (1.5×10−12
2.7×10−15)
=log10(555.56)
= 2.74
Therefore, the
LOD score for this putative QTL is 2.74.
1.72 PROBLEM 2: HERITABILITY ESTIMATION
In a QTL mapping study for plant height in rice, you’ve identified 5 QTLs with the following
effect sizes:
• QTL 1: 15% of phenotypic variance
• QTL 2: 10% of phenotypic variance
• QTL 3: 8% of phenotypic variance
• QTL 4: 5% of phenotypic variance
• QTL 5: 3% of phenotypic variance
Calculate the total heritability explained by these QTLs. If the total heritability of plant height is
estimated to be 0.65, what percentage of the genetic variance remains unexplained?
Solution:
Step 1: Calculate total heritability explained by QTLs
ℎ𝑄𝑇𝐿
2= 0.15 + 0.10 + 0.08 + 0.05 + 0.03
= 0.41
Step 2: Calculate unexplained heritability
ℎ𝑢𝑛𝑒𝑥𝑝𝑙𝑎𝑖𝑛𝑒𝑑
2= ℎ𝑡𝑜𝑡𝑎𝑙
2− ℎ𝑄𝑇𝐿
2
= 0.65 − 0.41
= 0.24
Step 3: Calculate percentage of unexplained genetic variance
Percentage unexplained =ℎ𝑢𝑛𝑒𝑥𝑝𝑙𝑎𝑖𝑛𝑒𝑑
2
ℎ𝑡𝑜𝑡𝑎𝑙
2×100%
=0.24
0.65 ×100%
=36.92%
Therefore, 36.92% of the genetic variance remains unexplained.
1.73 PROBLEM 3: RECOMBINATION FRACTION ESTIMATION
In a QTL mapping study, you observe the following genotype frequencies in an F2 population of
1000 individuals:
• AA BB: 240
• AA Bb: 260
• AA bb: 10
• Aa BB: 255
• Aa Bb: 220
• Aa bb: 5
• aa BB: 8
• aa Bb: 2
• aa bb: 0
Estimate the recombination fraction between loci A and B.
Solution:
Step 1: Calculate the number of recombinant gametes Recombinant gametes are AB and ab.
𝑁𝐴𝐵 = 2(240)+255 +260 + 8 = 1003
𝑁𝑎𝑏 = 2(0)+5+2+10 =17
Total recombinant =1003 +17 =1020
Step 2: Calculate the total number of gametes
Total gametes = 2 × 1000 =2000
Step 3: Calculate the recombination fraction
𝑟 = Number of recombinant gametes
Total number of gametes
=1020
2000
= 0.51
The estimated recombination fraction between loci A and B is 0.51, suggesting that these loci
are unlinked (as r ≈ 0.5).
1.74 PROBLEM 4: QTL EFFECT SIZE CALCULATION
In a QTL mapping study for seed weight in soybeans, you’ve identified a QTL with the following
genotypic values:
• AA: 25g
• Aa: 22g
• aa: 18g
Calculate the additive (a) and dominance (d) effects of this QTL.
Solution:
Step 1: Calculate the additive effect (a) The additive effect is half the difference between the
homozygous genotypes:
𝑎 = 1
2(𝐴𝐴 −𝑎𝑎)
=1
2(25𝑔 − 18𝑔)
= 3.5𝑔
Step 2: Calculate the dominance effect (d) The dominance effect is the deviation of the
heterozygote from the average of the homozygotes:
𝑑 = 𝐴𝑎 −1
2(𝐴𝐴 +𝑎𝑎)
=22𝑔 − 1
2(25𝑔 + 18𝑔)
=22𝑔 − 21.5𝑔
= 0.5𝑔
Therefore, the additive effect (a) is 3.5g and the dominance effect (d) is 0.5g.
1.75 PROBLEM 5: POWER CALCULATION
You’re planning a QTL mapping study and want to determine the sample size needed to detect
a QTL that explains 5% of the phenotypic variance with 80% power at a significance level of
0.05. Assume you’re using an F2 population and a single marker analysis. Use the
approximation formula:
𝑛 = 2(𝑧𝛼/2 + 𝑧𝛽)2
ℎ𝑞
2
where 𝑛 is the required sample size, 𝑧𝛼/2 is the critical value of the standard normal distribution
for 𝛼/2, 𝑧𝛽 is the critical value for 𝛽 (1 - power), and ℎ𝑞
2 is the heritability explained by the QTL.
Solution:
Given:
• ℎ𝑞
2= 0.05 (5% of phenotypic variance)
• Power = 80%, so 𝛽 = 0.2
• Significance level 𝛼 = 0.05
Step 1: Find critical values
𝑧𝛼/2 = 𝑧0.025 = 1.96 (from standard normal table)
𝑧𝛽= 𝑧0.2 = 0.84 (from standard normal table)
Step 2: Apply the formula
𝑛 = 2(𝑧𝛼/2 + 𝑧𝛽)2
ℎ𝑞
2
=2(1.96 + 0.84)2
0.05
=2(2.8)2
0.05
=15.68
0.05
=313.6
Rounding up, you need a sample size of at least 314 F2 individuals to detect this QTL with 80%
power at a significance level of 0.05.
1.76 PROBLEM 1: LOD SCORE CALCULATION
Calculate the LOD score for a putative QTL given the following information:
• Likelihood under the alternative hypothesis (QTL present): 1.5 × 10−12
• Likelihood under the null hypothesis (no QTL): 2.7 × 10−15
Solution: The LOD score is calculated as:
𝐿𝑂𝐷 = log10 (Likelihood (QTL present)
Likelihood (no QTL) )
=log10 (1.5×10−12
2.7×10−15)
=log10(555.56)
= 2.74
Therefore, the
LOD score for this putative QTL is 2.74.
1.77 PROBLEM 2: HERITABILITY ESTIMATION
In a QTL mapping study for plant height in rice, you’ve identified 5 QTLs with the following
effect sizes:
• QTL 1: 15% of phenotypic variance
• QTL 2: 10% of phenotypic variance
• QTL 3: 8% of phenotypic variance
• QTL 4: 5% of phenotypic variance
• QTL 5: 3% of phenotypic variance
Calculate the total heritability explained by these QTLs. If the total heritability of plant height is
estimated to be 0.65, what percentage of the genetic variance remains unexplained?
Solution:
Step 1: Calculate total heritability explained by QTLs
ℎ𝑄𝑇𝐿
2= 0.15 + 0.10 + 0.08 + 0.05 + 0.03
= 0.41
Step 2: Calculate unexplained heritability
ℎ𝑢𝑛𝑒𝑥𝑝𝑙𝑎𝑖𝑛𝑒𝑑
2= ℎ𝑡𝑜𝑡𝑎𝑙
2− ℎ𝑄𝑇𝐿
2
= 0.65 − 0.41
= 0.24
Step 3: Calculate percentage of unexplained genetic variance
Percentage unexplained =ℎ𝑢𝑛𝑒𝑥𝑝𝑙𝑎𝑖𝑛𝑒𝑑
2
ℎ𝑡𝑜𝑡𝑎𝑙
2×100%
=0.24
0.65 ×100%
=36.92%
Therefore, 36.92% of the genetic variance remains unexplained.
1.78 PROBLEM 3: RECOMBINATION FRACTION ESTIMATION
In a QTL mapping study, you observe the following genotype frequencies in an F2 population of
1000 individuals:
• AA BB: 240
• AA Bb: 260
• AA bb: 10
• Aa BB: 255
• Aa Bb: 220
• Aa bb: 5
• aa BB: 8
• aa Bb: 2
• aa bb: 0
Estimate the recombination fraction between loci A and B.
Solution:
Step 1: Calculate the number of recombinant gametes Recombinant gametes are AB and ab.
𝑁𝐴𝐵 = 2(240)+255 +260 + 8 = 1003
𝑁𝑎𝑏 = 2(0)+5+2+10 =17
Total recombinant =1003 +17 =1020
Step 2: Calculate the total number of gametes
Total gametes = 2 × 1000 =2000
Step 3: Calculate the recombination fraction
𝑟 = Number of recombinant gametes
Total number of gametes
=1020
2000
= 0.51
The estimated recombination fraction between loci A and B is 0.51, suggesting that these loci
are unlinked (as r ≈ 0.5).
1.79 PROBLEM 4: QTL EFFECT SIZE CALCULATION
In a QTL mapping study for seed weight in soybeans, you’ve identified a QTL with the following
genotypic values:
• AA: 25g
• Aa: 22g
• aa: 18g
Calculate the additive (a) and dominance (d) effects of this QTL.
Solution:
Step 1: Calculate the additive effect (a) The additive effect is half the difference between the
homozygous genotypes:
𝑎 = 1
2(𝐴𝐴 −𝑎𝑎)
=1
2(25𝑔 − 18𝑔)
= 3.5𝑔
Step 2: Calculate the dominance effect (d) The dominance effect is the deviation of the
heterozygote from the average of the homozygotes:
𝑑 = 𝐴𝑎 −1
2(𝐴𝐴 +𝑎𝑎)
=22𝑔 − 1
2(25𝑔 + 18𝑔)
=22𝑔 − 21.5𝑔
= 0.5𝑔
Therefore, the additive effect (a) is 3.5g and the dominance effect (d) is 0.5g.
1.80 PROBLEM 5: POWER CALCULATION
You’re planning a QTL mapping study and want to determine the sample size needed to detect
a QTL that explains 5% of the phenotypic variance with 80% power at a significance level of
0.05. Assume you’re using an F2 population and a single marker analysis. Use the
approximation formula:
𝑛 = 2(𝑧𝛼/2 + 𝑧𝛽)2
ℎ𝑞
2
where 𝑛 is the required sample size, 𝑧𝛼/2 is the critical value of the standard normal distribution
for 𝛼/2, 𝑧𝛽 is the critical value for 𝛽 (1 - power), and ℎ𝑞
2 is the heritability explained by the QTL.
Solution:
Given:
• ℎ𝑞
2= 0.05 (5% of phenotypic variance)
• Power = 80%, so 𝛽 = 0.2
• Significance level 𝛼 = 0.05
Step 1: Find critical values
𝑧𝛼/2 = 𝑧0.025 = 1.96 (from standard normal table)
𝑧𝛽= 𝑧0.2 = 0.84 (from standard normal table)
Step 2: Apply the formula
𝑛 = 2(𝑧𝛼/2 + 𝑧𝛽)2
ℎ𝑞
2
=2(1.96 + 0.84)2
0.05
=2(2.8)2
0.05
=15.68
0.05
=313.6
Rounding up, you need a sample size of at least 314 F2 individuals to detect this QTL with 80%
power at a significance level of 0.05.
1.81 PROBLEM 1: LOD SCORE CALCULATION
Calculate the LOD score for a putative QTL given the following information:
• Likelihood under the alternative hypothesis (QTL present): 1.5 × 10−12
• Likelihood under the null hypothesis (no QTL): 2.7 × 10−15
Solution: The LOD score is calculated as:
𝐿𝑂𝐷 = log10 (Likelihood (QTL present)
Likelihood (no QTL) )
=log10 (1.5×10−12
2.7×10−15)
=log10(555.56)
= 2.74
Therefore, the
LOD score for this putative QTL is 2.74.
1.82 PROBLEM 2: HERITABILITY ESTIMATION
In a QTL mapping study for plant height in rice, you’ve identified 5 QTLs with the following
effect sizes:
• QTL 1: 15% of phenotypic variance
• QTL 2: 10% of phenotypic variance
• QTL 3: 8% of phenotypic variance
• QTL 4: 5% of phenotypic variance
• QTL 5: 3% of phenotypic variance
Calculate the total heritability explained by these QTLs. If the total heritability of plant height is
estimated to be 0.65, what percentage of the genetic variance remains unexplained?
Solution:
Step 1: Calculate total heritability explained by QTLs
ℎ𝑄𝑇𝐿
2= 0.15 + 0.10 + 0.08 + 0.05 + 0.03
= 0.41
Step 2: Calculate unexplained heritability
ℎ𝑢𝑛𝑒𝑥𝑝𝑙𝑎𝑖𝑛𝑒𝑑
2= ℎ𝑡𝑜𝑡𝑎𝑙
2− ℎ𝑄𝑇𝐿
2
= 0.65 − 0.41
= 0.24
Step 3: Calculate percentage of unexplained genetic variance
Percentage unexplained =ℎ𝑢𝑛𝑒𝑥𝑝𝑙𝑎𝑖𝑛𝑒𝑑
2
ℎ𝑡𝑜𝑡𝑎𝑙
2×100%
=0.24
0.65 ×100%
=36.92%
Therefore, 36.92% of the genetic variance remains unexplained.
1.83 PROBLEM 3: RECOMBINATION FRACTION ESTIMATION
In a QTL mapping study, you observe the following genotype frequencies in an F2 population of
1000 individuals:
• AA BB: 240
• AA Bb: 260
• AA bb: 10
• Aa BB: 255
• Aa Bb: 220
• Aa bb: 5
• aa BB: 8
• aa Bb: 2
• aa bb: 0
Estimate the recombination fraction between loci A and B.
Solution:
Step 1: Calculate the number of recombinant gametes Recombinant gametes are AB and ab.
𝑁𝐴𝐵 = 2(240)+255 +260 + 8 = 1003
𝑁𝑎𝑏 = 2(0)+5+2+10 =17
Total recombinant =1003 +17 =1020
Step 2: Calculate the total number of gametes
Total gametes = 2 × 1000 =2000
Step 3: Calculate the recombination fraction
𝑟 = Number of recombinant gametes
Total number of gametes
=1020
2000
= 0.51
The estimated recombination fraction between loci A and B is 0.51, suggesting that these loci
are unlinked (as r ≈ 0.5).
1.84 PROBLEM 4: QTL EFFECT SIZE CALCULATION
In a QTL mapping study for seed weight in soybeans, you’ve identified a QTL with the following
genotypic values:
• AA: 25g
• Aa: 22g
• aa: 18g
Calculate the additive (a) and dominance (d) effects of this QTL.
Solution:
Step 1: Calculate the additive effect (a) The additive effect is half the difference between the
homozygous genotypes:
𝑎 = 1
2(𝐴𝐴 −𝑎𝑎)
=1
2(25𝑔 − 18𝑔)
= 3.5𝑔
Step 2: Calculate the dominance effect (d) The dominance effect is the deviation of the
heterozygote from the average of the homozygotes:
𝑑 = 𝐴𝑎 −1
2(𝐴𝐴 +𝑎𝑎)
=22𝑔 − 1
2(25𝑔 + 18𝑔)
=22𝑔 − 21.5𝑔
= 0.5𝑔
Therefore, the additive effect (a) is 3.5g and the dominance effect (d) is 0.5g.
1.85 PROBLEM 5: POWER CALCULATION
You’re planning a QTL mapping study and want to determine the sample size needed to detect
a QTL that explains 5% of the phenotypic variance with 80% power at a significance level of
0.05. Assume you’re using an F2 population and a single marker analysis. Use the
approximation formula:
𝑛 = 2(𝑧𝛼/2 + 𝑧𝛽)2
ℎ𝑞
2
where 𝑛 is the required sample size, 𝑧𝛼/2 is the critical value of the standard normal distribution
for 𝛼/2, 𝑧𝛽 is the critical value for 𝛽 (1 - power), and ℎ𝑞
2 is the heritability explained by the QTL.
Solution:
Given:
• ℎ𝑞
2= 0.05 (5% of phenotypic variance)
• Power = 80%, so 𝛽 = 0.2
• Significance level 𝛼 = 0.05
Step 1: Find critical values
𝑧𝛼/2 = 𝑧0.025 = 1.96 (from standard normal table)
𝑧𝛽= 𝑧0.2 = 0.84 (from standard normal table)
Step 2: Apply the formula
𝑛 = 2(𝑧𝛼/2 + 𝑧𝛽)2
ℎ𝑞
2
=2(1.96 + 0.84)2
0.05
=2(2.8)2
0.05
=15.68
0.05
=313.6
Rounding up, you need a sample size of at least 314 F2 individuals to detect this QTL with 80%
power at a significance level of 0.05.
1.86 PROBLEM 1: LOD SCORE CALCULATION
Calculate the LOD score for a putative QTL given the following information:
• Likelihood under the alternative hypothesis (QTL present): 1.5 × 10−12
• Likelihood under the null hypothesis (no QTL): 2.7 × 10−15
Solution: The LOD score is calculated as:
𝐿𝑂𝐷 = log10 (Likelihood (QTL present)
Likelihood (no QTL) )
=log10 (1.5×10−12
2.7×10−15)
=log10(555.56)
= 2.74
Therefore, the
LOD score for this putative QTL is 2.74.
1.87 PROBLEM 2: HERITABILITY ESTIMATION
In a QTL mapping study for plant height in rice, you’ve identified 5 QTLs with the following
effect sizes:
• QTL 1: 15% of phenotypic variance
• QTL 2: 10% of phenotypic variance
• QTL 3: 8% of phenotypic variance
• QTL 4: 5% of phenotypic variance
• QTL 5: 3% of phenotypic variance
Calculate the total heritability explained by these QTLs. If the total heritability of plant height is
estimated to be 0.65, what percentage of the genetic variance remains unexplained?
Solution:
Step 1: Calculate total heritability explained by QTLs
ℎ𝑄𝑇𝐿
2= 0.15 + 0.10 + 0.08 + 0.05 + 0.03
= 0.41
Step 2: Calculate unexplained heritability
ℎ𝑢𝑛𝑒𝑥𝑝𝑙𝑎𝑖𝑛𝑒𝑑
2= ℎ𝑡𝑜𝑡𝑎𝑙
2− ℎ𝑄𝑇𝐿
2
= 0.65 − 0.41
= 0.24
Step 3: Calculate percentage of unexplained genetic variance
Percentage unexplained =ℎ𝑢𝑛𝑒𝑥𝑝𝑙𝑎𝑖𝑛𝑒𝑑
2
ℎ𝑡𝑜𝑡𝑎𝑙
2×100%
=0.24
0.65 ×100%
=36.92%
Therefore, 36.92% of the genetic variance remains unexplained.
1.88 PROBLEM 3: RECOMBINATION FRACTION ESTIMATION
In a QTL mapping study, you observe the following genotype frequencies in an F2 population of
1000 individuals:
• AA BB: 240
• AA Bb: 260
• AA bb: 10
• Aa BB: 255
• Aa Bb: 220
• Aa bb: 5
• aa BB: 8
• aa Bb: 2
• aa bb: 0
Estimate the recombination fraction between loci A and B.
Solution:
Step 1: Calculate the number of recombinant gametes Recombinant gametes are AB and ab.
𝑁𝐴𝐵 = 2(240)+255 +260 + 8 = 1003
𝑁𝑎𝑏 = 2(0)+5+2+10 =17
Total recombinant =1003 +17 =1020
Step 2: Calculate the total number of gametes
Total gametes = 2 × 1000 =2000
Step 3: Calculate the recombination fraction
𝑟 = Number of recombinant gametes
Total number of gametes
=1020
2000
= 0.51
The estimated recombination fraction between loci A and B is 0.51, suggesting that these loci
are unlinked (as r ≈ 0.5).
1.89 PROBLEM 4: QTL EFFECT SIZE CALCULATION
In a QTL mapping study for seed weight in soybeans, you’ve identified a QTL with the following
genotypic values:
• AA: 25g
• Aa: 22g
• aa: 18g
Calculate the additive (a) and dominance (d) effects of this QTL.
Solution:
Step 1: Calculate the additive effect (a) The additive effect is half the difference between the
homozygous genotypes:
𝑎 = 1
2(𝐴𝐴 −𝑎𝑎)
=1
2(25𝑔 − 18𝑔)
= 3.5𝑔
Step 2: Calculate the dominance effect (d) The dominance effect is the deviation of the
heterozygote from the average of the homozygotes:
𝑑 = 𝐴𝑎 −1
2(𝐴𝐴 +𝑎𝑎)
=22𝑔 − 1
2(25𝑔 + 18𝑔)
=22𝑔 − 21.5𝑔
= 0.5𝑔
Therefore, the additive effect (a) is 3.5g and the dominance effect (d) is 0.5g.
1.90 PROBLEM 5: POWER CALCULATION
You’re planning a QTL mapping study and want to determine the sample size needed to detect
a QTL that explains 5% of the phenotypic variance with 80% power at a significance level of
0.05. Assume you’re using an F2 population and a single marker analysis. Use the
approximation formula:
𝑛 = 2(𝑧𝛼/2 + 𝑧𝛽)2
ℎ𝑞
2
where 𝑛 is the required sample size, 𝑧𝛼/2 is the critical value of the standard normal distribution
for 𝛼/2, 𝑧𝛽 is the critical value for 𝛽 (1 - power), and ℎ𝑞
2 is the heritability explained by the QTL.
Solution:
Given:
• ℎ𝑞
2= 0.05 (5% of phenotypic variance)
• Power = 80%, so 𝛽 = 0.2
• Significance level 𝛼 = 0.05
Step 1: Find critical values
𝑧𝛼/2 = 𝑧0.025 = 1.96 (from standard normal table)
𝑧𝛽= 𝑧0.2 = 0.84 (from standard normal table)
Step 2: Apply the formula
𝑛 = 2(𝑧𝛼/2 + 𝑧𝛽)2
ℎ𝑞
2
=2(1.96 + 0.84)2
0.05
=2(2.8)2
0.05
=15.68
0.05
=313.6
Rounding up, you need a sample size of at least 314 F2 individuals to detect this QTL with 80%
power at a significance level of 0.05.
1.91 PROBLEM 1: LOD SCORE CALCULATION
Calculate the LOD score for a putative QTL given the following information:
• Likelihood under the alternative hypothesis (QTL present): 1.5 × 10−12
• Likelihood under the null hypothesis (no QTL): 2.7 × 10−15
Solution: The LOD score is calculated as:
𝐿𝑂𝐷 = log10 (Likelihood (QTL present)
Likelihood (no QTL) )
=log10 (1.5×10−12
2.7×10−15)
=log10(555.56)
= 2.74
Therefore, the
LOD score for this putative QTL is 2.74.
1.92 PROBLEM 2: HERITABILITY ESTIMATION
In a QTL mapping study for plant height in rice, you’ve identified 5 QTLs with the following
effect sizes:
• QTL 1: 15% of phenotypic variance
• QTL 2: 10% of phenotypic variance
• QTL 3: 8% of phenotypic variance
• QTL 4: 5% of phenotypic variance
• QTL 5: 3% of phenotypic variance
Calculate the total heritability explained by these QTLs. If the total heritability of plant height is
estimated to be 0.65, what percentage of the genetic variance remains unexplained?
Solution:
Step 1: Calculate total heritability explained by QTLs
ℎ𝑄𝑇𝐿
2= 0.15 + 0.10 + 0.08 + 0.05 + 0.03
= 0.41
Step 2: Calculate unexplained heritability
ℎ𝑢𝑛𝑒𝑥𝑝𝑙𝑎𝑖𝑛𝑒𝑑
2= ℎ𝑡𝑜𝑡𝑎𝑙
2− ℎ𝑄𝑇𝐿
2
= 0.65 − 0.41
= 0.24
Step 3: Calculate percentage of unexplained genetic variance
Percentage unexplained =ℎ𝑢𝑛𝑒𝑥𝑝𝑙𝑎𝑖𝑛𝑒𝑑
2
ℎ𝑡𝑜𝑡𝑎𝑙
2×100%
=0.24
0.65 ×100%
=36.92%
Therefore, 36.92% of the genetic variance remains unexplained.
1.93 PROBLEM 3: RECOMBINATION FRACTION ESTIMATION
In a QTL mapping study, you observe the following genotype frequencies in an F2 population of
1000 individuals:
• AA BB: 240
• AA Bb: 260
• AA bb: 10
• Aa BB: 255
• Aa Bb: 220
• Aa bb: 5
• aa BB: 8
• aa Bb: 2
• aa bb: 0
Estimate the recombination fraction between loci A and B.
Solution:
Step 1: Calculate the number of recombinant gametes Recombinant gametes are AB and ab.
𝑁𝐴𝐵 = 2(240)+255 +260 + 8 = 1003
𝑁𝑎𝑏 = 2(0)+5+2+10 =17
Total recombinant =1003 +17 =1020
Step 2: Calculate the total number of gametes
Total gametes = 2 × 1000 =2000
Step 3: Calculate the recombination fraction
𝑟 = Number of recombinant gametes
Total number of gametes
=1020
2000
= 0.51
The estimated recombination fraction between loci A and B is 0.51, suggesting that these loci
are unlinked (as r ≈ 0.5).
1.94 PROBLEM 4: QTL EFFECT SIZE CALCULATION
In a QTL mapping study for seed weight in soybeans, you’ve identified a QTL with the following
genotypic values:
• AA: 25g
• Aa: 22g
• aa: 18g
Calculate the additive (a) and dominance (d) effects of this QTL.
Solution:
Step 1: Calculate the additive effect (a) The additive effect is half the difference between the
homozygous genotypes:
𝑎 = 1
2(𝐴𝐴 −𝑎𝑎)
=1
2(25𝑔 − 18𝑔)
= 3.5𝑔
Step 2: Calculate the dominance effect (d) The dominance effect is the deviation of the
heterozygote from the average of the homozygotes:
𝑑 = 𝐴𝑎 −1
2(𝐴𝐴 +𝑎𝑎)
=22𝑔 − 1
2(25𝑔 + 18𝑔)
=22𝑔 − 21.5𝑔
= 0.5𝑔
Therefore, the additive effect (a) is 3.5g and the dominance effect (d) is 0.5g.
1.95 PROBLEM 5: POWER CALCULATION
You’re planning a QTL mapping study and want to determine the sample size needed to detect
a QTL that explains 5% of the phenotypic variance with 80% power at a significance level of
0.05. Assume you’re using an F2 population and a single marker analysis. Use the
approximation formula:
𝑛 = 2(𝑧𝛼/2 + 𝑧𝛽)2
ℎ𝑞
2
where 𝑛 is the required sample size, 𝑧𝛼/2 is the critical value of the standard normal distribution
for 𝛼/2, 𝑧𝛽 is the critical value for 𝛽 (1 - power), and ℎ𝑞
2 is the heritability explained by the QTL.
Solution:
Given:
• ℎ𝑞
2= 0.05 (5% of phenotypic variance)
• Power = 80%, so 𝛽 = 0.2
• Significance level 𝛼 = 0.05
Step 1: Find critical values
𝑧𝛼/2 = 𝑧0.025 = 1.96 (from standard normal table)
𝑧𝛽= 𝑧0.2 = 0.84 (from standard normal table)
Step 2: Apply the formula
𝑛 = 2(𝑧𝛼/2 + 𝑧𝛽)2
ℎ𝑞
2
=2(1.96 + 0.84)2
0.05
=2(2.8)2
0.05
=15.68
0.05
=313.6
Rounding up, you need a sample size of at least 314 F2 individuals to detect this QTL with 80%
power at a significance level of 0.05.
1.96 PROBLEM 1: LOD SCORE CALCULATION
Calculate the LOD score for a putative QTL given the following information:
• Likelihood under the alternative hypothesis (QTL present): 1.5 × 10−12
• Likelihood under the null hypothesis (no QTL): 2.7 × 10−15
Solution: The LOD score is calculated as:
𝐿𝑂𝐷 = log10 (Likelihood (QTL present)
Likelihood (no QTL) )
=log10 (1.5×10−12
2.7×10−15)
=log10(555.56)
= 2.74
Therefore, the
LOD score for this putative QTL is 2.74.
1.97 PROBLEM 2: HERITABILITY ESTIMATION
In a QTL mapping study for plant height in rice, you’ve identified 5 QTLs with the following
effect sizes:
• QTL 1: 15% of phenotypic variance
• QTL 2: 10% of phenotypic variance
• QTL 3: 8% of phenotypic variance
• QTL 4: 5% of phenotypic variance
• QTL 5: 3% of phenotypic variance
Calculate the total heritability explained by these QTLs. If the total heritability of plant height is
estimated to be 0.65, what percentage of the genetic variance remains unexplained?
Solution:
Step 1: Calculate total heritability explained by QTLs
ℎ𝑄𝑇𝐿
2= 0.15 + 0.10 + 0.08 + 0.05 + 0.03
= 0.41
Step 2: Calculate unexplained heritability
ℎ𝑢𝑛𝑒𝑥𝑝𝑙𝑎𝑖𝑛𝑒𝑑
2= ℎ𝑡𝑜𝑡𝑎𝑙
2− ℎ𝑄𝑇𝐿
2
= 0.65 − 0.41
= 0.24
Step 3: Calculate percentage of unexplained genetic variance
Percentage unexplained =ℎ𝑢𝑛𝑒𝑥𝑝𝑙𝑎𝑖𝑛𝑒𝑑
2
ℎ𝑡𝑜𝑡𝑎𝑙
2×100%
=0.24
0.65 ×100%
=36.92%
Therefore, 36.92% of the genetic variance remains unexplained.
1.98 PROBLEM 3: RECOMBINATION FRACTION ESTIMATION
In a QTL mapping study, you observe the following genotype frequencies in an F2 population of
1000 individuals:
• AA BB: 240
• AA Bb: 260
• AA bb: 10
• Aa BB: 255
• Aa Bb: 220
• Aa bb: 5
• aa BB: 8
• aa Bb: 2
• aa bb: 0
Estimate the recombination fraction between loci A and B.
Solution:
Step 1: Calculate the number of recombinant gametes Recombinant gametes are AB and ab.
𝑁𝐴𝐵 = 2(240)+255 +260 + 8 = 1003
𝑁𝑎𝑏 = 2(0)+5+2+10 =17
Total recombinant =1003 +17 =1020
Step 2: Calculate the total number of gametes
Total gametes = 2 × 1000 =2000
Step 3: Calculate the recombination fraction
𝑟 = Number of recombinant gametes
Total number of gametes
=1020
2000
= 0.51
The estimated recombination fraction between loci A and B is 0.51, suggesting that these loci
are unlinked (as r ≈ 0.5).
1.99 PROBLEM 4: QTL EFFECT SIZE CALCULATION
In a QTL mapping study for seed weight in soybeans, you’ve identified a QTL with the following
genotypic values:
• AA: 25g
• Aa: 22g
• aa: 18g
Calculate the additive (a) and dominance (d) effects of this QTL.
Solution:
Step 1: Calculate the additive effect (a) The additive effect is half the difference between the
homozygous genotypes:
𝑎 = 1
2(𝐴𝐴 −𝑎𝑎)
=1
2(25𝑔 − 18𝑔)
= 3.5𝑔
Step 2: Calculate the dominance effect (d) The dominance effect is the deviation of the
heterozygote from the average of the homozygotes:
𝑑 = 𝐴𝑎 −1
2(𝐴𝐴 +𝑎𝑎)
=22𝑔 − 1
2(25𝑔 + 18𝑔)
=22𝑔 − 21.5𝑔
= 0.5𝑔
Therefore, the additive effect (a) is 3.5g and the dominance effect (d) is 0.5g.
1.100 PROBLEM 5: POWER CALCULATION
You’re planning a QTL mapping study and want to determine the sample size needed to detect
a QTL that explains 5% of the phenotypic variance with 80% power at a significance level of
0.05. Assume you’re using an F2 population and a single marker analysis. Use the
approximation formula:
𝑛 = 2(𝑧𝛼/2 + 𝑧𝛽)2
ℎ𝑞
2
where 𝑛 is the required sample size, 𝑧𝛼/2 is the critical value of the standard normal distribution
for 𝛼/2, 𝑧𝛽 is the critical value for 𝛽 (1 - power), and ℎ𝑞
2 is the heritability explained by the QTL.
Solution:
Given:
• ℎ𝑞
2= 0.05 (5% of phenotypic variance)
• Power = 80%, so 𝛽 = 0.2
• Significance level 𝛼 = 0.05
Step 1: Find critical values
𝑧𝛼/2 = 𝑧0.025 = 1.96 (from standard normal table)
𝑧𝛽= 𝑧0.2 = 0.84 (from standard normal table)
Step 2: Apply the formula
𝑛 = 2(𝑧𝛼/2 + 𝑧𝛽)2
ℎ𝑞
2
=2(1.96 + 0.84)2
0.05
=2(2.8)2
0.05
=15.68
0.05
=313.6
Rounding up, you need a sample size of at least 314 F2 individuals to detect this QTL with 80%
power at a significance level of 0.05.
1.101 PROBLEM 1: LOD SCORE CALCULATION
Calculate the LOD score for a putative QTL given the following information:
• Likelihood under the alternative hypothesis (QTL present): 1.5 × 10−12
• Likelihood under the null hypothesis (no QTL): 2.7 × 10−15
Solution: The LOD score is calculated as:
𝐿𝑂𝐷 = log10 (Likelihood (QTL present)
Likelihood (no QTL) )
=log10 (1.5×10−12
2.7×10−15)
=log10(555.56)
= 2.74
Therefore, the
LOD score for this putative QTL is 2.74.
1.102 PROBLEM 2: HERITABILITY ESTIMATION
In a QTL mapping study for plant height in rice, you’ve identified 5 QTLs with the following
effect sizes:
• QTL 1: 15% of phenotypic variance
• QTL 2: 10% of phenotypic variance
• QTL 3: 8% of phenotypic variance
• QTL 4: 5% of phenotypic variance
• QTL 5: 3% of phenotypic variance
Calculate the total heritability explained by these QTLs. If the total heritability of plant height is
estimated to be 0.65, what percentage of the genetic variance remains unexplained?
Solution:
Step 1: Calculate total heritability explained by QTLs
ℎ𝑄𝑇𝐿
2= 0.15 + 0.10 + 0.08 + 0.05 + 0.03
= 0.41
Step 2: Calculate unexplained heritability
ℎ𝑢𝑛𝑒𝑥𝑝𝑙𝑎𝑖𝑛𝑒𝑑
2= ℎ𝑡𝑜𝑡𝑎𝑙
2− ℎ𝑄𝑇𝐿
2
= 0.65 − 0.41
= 0.24
Step 3: Calculate percentage of unexplained genetic variance
Percentage unexplained =ℎ𝑢𝑛𝑒𝑥𝑝𝑙𝑎𝑖𝑛𝑒𝑑
2
ℎ𝑡𝑜𝑡𝑎𝑙
2×100%
=0.24
0.65 ×100%
=36.92%
Therefore, 36.92% of the genetic variance remains unexplained.
1.103 PROBLEM 3: RECOMBINATION FRACTION ESTIMATION
In a QTL mapping study, you observe the following genotype frequencies in an F2 population of
1000 individuals:
• AA BB: 240
• AA Bb: 260
• AA bb: 10
• Aa BB: 255
• Aa Bb: 220
• Aa bb: 5
• aa BB: 8
• aa Bb: 2
• aa bb: 0
Estimate the recombination fraction between loci A and B.
Solution:
Step 1: Calculate the number of recombinant gametes Recombinant gametes are AB and ab.
𝑁𝐴𝐵 = 2(240)+255 +260 + 8 = 1003
𝑁𝑎𝑏 = 2(0)+5+2+10 =17
Total recombinant =1003 +17 =1020
Step 2: Calculate the total number of gametes
Total gametes = 2 × 1000 =2000
Step 3: Calculate the recombination fraction
𝑟 = Number of recombinant gametes
Total number of gametes
=1020
2000
= 0.51
The estimated recombination fraction between loci A and B is 0.51, suggesting that these loci
are unlinked (as r ≈ 0.5).
1.104 PROBLEM 4: QTL EFFECT SIZE CALCULATION
In a QTL mapping study for seed weight in soybeans, you’ve identified a QTL with the following
genotypic values:
• AA: 25g
• Aa: 22g
• aa: 18g
Calculate the additive (a) and dominance (d) effects of this QTL.
Solution:
Step 1: Calculate the additive effect (a) The additive effect is half the difference between the
homozygous genotypes:
𝑎 = 1
2(𝐴𝐴 −𝑎𝑎)
=1
2(25𝑔 − 18𝑔)
= 3.5𝑔
Step 2: Calculate the dominance effect (d) The dominance effect is the deviation of the
heterozygote from the average of the homozygotes:
𝑑 = 𝐴𝑎 −1
2(𝐴𝐴 +𝑎𝑎)
=22𝑔 − 1
2(25𝑔 + 18𝑔)
=22𝑔 − 21.5𝑔
= 0.5𝑔
Therefore, the additive effect (a) is 3.5g and the dominance effect (d) is 0.5g.
1.105 PROBLEM 5: POWER CALCULATION
You’re planning a QTL mapping study and want to determine the sample size needed to detect
a QTL that explains 5% of the phenotypic variance with 80% power at a significance level of
0.05. Assume you’re using an F2 population and a single marker analysis. Use the
approximation formula:
𝑛 = 2(𝑧𝛼/2 + 𝑧𝛽)2
ℎ𝑞
2
where 𝑛 is the required sample size, 𝑧𝛼/2 is the critical value of the standard normal distribution
for 𝛼/2, 𝑧𝛽 is the critical value for 𝛽 (1 - power), and ℎ𝑞
2 is the heritability explained by the QTL.
Solution:
Given:
• ℎ𝑞
2= 0.05 (5% of phenotypic variance)
• Power = 80%, so 𝛽 = 0.2
• Significance level 𝛼 = 0.05
Step 1: Find critical values
𝑧𝛼/2 = 𝑧0.025 = 1.96 (from standard normal table)
𝑧𝛽= 𝑧0.2 = 0.84 (from standard normal table)
Step 2: Apply the formula
𝑛 = 2(𝑧𝛼/2 + 𝑧𝛽)2
ℎ𝑞
2
=2(1.96 + 0.84)2
0.05
=2(2.8)2
0.05
=15.68
0.05
=313.6
Rounding up, you need a sample size of at least 314 F2 individuals to detect this QTL with 80%
power at a significance level of 0.05.
1.106 PROBLEM 1: LOD SCORE CALCULATION
Calculate the LOD score for a putative QTL given the following information:
• Likelihood under the alternative hypothesis (QTL present): 1.5 × 10−12
• Likelihood under the null hypothesis (no QTL): 2.7 × 10−15
Solution: The LOD score is calculated as:
𝐿𝑂𝐷 = log10 (Likelihood (QTL present)
Likelihood (no QTL) )
=log10 (1.5×10−12
2.7×10−15)
=log10(555.56)
= 2.74
Therefore, the
LOD score for this putative QTL is 2.74.
1.107 PROBLEM 2: HERITABILITY ESTIMATION
In a QTL mapping study for plant height in rice, you’ve identified 5 QTLs with the following
effect sizes:
• QTL 1: 15% of phenotypic variance
• QTL 2: 10% of phenotypic variance
• QTL 3: 8% of phenotypic variance
• QTL 4: 5% of phenotypic variance
• QTL 5: 3% of phenotypic variance
Calculate the total heritability explained by these QTLs. If the total heritability of plant height is
estimated to be 0.65, what percentage of the genetic variance remains unexplained?
Solution:
Step 1: Calculate total heritability explained by QTLs
ℎ𝑄𝑇𝐿
2= 0.15 + 0.10 + 0.08 + 0.05 + 0.03
= 0.41
Step 2: Calculate unexplained heritability
ℎ𝑢𝑛𝑒𝑥𝑝𝑙𝑎𝑖𝑛𝑒𝑑
2= ℎ𝑡𝑜𝑡𝑎𝑙
2− ℎ𝑄𝑇𝐿
2
= 0.65 − 0.41
= 0.24
Step 3: Calculate percentage of unexplained genetic variance
Percentage unexplained =ℎ𝑢𝑛𝑒𝑥𝑝𝑙𝑎𝑖𝑛𝑒𝑑
2
ℎ𝑡𝑜𝑡𝑎𝑙
2×100%
=0.24
0.65 ×100%
=36.92%
Therefore, 36.92% of the genetic variance remains unexplained.
1.108 PROBLEM 3: RECOMBINATION FRACTION ESTIMATION
In a QTL mapping study, you observe the following genotype frequencies in an F2 population of
1000 individuals:
• AA BB: 240
• AA Bb: 260
• AA bb: 10
• Aa BB: 255
• Aa Bb: 220
• Aa bb: 5
• aa BB: 8
• aa Bb: 2
• aa bb: 0
Estimate the recombination fraction between loci A and B.
Solution:
Step 1: Calculate the number of recombinant gametes Recombinant gametes are AB and ab.
𝑁𝐴𝐵 = 2(240)+255 +260 + 8 = 1003
𝑁𝑎𝑏 = 2(0)+5+2+10 =17
Total recombinant =1003 +17 =1020
Step 2: Calculate the total number of gametes
Total gametes = 2 × 1000 =2000
Step 3: Calculate the recombination fraction
𝑟 = Number of recombinant gametes
Total number of gametes
=1020
2000
= 0.51
The estimated recombination fraction between loci A and B is 0.51, suggesting that these loci
are unlinked (as r ≈ 0.5).
1.109 PROBLEM 4: QTL EFFECT SIZE CALCULATION
In a QTL mapping study for seed weight in soybeans, you’ve identified a QTL with the following
genotypic values:
• AA: 25g
• Aa: 22g
• aa: 18g
Calculate the additive (a) and dominance (d) effects of this QTL.
Solution:
Step 1: Calculate the additive effect (a) The additive effect is half the difference between the
homozygous genotypes:
𝑎 = 1
2(𝐴𝐴 −𝑎𝑎)
=1
2(25𝑔 − 18𝑔)
= 3.5𝑔
Step 2: Calculate the dominance effect (d) The dominance effect is the deviation of the
heterozygote from the average of the homozygotes:
𝑑 = 𝐴𝑎 −1
2(𝐴𝐴 +𝑎𝑎)
=22𝑔 − 1
2(25𝑔 + 18𝑔)
=22𝑔 − 21.5𝑔
= 0.5𝑔
Therefore, the additive effect (a) is 3.5g and the dominance effect (d) is 0.5g.
1.110 PROBLEM 5: POWER CALCULATION
You’re planning a QTL mapping study and want to determine the sample size needed to detect
a QTL that explains 5% of the phenotypic variance with 80% power at a significance level of
0.05. Assume you’re using an F2 population and a single marker analysis. Use the
approximation formula:
𝑛 = 2(𝑧𝛼/2 + 𝑧𝛽)2
ℎ𝑞
2
where 𝑛 is the required sample size, 𝑧𝛼/2 is the critical value of the standard normal distribution
for 𝛼/2, 𝑧𝛽 is the critical value for 𝛽 (1 - power), and ℎ𝑞
2 is the heritability explained by the QTL.
Solution:
Given:
• ℎ𝑞
2= 0.05 (5% of phenotypic variance)
• Power = 80%, so 𝛽 = 0.2
• Significance level 𝛼 = 0.05
Step 1: Find critical values
𝑧𝛼/2 = 𝑧0.025 = 1.96 (from standard normal table)
𝑧𝛽= 𝑧0.2 = 0.84 (from standard normal table)
Step 2: Apply the formula
𝑛 = 2(𝑧𝛼/2 + 𝑧𝛽)2
ℎ𝑞
2
=2(1.96 + 0.84)2
0.05
=2(2.8)2
0.05
=15.68
0.05
=313.6
Rounding up, you need a sample size of at least 314 F2 individuals to detect this QTL with 80%
power at a significance level of 0.05.
1.111 PROBLEM 1: LOD SCORE CALCULATION
Calculate the LOD score for a putative QTL given the following information:
• Likelihood under the alternative hypothesis (QTL present): 1.5 × 10−12
• Likelihood under the null hypothesis (no QTL): 2.7 × 10−15
Solution: The LOD score is calculated as:
𝐿𝑂𝐷 = log10 (Likelihood (QTL present)
Likelihood (no QTL) )
=log10 (1.5×10−12
2.7×10−15)
=log10(555.56)
= 2.74
Therefore, the
LOD score for this putative QTL is 2.74.
1.112 PROBLEM 2: HERITABILITY ESTIMATION
In a QTL mapping study for plant height in rice, you’ve identified 5 QTLs with the following
effect sizes:
• QTL 1: 15% of phenotypic variance
• QTL 2: 10% of phenotypic variance
• QTL 3: 8% of phenotypic variance
• QTL 4: 5% of phenotypic variance
• QTL 5: 3% of phenotypic variance
Calculate the total heritability explained by these QTLs. If the total heritability of plant height is
estimated to be 0.65, what percentage of the genetic variance remains unexplained?
Solution:
Step 1: Calculate total heritability explained by QTLs
ℎ𝑄𝑇𝐿
2= 0.15 + 0.10 + 0.08 + 0.05 + 0.03
= 0.41
Step 2: Calculate unexplained heritability
ℎ𝑢𝑛𝑒𝑥𝑝𝑙𝑎𝑖𝑛𝑒𝑑
2= ℎ𝑡𝑜𝑡𝑎𝑙
2− ℎ𝑄𝑇𝐿
2
= 0.65 − 0.41
= 0.24
Step 3: Calculate percentage of unexplained genetic variance
Percentage unexplained =ℎ𝑢𝑛𝑒𝑥𝑝𝑙𝑎𝑖𝑛𝑒𝑑
2
ℎ𝑡𝑜𝑡𝑎𝑙
2×100%
=0.24
0.65 ×100%
=36.92%
Therefore, 36.92% of the genetic variance remains unexplained.
1.113 PROBLEM 3: RECOMBINATION FRACTION ESTIMATION
In a QTL mapping study, you observe the following genotype frequencies in an F2 population of
1000 individuals:
• AA BB: 240
• AA Bb: 260
• AA bb: 10
• Aa BB: 255
• Aa Bb: 220
• Aa bb: 5
• aa BB: 8
• aa Bb: 2
• aa bb: 0
Estimate the recombination fraction between loci A and B.
Solution:
Step 1: Calculate the number of recombinant gametes Recombinant gametes are AB and ab.
𝑁𝐴𝐵 = 2(240)+255 +260 + 8 = 1003
𝑁𝑎𝑏 = 2(0)+5+2+10 =17
Total recombinant =1003 +17 =1020
Step 2: Calculate the total number of gametes
Total gametes = 2 × 1000 =2000
Step 3: Calculate the recombination fraction
𝑟 = Number of recombinant gametes
Total number of gametes
=1020
2000
= 0.51
The estimated recombination fraction between loci A and B is 0.51, suggesting that these loci
are unlinked (as r ≈ 0.5).
1.114 PROBLEM 4: QTL EFFECT SIZE CALCULATION
In a QTL mapping study for seed weight in soybeans, you’ve identified a QTL with the following
genotypic values:
• AA: 25g
• Aa: 22g
• aa: 18g
Calculate the additive (a) and dominance (d) effects of this QTL.
Solution:
Step 1: Calculate the additive effect (a) The additive effect is half the difference between the
homozygous genotypes:
𝑎 = 1
2(𝐴𝐴 −𝑎𝑎)
=1
2(25𝑔 − 18𝑔)
= 3.5𝑔
Step 2: Calculate the dominance effect (d) The dominance effect is the deviation of the
heterozygote from the average of the homozygotes:
𝑑 = 𝐴𝑎 −1
2(𝐴𝐴 +𝑎𝑎)
=22𝑔 − 1
2(25𝑔 + 18𝑔)
=22𝑔 − 21.5𝑔
= 0.5𝑔
Therefore, the additive effect (a) is 3.5g and the dominance effect (d) is 0.5g.
1.115 PROBLEM 5: POWER CALCULATION
You’re planning a QTL mapping study and want to determine the sample size needed to detect
a QTL that explains 5% of the phenotypic variance with 80% power at a significance level of
0.05. Assume you’re using an F2 population and a single marker analysis. Use the
approximation formula:
𝑛 = 2(𝑧𝛼/2 + 𝑧𝛽)2
ℎ𝑞
2
where 𝑛 is the required sample size, 𝑧𝛼/2 is the critical value of the standard normal distribution
for 𝛼/2, 𝑧𝛽 is the critical value for 𝛽 (1 - power), and ℎ𝑞
2 is the heritability explained by the QTL.
Solution:
Given:
• ℎ𝑞
2= 0.05 (5% of phenotypic variance)
• Power = 80%, so 𝛽 = 0.2
• Significance level 𝛼 = 0.05
Step 1: Find critical values
𝑧𝛼/2 = 𝑧0.025 = 1.96 (from standard normal table)
𝑧𝛽= 𝑧0.2 = 0.84 (from standard normal table)
Step 2: Apply the formula
𝑛 = 2(𝑧𝛼/2 + 𝑧𝛽)2
ℎ𝑞
2
=2(1.96 + 0.84)2
0.05
=2(2.8)2
0.05
=15.68
0.05
=313.6
Rounding up, you need a sample size of at least 314 F2 individuals to detect this QTL with 80%
power at a significance level of 0.05.
1.116 PROBLEM 1: LOD SCORE CALCULATION
Calculate the LOD score for a putative QTL given the following information:
• Likelihood under the alternative hypothesis (QTL present): 1.5 × 10−12
• Likelihood under the null hypothesis (no QTL): 2.7 × 10−15
Solution: The LOD score is calculated as:
𝐿𝑂𝐷 = log10 (Likelihood (QTL present)
Likelihood (no QTL) )
=log10 (1.5×10−12
2.7×10−15)
=log10(555.56)
= 2.74
Therefore, the
LOD score for this putative QTL is 2.74.
1.117 PROBLEM 2: HERITABILITY ESTIMATION
In a QTL mapping study for plant height in rice, you’ve identified 5 QTLs with the following
effect sizes:
• QTL 1: 15% of phenotypic variance
• QTL 2: 10% of phenotypic variance
• QTL 3: 8% of phenotypic variance
• QTL 4: 5% of phenotypic variance
• QTL 5: 3% of phenotypic variance
Calculate the total heritability explained by these QTLs. If the total heritability of plant height is
estimated to be 0.65, what percentage of the genetic variance remains unexplained?
Solution:
Step 1: Calculate total heritability explained by QTLs
ℎ𝑄𝑇𝐿
2= 0.15 + 0.10 + 0.08 + 0.05 + 0.03
= 0.41
Step 2: Calculate unexplained heritability
ℎ𝑢𝑛𝑒𝑥𝑝𝑙𝑎𝑖𝑛𝑒𝑑
2= ℎ𝑡𝑜𝑡𝑎𝑙
2− ℎ𝑄𝑇𝐿
2
= 0.65 − 0.41
= 0.24
Step 3: Calculate percentage of unexplained genetic variance
Percentage unexplained =ℎ𝑢𝑛𝑒𝑥𝑝𝑙𝑎𝑖𝑛𝑒𝑑
2
ℎ𝑡𝑜𝑡𝑎𝑙
2×100%
=0.24
0.65 ×100%
=36.92%
Therefore, 36.92% of the genetic variance remains unexplained.
1.118 PROBLEM 3: RECOMBINATION FRACTION ESTIMATION
In a QTL mapping study, you observe the following genotype frequencies in an F2 population of
1000 individuals:
• AA BB: 240
• AA Bb: 260
• AA bb: 10
• Aa BB: 255
• Aa Bb: 220
• Aa bb: 5
• aa BB: 8
• aa Bb: 2
• aa bb: 0
Estimate the recombination fraction between loci A and B.
Solution:
Step 1: Calculate the number of recombinant gametes Recombinant gametes are AB and ab.
𝑁𝐴𝐵 = 2(240)+255 +260 + 8 = 1003
𝑁𝑎𝑏 = 2(0)+5+2+10 =17
Total recombinant =1003 +17 =1020
Step 2: Calculate the total number of gametes
Total gametes = 2 × 1000 =2000
Step 3: Calculate the recombination fraction
𝑟 = Number of recombinant gametes
Total number of gametes
=1020
2000
= 0.51
The estimated recombination fraction between loci A and B is 0.51, suggesting that these loci
are unlinked (as r ≈ 0.5).
1.119 PROBLEM 4: QTL EFFECT SIZE CALCULATION
In a QTL mapping study for seed weight in soybeans, you’ve identified a QTL with the following
genotypic values:
• AA: 25g
• Aa: 22g
• aa: 18g
Calculate the additive (a) and dominance (d) effects of this QTL.
Solution:
Step 1: Calculate the additive effect (a) The additive effect is half the difference between the
homozygous genotypes:
𝑎 = 1
2(𝐴𝐴 −𝑎𝑎)
=1
2(25𝑔 − 18𝑔)
= 3.5𝑔
Step 2: Calculate the dominance effect (d) The dominance effect is the deviation of the
heterozygote from the average of the homozygotes:
𝑑 = 𝐴𝑎 −1
2(𝐴𝐴 +𝑎𝑎)
=22𝑔 − 1
2(25𝑔 + 18𝑔)
=22𝑔 − 21.5𝑔
= 0.5𝑔
Therefore, the additive effect (a) is 3.5g and the dominance effect (d) is 0.5g.
1.120 PROBLEM 5: POWER CALCULATION
You’re planning a QTL mapping study and want to determine the sample size needed to detect
a QTL that explains 5% of the phenotypic variance with 80% power at a significance level of
0.05. Assume you’re using an F2 population and a single marker analysis. Use the
approximation formula:
𝑛 = 2(𝑧𝛼/2 + 𝑧𝛽)2
ℎ𝑞
2
where 𝑛 is the required sample size, 𝑧𝛼/2 is the critical value of the standard normal distribution
for 𝛼/2, 𝑧𝛽 is the critical value for 𝛽 (1 - power), and ℎ𝑞
2 is the heritability explained by the QTL.
Solution:
Given:
• ℎ𝑞
2= 0.05 (5% of phenotypic variance)
• Power = 80%, so 𝛽 = 0.2
• Significance level 𝛼 = 0.05
Step 1: Find critical values
𝑧𝛼/2 = 𝑧0.025 = 1.96 (from standard normal table)
𝑧𝛽= 𝑧0.2 = 0.84 (from standard normal table)
Step 2: Apply the formula
𝑛 = 2(𝑧𝛼/2 + 𝑧𝛽)2
ℎ𝑞
2
=2(1.96 + 0.84)2
0.05
=2(2.8)2
0.05
=15.68
0.05
=313.6
Rounding up, you need a sample size of at least 314 F2 individuals to detect this QTL with 80%
power at a significance level of 0.05.
1.121 PROBLEM 1: LOD SCORE CALCULATION
Calculate the LOD score for a putative QTL given the following information:
• Likelihood under the alternative hypothesis (QTL present): 1.5 × 10−12
• Likelihood under the null hypothesis (no QTL): 2.7 × 10−15
Solution: The LOD score is calculated as:
𝐿𝑂𝐷 = log10 (Likelihood (QTL present)
Likelihood (no QTL) )
=log10 (1.5×10−12
2.7×10−15)
=log10(555.56)
= 2.74
Therefore, the
LOD score for this putative QTL is 2.74.
1.122 PROBLEM 2: HERITABILITY ESTIMATION
In a QTL mapping study for plant height in rice, you’ve identified 5 QTLs with the following
effect sizes:
• QTL 1: 15% of phenotypic variance
• QTL 2: 10% of phenotypic variance
• QTL 3: 8% of phenotypic variance
• QTL 4: 5% of phenotypic variance
• QTL 5: 3% of phenotypic variance
Calculate the total heritability explained by these QTLs. If the total heritability of plant height is
estimated to be 0.65, what percentage of the genetic variance remains unexplained?
Solution:
Step 1: Calculate total heritability explained by QTLs
ℎ𝑄𝑇𝐿
2= 0.15 + 0.10 + 0.08 + 0.05 + 0.03
= 0.41
Step 2: Calculate unexplained heritability
ℎ𝑢𝑛𝑒𝑥𝑝𝑙𝑎𝑖𝑛𝑒𝑑
2= ℎ𝑡𝑜𝑡𝑎𝑙
2− ℎ𝑄𝑇𝐿
2
= 0.65 − 0.41
= 0.24
Step 3: Calculate percentage of unexplained genetic variance
Percentage unexplained =ℎ𝑢𝑛𝑒𝑥𝑝𝑙𝑎𝑖𝑛𝑒𝑑
2
ℎ𝑡𝑜𝑡𝑎𝑙
2×100%
=0.24
0.65 ×100%
=36.92%
Therefore, 36.92% of the genetic variance remains unexplained.
1.123 PROBLEM 3: RECOMBINATION FRACTION ESTIMATION
In a QTL mapping study, you observe the following genotype frequencies in an F2 population of
1000 individuals:
• AA BB: 240
• AA Bb: 260
• AA bb: 10
• Aa BB: 255
• Aa Bb: 220
• Aa bb: 5
• aa BB: 8
• aa Bb: 2
• aa bb: 0
Estimate the recombination fraction between loci A and B.
Solution:
Step 1: Calculate the number of recombinant gametes Recombinant gametes are AB and ab.
𝑁𝐴𝐵 = 2(240)+255 +260 + 8 = 1003
𝑁𝑎𝑏 = 2(0)+5+2+10 =17
Total recombinant =1003 +17 =1020
Step 2: Calculate the total number of gametes
Total gametes = 2 × 1000 =2000
Step 3: Calculate the recombination fraction
𝑟 = Number of recombinant gametes
Total number of gametes
=1020
2000
= 0.51
The estimated recombination fraction between loci A and B is 0.51, suggesting that these loci
are unlinked (as r ≈ 0.5).
1.124 PROBLEM 4: QTL EFFECT SIZE CALCULATION
In a QTL mapping study for seed weight in soybeans, you’ve identified a QTL with the following
genotypic values:
• AA: 25g
• Aa: 22g
• aa: 18g
Calculate the additive (a) and dominance (d) effects of this QTL.
Solution:
Step 1: Calculate the additive effect (a) The additive effect is half the difference between the
homozygous genotypes:
𝑎 = 1
2(𝐴𝐴 −𝑎𝑎)
=1
2(25𝑔 − 18𝑔)
= 3.5𝑔
Step 2: Calculate the dominance effect (d) The dominance effect is the deviation of the
heterozygote from the average of the homozygotes:
𝑑 = 𝐴𝑎 −1
2(𝐴𝐴 +𝑎𝑎)
=22𝑔 − 1
2(25𝑔 + 18𝑔)
=22𝑔 − 21.5𝑔
= 0.5𝑔
Therefore, the additive effect (a) is 3.5g and the dominance effect (d) is 0.5g.
1.125 PROBLEM 5: POWER CALCULATION
You’re planning a QTL mapping study and want to determine the sample size needed to detect
a QTL that explains 5% of the phenotypic variance with 80% power at a significance level of
0.05. Assume you’re using an F2 population and a single marker analysis. Use the
approximation formula:
𝑛 = 2(𝑧𝛼/2 + 𝑧𝛽)2
ℎ𝑞
2
where 𝑛 is the required sample size, 𝑧𝛼/2 is the critical value of the standard normal distribution
for 𝛼/2, 𝑧𝛽 is the critical value for 𝛽 (1 - power), and ℎ𝑞
2 is the heritability explained by the QTL.
Solution:
Given:
• ℎ𝑞
2= 0.05 (5% of phenotypic variance)
• Power = 80%, so 𝛽 = 0.2
• Significance level 𝛼 = 0.05
Step 1: Find critical values
𝑧𝛼/2 = 𝑧0.025 = 1.96 (from standard normal table)
𝑧𝛽= 𝑧0.2 = 0.84 (from standard normal table)
Step 2: Apply the formula
𝑛 = 2(𝑧𝛼/2 + 𝑧𝛽)2
ℎ𝑞
2
=2(1.96 + 0.84)2
0.05
=2(2.8)2
0.05
=15.68
0.05
=313.6
Rounding up, you need a sample size of at least 314 F2 individuals to detect this QTL with 80%
power at a significance level of 0.05.
1.126 PROBLEM 1: LOD SCORE CALCULATION
Calculate the LOD score for a putative QTL given the following information:
• Likelihood under the alternative hypothesis (QTL present): 1.5 × 10−12
• Likelihood under the null hypothesis (no QTL): 2.7 × 10−15
Solution: The LOD score is calculated as:
𝐿𝑂𝐷 = log10 (Likelihood (QTL present)
Likelihood (no QTL) )
=log10 (1.5×10−12
2.7×10−15)
=log10(555.56)
= 2.74
Therefore, the
LOD score for this putative QTL is 2.74.
1.127 PROBLEM 2: HERITABILITY ESTIMATION
In a QTL mapping study for plant height in rice, you’ve identified 5 QTLs with the following
effect sizes:
• QTL 1: 15% of phenotypic variance
• QTL 2: 10% of phenotypic variance
• QTL 3: 8% of phenotypic variance
• QTL 4: 5% of phenotypic variance
• QTL 5: 3% of phenotypic variance
Calculate the total heritability explained by these QTLs. If the total heritability of plant height is
estimated to be 0.65, what percentage of the genetic variance remains unexplained?
Solution:
Step 1: Calculate total heritability explained by QTLs
ℎ𝑄𝑇𝐿
2= 0.15 + 0.10 + 0.08 + 0.05 + 0.03
= 0.41
Step 2: Calculate unexplained heritability
ℎ𝑢𝑛𝑒𝑥𝑝𝑙𝑎𝑖𝑛𝑒𝑑
2= ℎ𝑡𝑜𝑡𝑎𝑙
2− ℎ𝑄𝑇𝐿
2
= 0.65 − 0.41
= 0.24
Step 3: Calculate percentage of unexplained genetic variance
Percentage unexplained =ℎ𝑢𝑛𝑒𝑥𝑝𝑙𝑎𝑖𝑛𝑒𝑑
2
ℎ𝑡𝑜𝑡𝑎𝑙
2×100%
=0.24
0.65 ×100%
=36.92%
Therefore, 36.92% of the genetic variance remains unexplained.
1.128 PROBLEM 3: RECOMBINATION FRACTION ESTIMATION
In a QTL mapping study, you observe the following genotype frequencies in an F2 population of
1000 individuals:
• AA BB: 240
• AA Bb: 260
• AA bb: 10
• Aa BB: 255
• Aa Bb: 220
• Aa bb: 5
• aa BB: 8
• aa Bb: 2
• aa bb: 0
Estimate the recombination fraction between loci A and B.
Solution:
Step 1: Calculate the number of recombinant gametes Recombinant gametes are AB and ab.
𝑁𝐴𝐵 = 2(240)+255 +260 + 8 = 1003
𝑁𝑎𝑏 = 2(0)+5+2+10 =17
Total recombinant =1003 +17 =1020
Step 2: Calculate the total number of gametes
Total gametes = 2 × 1000 =2000
Step 3: Calculate the recombination fraction
𝑟 = Number of recombinant gametes
Total number of gametes
=1020
2000
= 0.51
The estimated recombination fraction between loci A and B is 0.51, suggesting that these loci
are unlinked (as r ≈ 0.5).
1.129 PROBLEM 4: QTL EFFECT SIZE CALCULATION
In a QTL mapping study for seed weight in soybeans, you’ve identified a QTL with the following
genotypic values:
• AA: 25g
• Aa: 22g
• aa: 18g
Calculate the additive (a) and dominance (d) effects of this QTL.
Solution:
Step 1: Calculate the additive effect (a) The additive effect is half the difference between the
homozygous genotypes:
𝑎 = 1
2(𝐴𝐴 −𝑎𝑎)
=1
2(25𝑔 − 18𝑔)
= 3.5𝑔
Step 2: Calculate the dominance effect (d) The dominance effect is the deviation of the
heterozygote from the average of the homozygotes:
𝑑 = 𝐴𝑎 −1
2(𝐴𝐴 +𝑎𝑎)
=22𝑔 − 1
2(25𝑔 + 18𝑔)
=22𝑔 − 21.5𝑔
= 0.5𝑔
Therefore, the additive effect (a) is 3.5g and the dominance effect (d) is 0.5g.
1.130 PROBLEM 5: POWER CALCULATION
You’re planning a QTL mapping study and want to determine the sample size needed to detect
a QTL that explains 5% of the phenotypic variance with 80% power at a significance level of
0.05. Assume you’re using an F2 population and a single marker analysis. Use the
approximation formula:
𝑛 = 2(𝑧𝛼/2 + 𝑧𝛽)2
ℎ𝑞
2
where 𝑛 is the required sample size, 𝑧𝛼/2 is the critical value of the standard normal distribution
for 𝛼/2, 𝑧𝛽 is the critical value for 𝛽 (1 - power), and ℎ𝑞
2 is the heritability explained by the QTL.
Solution:
Given:
• ℎ𝑞
2= 0.05 (5% of phenotypic variance)
• Power = 80%, so 𝛽 = 0.2
• Significance level 𝛼 = 0.05
Step 1: Find critical values
𝑧𝛼/2 = 𝑧0.025 = 1.96 (from standard normal table)
𝑧𝛽= 𝑧0.2 = 0.84 (from standard normal table)
Step 2: Apply the formula
𝑛 = 2(𝑧𝛼/2 + 𝑧𝛽)2
ℎ𝑞
2
=2(1.96 + 0.84)2
0.05
=2(2.8)2
0.05
=15.68
0.05
=313.6
Rounding up, you need a sample size of at least 314 F2 individuals to detect this QTL with 80%
power at a significance level of 0.05.
1.131 PROBLEM 1: LOD SCORE CALCULATION
Calculate the LOD score for a putative QTL given the following information:
• Likelihood under the alternative hypothesis (QTL present): 1.5 × 10−12
• Likelihood under the null hypothesis (no QTL): 2.7 × 10−15
Solution: The LOD score is calculated as:
𝐿𝑂𝐷 = log10 (Likelihood (QTL present)
Likelihood (no QTL) )
=log10 (1.5×10−12
2.7×10−15)
=log10(555.56)
= 2.74
Therefore, the
LOD score for this putative QTL is 2.74.
1.132 PROBLEM 2: HERITABILITY ESTIMATION
In a QTL mapping study for plant height in rice, you’ve identified 5 QTLs with the following
effect sizes:
• QTL 1: 15% of phenotypic variance
• QTL 2: 10% of phenotypic variance
• QTL 3: 8% of phenotypic variance
• QTL 4: 5% of phenotypic variance
• QTL 5: 3% of phenotypic variance
Calculate the total heritability explained by these QTLs. If the total heritability of plant height is
estimated to be 0.65, what percentage of the genetic variance remains unexplained?
Solution:
Step 1: Calculate total heritability explained by QTLs
ℎ𝑄𝑇𝐿
2= 0.15 + 0.10 + 0.08 + 0.05 + 0.03
= 0.41
Step 2: Calculate unexplained heritability
ℎ𝑢𝑛𝑒𝑥𝑝𝑙𝑎𝑖𝑛𝑒𝑑
2= ℎ𝑡𝑜𝑡𝑎𝑙
2− ℎ𝑄𝑇𝐿
2
= 0.65 − 0.41
= 0.24
Step 3: Calculate percentage of unexplained genetic variance
Percentage unexplained =ℎ𝑢𝑛𝑒𝑥𝑝𝑙𝑎𝑖𝑛𝑒𝑑
2
ℎ𝑡𝑜𝑡𝑎𝑙
2×100%
=0.24
0.65 ×100%
=36.92%
Therefore, 36.92% of the genetic variance remains unexplained.
1.133 PROBLEM 3: RECOMBINATION FRACTION ESTIMATION
In a QTL mapping study, you observe the following genotype frequencies in an F2 population of
1000 individuals:
• AA BB: 240
• AA Bb: 260
• AA bb: 10
• Aa BB: 255
• Aa Bb: 220
• Aa bb: 5
• aa BB: 8
• aa Bb: 2
• aa bb: 0
Estimate the recombination fraction between loci A and B.
Solution:
Step 1: Calculate the number of recombinant gametes Recombinant gametes are AB and ab.
𝑁𝐴𝐵 = 2(240)+255 +260 + 8 = 1003
𝑁𝑎𝑏 = 2(0)+5+2+10 =17
Total recombinant =1003 +17 =1020
Step 2: Calculate the total number of gametes
Total gametes = 2 × 1000 =2000
Step 3: Calculate the recombination fraction
𝑟 = Number of recombinant gametes
Total number of gametes
=1020
2000
= 0.51
The estimated recombination fraction between loci A and B is 0.51, suggesting that these loci
are unlinked (as r ≈ 0.5).
1.134 PROBLEM 4: QTL EFFECT SIZE CALCULATION
In a QTL mapping study for seed weight in soybeans, you’ve identified a QTL with the following
genotypic values:
• AA: 25g
• Aa: 22g
• aa: 18g
Calculate the additive (a) and dominance (d) effects of this QTL.
Solution:
Step 1: Calculate the additive effect (a) The additive effect is half the difference between the
homozygous genotypes:
𝑎 = 1
2(𝐴𝐴 −𝑎𝑎)
=1
2(25𝑔 − 18𝑔)
= 3.5𝑔
Step 2: Calculate the dominance effect (d) The dominance effect is the deviation of the
heterozygote from the average of the homozygotes:
𝑑 = 𝐴𝑎 −1
2(𝐴𝐴 +𝑎𝑎)
=22𝑔 − 1
2(25𝑔 + 18𝑔)
=22𝑔 − 21.5𝑔
= 0.5𝑔
Therefore, the additive effect (a) is 3.5g and the dominance effect (d) is 0.5g.
1.135 PROBLEM 5: POWER CALCULATION
You’re planning a QTL mapping study and want to determine the sample size needed to detect
a QTL that explains 5% of the phenotypic variance with 80% power at a significance level of
0.05. Assume you’re using an F2 population and a single marker analysis. Use the
approximation formula:
𝑛 = 2(𝑧𝛼/2 + 𝑧𝛽)2
ℎ𝑞
2
where 𝑛 is the required sample size, 𝑧𝛼/2 is the critical value of the standard normal distribution
for 𝛼/2, 𝑧𝛽 is the critical value for 𝛽 (1 - power), and ℎ𝑞
2 is the heritability explained by the QTL.
Solution:
Given:
• ℎ𝑞
2= 0.05 (5% of phenotypic variance)
• Power = 80%, so 𝛽 = 0.2
• Significance level 𝛼 = 0.05
Step 1: Find critical values
𝑧𝛼/2 = 𝑧0.025 = 1.96 (from standard normal table)
𝑧𝛽= 𝑧0.2 = 0.84 (from standard normal table)
Step 2: Apply the formula
𝑛 = 2(𝑧𝛼/2 + 𝑧𝛽)2
ℎ𝑞
2
=2(1.96 + 0.84)2
0.05
=2(2.8)2
0.05
=15.68
0.05
=313.6
Rounding up, you need a sample size of at least 314 F2 individuals to detect this QTL with 80%
power at a significance level of 0.05.
