lab report ASP
lulsab
02_22_17biology370l_4_drosophilageneticmapping.pptxRecombination and Linkage
A Three point test cross in Drosophila
Linkage
Relationship between two or more genes near each other on the same chromosome is known as Linkage
Law of independent assortment applies to genes on different chromosomes(non linked genes)
Linkage
In a three point cross if there is no linkage what would be the expected ratio for all classes ?
1:1:1:1:1:1:1:1
Aim for three point linkage experiment
Locate the respective position of the three linked gene loci on the same chromosome
Measure distances between genes
Construct the genetic map which estimate the physical distances between gene loci
Drosophila Genes & Phenotypes
Genotype (alleles)
y cv f mutant and + + + (Y CV F) wildtype
Phenotype (appearance)
Mutant: wild type:
y: yellow body grey body
cv: crossveinless wings crossvein wings
f: forked bristles straight bristles
Genes are located on the X chromosome
Parental Cross:
+++ / Y X ycvf / ycvf
Males Females
Phenotype (P): [WT] [yellow, crossveinless, forked]
F1 Generation:
ycvf / Y and +++ / ycvf
Males Females
P: [yellow, crossveinless, forked] [WT]
F1 trihybrid test cross:
ycvf / Y X +++ / ycvf
4 Males 6 Females
First day lab set up
Test Cross-Testing genotype of parents showing dominant phenotype
Test cross- Cross between F1 progeny and recessive parents.
F1 trihybrid test cross:
ycvf / Y x +++ / ycvf
4 Males 6 Females
During meiosis of spermatogenesis male Drosophila flies do not exhibit crossing over between homologous pair - no recombinant product.
X chromosome and Y chromosome do have regions of pairing, and do form tetrads, but don’t recombine.
Male show complete linkage, advantageous for our experiment.
Tri-hybrid females had crossing over .
8 Phenotypes You will be expecting today!!
Grey, crossvein,straight (+ + + )
Yellow,crossveinless,forked(y cv f )
Grey,crossveinless,forked (+cv f )
Yellow,crossvein,straight(y + +)
Grey,crossvein,forked (+ + f)
Yellow,crossveinless,straight (y cv +)
Grey, crossveinless,straight(+cv +)
Yellow,crossvein,forked(y + f)
Place 10 -12 flies at a time to get the phenotypes.
Non-recombinant
Single CO- region I
Single CO-region II
Double CO
Our results
pool out the whole class data to calculate the genetic distance more accurately
Today’s Goal: Construct a crude genetic map
Find the gene position
calculate c.o.c., interference
Perform X²(to determine if the map distance from the pooled class data are statistically same with the published map distance)
F2 Trihybrid Test Cross example
axp / +++ X axp / Y
| gametes | a x p Y | Class | #F2 | |
| + + + | + + +/ a x p | + + +/ Y | 1 | 62 |
| a x p | a x p/ a x p | a x p/ Y | 2 | 52 |
| + x p | + x p/ a x p | + x p/ Y | 3 | 14 |
| a + + | a + +/ a x p | a + + / Y | 4 | 16 |
| + + p | + + p/ a x p | + + p/ Y | 5 | 4 |
| a x + | a x +/ a x p | a x +/ Y | 6 | 2 |
| + x + | + x +/ a x p | + x +/ Y | 7 | 27 |
| a + p | a + p/ a x p | a + p/ Y | 8 | 23 |
** phenotypes of both sexes are similar, gender does not matter
F1 Parents-
F2 Data Analysis: Assigning Classes
| phenotypes | Class | #F2 |
| + + + | 1 | 62 |
| a x p | 2 | 52 |
| + x p | 3 | 14 |
| a + + | 4 | 16 |
| + + p | 5 | 4 |
| a x + | 6 | 2 |
| + x + | 7 | 27 |
| a + p | 8 | 23 |
Non-recombinants (parental) usually have greatest #
Double crossover classes usually have smallest #
Single crossover I
Reciprocal classes have similar numbers
Single crossover II
F2 Data Analysis: 1st goal- determine the order of loci
We will be comparing P & DCO classes
| phenotypes | Class | #F2 |
| + + + | 1 | 62 |
| a x p | 2 | 52 |
| + x p | 3 | 14 |
| a + + | 4 | 16 |
| + + p | 5 | 4 |
| a x + | 6 | 2 |
| + x + | 7 | 27 |
| a + p | 8 | 23 |
200
+ + +
a x p
+ + p
a x +
a+ p x+
a p+ x
a+ p+ x+
a p x
In DC classes ,middle gene gets inverted from its initial position in the parental chromosome
F2 Data Analysis: Rearrange and Calculate Recombination percentage
RF =
| phenotypes | Class | #F2 |
| + + + | 1 | 62 |
| a p x | 2 | 52 |
| + p x | 3 | 14 |
| a + + | 4 | 16 |
| + + x | 5 | 27 |
| a p + | 6 | 23 |
| + p + | 7 | 4 |
| a + x | 8 | 2 |
200
RF I = = 0.18
Recombination % =(RF X 100)=0.18x100= 18% 18 map units
Between a and p
RF II = = 0.28
Recombination % = 28% 28 map units
Between p and x
Note that F2# is greater for CO between p and x
F2 Data Analysis: Drawing the Map based on recombinant frequency
1 map unit = 1% recombination
Recombination % = 18% 18 map units in Region I (a and p)
Recombination % = 28% 28 map units in Region II (p and x)
a p x
18 m.u.
28 m.u.
2nd step: Calculate C.O.C and interference
Coefficient of coincidence(c.o.c.) =
Expected DCO = (RF I)(RF II)(Total #)
Interference(I) = 1 – c.o.c
Exp DCO = (0.18)(0.28)(200) = 10.08
c.o.c. = = 0.595
I = 1 – 0.595 = 0.405
~ 40% interference
If I = 1 No DCO observed (complete interference)
If I = 0 # of DCO expected (no interference)
If I = between 0 & 1 intermediate amount of interference
Negative value means scoring is wrong
C.O.C-measure the amount the interference of one CO to reduce the formation of another CO
Third Goal:X² Analyses ( Observed vs published distance between 1st and 2nd gene)
Accept differences between observed and accepted are due to….
a
p
x
13.7 mu
43 mu
| X² for Region I | Observed | Expected | (f-f*)2/f* |
| Recomb. I | 14+16+4+2= 36 | (RF1 Pub) x total 0.137x200=27.4 | 2.7 |
| NonRecomb I | Total-recomb 200-36=164 | Total-recomb 200-27.4=172.6 | 0.43 |
X2 = 3.13
df = 1
X² Analyses: Observed vs Published for Region II
Reject differences between observed and accepted are NOT due to….
a
p
x
13.7 mu
43 mu
X2 = 18.4
df = 1
| X² for Region II | Observed | Expected | (f-f*)2/f* |
| Recomb. II | 27+23+4+2= 56 | (RF2 pub) x total 0.43x200=86 | 10.5 |
| NonRecomb II | Total-recomb 200-56=144 | Total-recomb 200-86=114 | 7.9 |
X² Analyses: Reciprocal Classes
Numbers for reciprocal classes should be 1:1 ratio
| X² Class 1&2 | Observed | Expected | (f-f*)2/f* |
| Class 1 | 62 | ½ (total) = ½(114) 57 | 0.44 |
| Class 2 | 52 | ½ (total) = ½ (114) 57 | 0.44 |
X2 = 0.88
df = 1
Accept differences between observed and expected are due to chance
Possible Sources of Error
Over-estimation of RF
Inadequate sample size
Misscoring of flies
Missing double crossovers
Under-estimation of RF
Some crossover flies are less viable and selected against
Ex. Triple mutants less viable than double/single mutants
Failure to detect multiple crossovers in a given region
**Do not want to hear the typical “standard experimental error”
Ways to Improve
Score phenotypes more carefully.
It is better to use more markers so RF is not being measured for distant genes that would have lots of crossovers, such as cv-f region, unlike the y-cv region, which is smaller, so less DCO occur.
Increase sample size
In summary
1. Determine the order of the three genes on the chromosome and find the middle gene
2. Calculate the recombination percentages for the two intervening regions
3. Construct the genetic map
4. Calculate the COC and Interference
