Genetic Inheritance

a. What are the phenotypes of the parents? Man: Woman:

b. What are the genotypes of the parents?

Man: Woman

c. What are the possible gametes for each parent? Fill in the Punnett square below to determine the genotypes and phenotypes of their children.

d. What is the genotypic ratio for this cross?

e. What is the phenotypic ratio for this cross?

II. Incomplete Dominance and Codominance

So far, we have been studying the inheritance of genes that show complete dominance.

In other words, one allele is completely dominant over another. In some cases such as incomplete dominance and codominance, one allele is not completely dominant over another or the alleles affect the phenotype equally.

( Incomplete dominance – neither allele is completely dominant over the other, and the phenotype of the heterozygote is intermediate between the phenotypes of the two homozygotes )

Examples of incomplete dominance include flower color in snapdragons and four-o clock plants, feather color in parrots (FF = blue, ff = yellow, Ff = green) and hair texture in humans.

Example: Human hair texture is an example of incomplete dominance. C = curly; C1 = straight; CC1 = wavy (i.e. intermediate between straight and curly).

If a straight haired man marries a curly haired woman, what are the predicted genotypes and phenotypes of their children?

C C

Genotypic ratio: all CC1 Phenotypic ratio: all wavy

If a wavy haired man marries a wavy haired woman, what are the predicted genotypes and phenotypes of their children?

( CC 1 CC 1 C 1 C 1 CC )Male CC1 X CC1 Female C C1

C Genotypic ratio: 1 CC: 2 CC1 : 1 C1C1

C1 Phenotypic ratio: 1 curly: 2 wavy: 1 straight

Practice problem 3:

a. If a straight haired man marries a wavy haired woman, what are the predicted genotypes and phenotypes of their children? Show all work including a Punnett square.

b. What color of flowers would be produced, if a plant with pink flowers is crossed with a plant that produces white flowers? (RR = red, rr = white, Rr = pink) Show all work including a Punnett square.

( phenotypes would be observed ect on the phenotype of each allele is equal ; in a heterozygote both Codominance )

Examples of codominance include ABO blood type and coat color in cattle.

Example: In short horn cattle coat color pattern is determined by two alleles (R = red; W = white) that are both equally expressed. RW cattle are roan (red and white patches); therefore, both alleles affect the phenotype equally.

RW X RW

Roan Roan

( RW RW WW RR )R W

R Genotypic ratio: 1 RR: 2 RW : 1 WW

W Phenotypic ratio: 1 red : 2 roan : 1 white

Practice problem 4:

The ABO system of blood typing exhibits codominance in humans. There are 3 alleles for the gene that determine ABO blood type (IA and IB are codominant alleles; allele i is

recessive to alleles IA and IB). IAIA and IAi are type A; IBIB and IBi are type B; IAIB is type AB; ii is type O

A. If a man with blood type O marries a woman that is blood type AB, what are the predicted genotypes and phenotypes of their children? Show all work including a Punnett square.

B. If a man with blood type A (IAi) marries a woman with blood type AB, what are the predicted genotypes and phenotypes of their children? Show all work including a Punnett square.

When several generations of crosses are done, the following terms are used:

( P 0 generation - the parent generation of a genetic cross F 1 generation F 2 generation )

A dihybrid cross is a genetic cross involving two organisms heterozygous for two different traits (di = two).

Example of a dihybrid cross:

A pea plant homozygous for both purple flowers (P = purple, p = white) and round seeds

(R = round, r = wrinkled) is crossed with a pea plant that produces white flowers and wrinkled seeds.

1/2

Meiosis

( PR ) ( PR )1/2

1/2

Meiosis

( pr ) ( pr )1/2

( PpRr Purple round )possible sperm possible eggs PR PR

pr

pr PpRr

Purple round

PpRr

Purple round PpRr

Purple round

F1 generation

All of the offspring in the F1 generation are PpRr and have purple flowers and round seeds.

( ****Always put alleles of same gene together and then if a dominant allele is present for a particular gene . )

What is the probability of producing each combination of alleles for the gametes according to Mendel’s Law of Independent Assortment?

( Mendel’s Law of Independent Assortment - during meiosis, homologous chromo- somes pair and their respective alleles assort into gametes independently of other erent chromosomes. )

In other words … say you are following the inheritance of two different genes on two different chromosomes. The two pairs of homologous chromosomes must pair up in the middle of the cell during metaphase of meiosis I. There are two different ways they can line up along the metaphase plate to produce the gametes as shown; this is why not only PR and pr gametes, but also Pr and pR gametes are produced by the PpRr parent as shown below.

PpRr X ppRr

Meiosis Meiosis

( PR Pr ) ( pR pr ) ( pR ) ( pr ) ( pR ) ( pr )

possible sperm possible eggs

First plant: ¼ PR : ¼ Pr : ¼ pR : ¼ pr or 1 PR : 1 Pr : 1 pR : 1 pr Second plant: ½ pR : ½ pr or 1 pR : 1 pr

Since the alignment at the metaphase plate of the two pairs of homologous chromosomes carrying the two different genes is random, the ratio of the gametes produced is equal.

( To determine the gametes that result from independent assortment: 1) Pair each allele of one gene with each allele of the other gene OR 2) , inner last from algebra?) F irst alleles: PR O uter alleles: Pr I nner alleles: pR L ast alleles: pr )

( T ombinations possible: Example: What is the # of possible t gametes from an individual with the following genotypes? ) Using the gametes from abov

PR Pr pR pr

( PpRR purple round PpRr purple round ppRR white round ppRr white roun PpRr purple round Pprr purple wrinkled ppRr white round pprr white wrink PpRR purple round PpRr purple round ppRR white round ppRr white round PpRr purple round Pprr purple wrinkled ppRr white round pprr white wrink )pR

pr

pR

pr

What is the phenotypic ratio of expected offspring? 6/16 P_R_ (i.e. purple round)

6/16 ppR_ (i.e. white round) 2/16 P_rr (i.e. purple wrinkled) 2/16 pprr (i.e. white wrinkled)

d

led

led

If you reduce the fractions, you get a 3/8 : 3/8 : 1/8 : 1/8 or a phenotypic ratio of 3 purple round: 3 white round: 1 purple wrinkled: 1 white wrinkled.

( Notice in Mendel’s classic experiment crossing two pea plants heterozygous for both yellow and round peas, he saw a 9:3:3:1 ratio for the phenotypes. However, you only see this ratio if you cross two individuals that are both heterozygous for each of two traits. (see your textbook) )

Practice problem 5:

List each different type of gamete that would be predicted to be produced from the following individuals?

a. DdEE

b. DdEe

c. ddEe

d. DDee

Practice problem 6:

Albinism is characterized by a lack of pigment in the skin, eyes and hair.

A = normal pigmentation; a = albinism. W = widow’s peak; w = straight hairline.

If Miguel is heterozygous for pigmentation and is heterozygous for widow’s peak and Lorena is heterozygous for pigmentation and has a straight hairline, what will be the phenotypic ratio for their children? Show all work including a Punnett square .

Miguel’s genotype: Lorena’s genotype:

( Another method you can use to determine the probability of getting a particular genotype and phenotype when following two traits: Cross each gene separately and then multiply the probabilities together . This method is especially useful when more than two genes are involved in a cross. )

Example: A pea plant heterozygous for both purple flowers and round seeds and a pea plant heterozygous for purple flowers and producing wrinkled seeds are crossed.

A. What is the probability of these plants producing a plant with the genotype PPrr?

PpRr X Pprr purple round purple wrinkled

Cross each gene separately:

( PP purple Pp purple Pp Purple pp white ) ( Rr round rr wrinkled Rr round rr wrinkled )P p R r

P r

p r

The probability of PP is 1/4, from left cross The probability of rr is 1/2, from right cross

To determine the probability of PPrr, multiply the probability of PP with the probability of rr:

¼ X ½ = 1/8

B. What is the probability of the parent plants above producing plants with purple flowers and round seeds?

Determine the probability of purple flowers using the left Punnett squares above: ¾ of the plants will have purple flowers

Determine the probability of round seeds using the right Punnett squares above: ½ of plants will produce round seeds

To determine the probability of producing plants with puple flowers and round seeds, multiply the probability of purple flowers with round seeds: ¾ X ½ = 3/8

( “or” = add “and” = multiply Hint : When you are determining probabilities, the word “ and ” indicates that you should multiply the probabilities, while the word “ or ” indicates that you should add the probabilities. )

IV. Multiple Genes –

What if you are studying the inheritance of more than two genes?

Again, cross each gene separately and then multiply the probabilities together.

Example: What is the probability of the following parents producing offspring with the genotype Aabbcc?

AABbCc x AaBbCc

( Aa AA Aa AA ) ( Bb Bb bb BB )A A B

A B

a b

b C c

( Cc Cc cc CC )C

c

AA x Aa ½ AA Bb x Bb ¼ bb Cc x Cc ¼ cc

Therefore, the probability of getting AAbbcc is : ½ x ¼ x ¼ = 1/32

( Determine the probability of the organisms producing each phenotype. Multiply the probabilities. in your Punnett squares. To determine the probability of a particular phenotype, )

Practice problem 7:

A. If the genotypes of the parent plants are as shown, what is the probability of these plants producing offspring with the following genotypes?

Parents: ttRrPp X TtRRPp

Show Punnett squares with predicted genotypes and phenotypes of offspring:

b. What is probability of producing a plant with genotype TtRRPp?

c. What is probability of producing a plant with genotype TtRrPP?

B. If gene T = Tall, t = dwarf; R = round seeds, r = wrinkled; P = purple flowers, p = white, fill in the phenotypes for each of the predicted genotypes in your Punnett squares on the previous page. What is the probability of producing plants with the following phenotypes?

a. tall plants with round seeds and purple flowers

b. dwarf plants with round seeds and purple flowers

c. tall plants with round seeds and white flowers

V. Genetics of Human Diseases and Disorders (recessive and dominant autosomal traits)

Many genetic diseases and disorders have been studied in humans. The majority of genes that when mutated cause these diseases are found on autosomes. We will discuss genes on the X chromosome in the section VI.

( Autosomes – non- sex chromosomes (chromosome 1 - 22 in humans; 22 pairs of autosomes or 44 autosomes total) Recessive autosomal disease – a recessive mutation in a gene that when present on both homologous chromosomes causes disease Two people heterozygous for a mutation (Aa X Aa) will have a 25% chance of having a child ected by the disease. ected = aa ted = AA or Aa Carrier – heterozygous individual, Aa (“carries” allele for disease and can pass on to children) Examples of recessive autosomal diseases or disorders in humans include : cy , albinism, Tay Sach’s disease, sickle-cell anemia, thalassemia, phenylketonuria, xeroderma pigmentosum XX = female, XY = male Sex chromosomes – X and Y )

Thalassemia is an autosomal recessive disorder causing mild to severe anemia. It is more prevalent in people with ancestors near the Mediterranean, including Southern Europe, North

Example: T = unaffected and t = thalassemia. Elizabeth has mild thalassemia. Bill is unaffected and so is his father, but his mother has thalassemia. What is the chance of Elizabeth and Bill having a child with thalassemia?

First, determine the genotypes of the parents. Elizabeth has thalassemia. So, Elizabeth’s genotype must be tt because the disease is autosomal recessive and therefore, only seen in homozygous recessive individuals.

Bill does not have thalassemia, so he could be TT or Tt. To determine Bill’s genotype, we must look at his parents since everyone receives one allele from their father and one allele from their mother for each gene. Bill’s mother had thalassemia (i.e. she’s tt); so, he must have received a "t" allele from her. Therefore, Bill must be heterozygous or Tt.

Bill Tt X tt Elizabeth

( Tt tt tt Tt )T t Genotypic ratio: 2/4 Tt : 2/4 tt or when reduce fraction t ½ Tt : ½ tt or 1 Tt : 1 tt ratio

t ected: 1 thalassemia

Therefore, ½ of their children are expected to be unaffected by the disease and ½ will have thalassemia.

Practice Problem 8:

Cystic fibrosis is an autosomal recessive disorder. Individuals with cystic fibrosis produce thick mucus in the lungs and pancreas that can severely compromise breathing.

a. How can two parents without Cystic Fibrosis have a child with this disease?

b. What is the chance of having an affected child? Show all work including a Punnett square.

( Dominant autosomal disease A person heterozygous for this gene will have the disease and his children will have a 50% chance ected. ected = Aa or AA ted = aa Examples of dominant autosomal diseases or disorders in humans include : Huntington’s disease, Marfan syndrome, achondroplasia, familial hypercholesterolemia, hypercalcemia. )

Example: Marfan syndrome is a dominant autosomal disease caused by a mutation in the fibrillin gene, which codes for a connective tissue protein. Individuals with Marfan syndrome are tall and thin with long arthritic looking fingers. They are subject to sudden death due to rupture of the aorta.

A man heterozygous for Marfan syndrome and an unaffected woman have a child. What is

the chance that this child will have Marfan syndrome?

Notice that for a dominant disorder an unaffected person is homozygous recessive. Therefore the woman’s genotype is mm.

Therefore, ½ of their children are expected to be unaffected by the disease and ½ will have Marfan syndrome.

Practice Problem 9:

Huntington’s disease is an autosomal dominant condition that affects and damages the nervous system. This neurological disorder usually strikes individuals later in life. A couple with a family history of Huntington’s disease is trying to decide if it would be wise to conceive a child. A series of tests determined that the wife does not have the Huntington’s allele and the husband is heterozygous for the disorder.

a. Which of them will eventually develop Huntington’s disease?

c. What is the chance that these parents can have a child that is normal? Show you work including a Punnett square.

VI. Sex - linked Traits

Some diseases and disorders occur more frequently in one sex than the other. The X chromosome in humans has approximately 1200 genes, whereas the Y chromosome has only about 80 genes. Consequently, most genes present on the X chromosome are not also present on the Y chromosome. Therefore, if a mutation or an allele that codes for an abnormal protein is present on the X chromosome in a male, there is no matching gene on the Y to supply the normal protein. Hence, males are more likely have from sex-linked mutations. Y-linked traits exist, but we will limit our studies to X-linked traits.

( ****NOTICE THAT NO ALLELE IS PRESENT ON THE Y CHROMOSOME Examples of recessive sex-linked diseases or disorders in humans include : Colorblindness, hemophilia, muscular dystrophy, ichthyosis ted female; carrier X H X h Usually the genotypes for sex-linked traits are written as follows: X h Y = male that inherits disorder )

Example: Muscular dystrophy is a fatal disease that causes muscle weakness and degeneration. A carrier female and an unaffected male have children.

A female carrier is by definition heterozygous: XD Xd

Since the male is unaffected, he must have the normal allele on the X chromosome (XD) but there is no allele on the Y chromosome. Therefore, his genotype is XDY.

XD Xd X XDY

unaffected female unaffected male (carrier)

A. What is the chance of these parents having a son with muscular dystrophy?

Male Gametes

Y

Female Gametes

( X D X D ected f )XD Xd

XDXd

ected female

XDY XdY

ected male male with muscular dystrophy

½ males are unaffected; ½ males have muscular dystrophy (notice the Y chromosome has no effect on the phenotype)

B. What is the chance of these parents having a daughter with muscular dystrophy? all females are unaffected (½ are carriers)

***Notice: 1) only sons have disorder; 2) sons inherit disorder from their mother Practice Problem 10:

Colorblindness is a recessive sex-linked disorder. A man with normal vision marries a woman who also has normal vision. They have a colorblind son. Use XN or Xn when writing genotypes.

Show all work

a. What are the genotypes of the father and mother?

Father: Mother:

b. Show all work including a Punnett square.

c. What are the phenotypes of the daughters?

d. What are the phenotypes of the sons?

e. What is the probability that they will have a colorblind son?

1) Hemophilia is a sex-linked recessive disease that causes reduced blood clotting, which can result in increased bruising and excessive bleeding. Monica who is a carrier for hemophilia marries Eric (he does not have hemophilia). Use the letter H when writing genotypes.

a. What is Monica’s genotype?

b. What is Eric’s genotype?

Show all work including a Punnett square below.

c. What are the phenotypes of the sons?

d. What are the phenotypes of the daughters?

e. What is the chance Eric and Monica will have a child with hemophilia?

2) In humans, earlobes are either “free” or “attached”. The free earlobe trait is dominant over attached earlobes. The ability to taste certain substances is also genetically controlled. Some people can taste Phenylthiocarbamate (PTC), a dominant trait, while others cannot (the recessive trait).

A man heterozygous for both free earlobes and ability to taste PTC marries a woman with attached earlobes who is heterozygous for the ability to taste PTC. Show all work including a Punnett square below. E= free earlobes; e = attached earlobes; P = PTC taster; p = non PTC taster.

a. What are the genotypes of the parents?

Male’s genotype: Female’s genotype:

b. What are the possible gametes for each parent? Draw a Punnett square to determine the genotypes and phenotypes of their children.

3) Phenylketonuria is an autosomal recessive disorder (see section V) that causes mental impairment and reduced pigmentation of hair and skin. A woman with phenylketonuria and a man heterozygous for phenylketonuria have children.

Female’s genotype: Male’s genotype:

a. Show all work including a Punnett square.

b. What is the probability of a child with phenylketonuria?

4) Alisha has blood type B. Her mother has blood type O. Alisha marries Jared, who has blood type AB.

a. What is Alisha’s genotype?

b. What is Jared’s genotype?

c. Using a Punnett square, determine the probability of the couple having a child with the following:

blood type B? blood type O? blood type AB? blood type A?

e. What is the chance that Alisha and Daniel will have a child with blood type O?

5) Achondroplasia is an autosomal dominant (see section V) condition that results in short stature or dwarfism due to lack of bone growth. Ndidi has achondroplasia and is concerned about passing it on to her children. Her father was of normal height, but her mother had achondroplasia. Use the letter A when writing genotypes.

a. What are the genotypes of Ndidi’s mother and father?

b. What is Ndidi’s genotype?

c. Ndidi marries Abubakar, who is heterozygous for achondroplasia. What is Abubakar’s genotype?

d. What is the chance that Ndidi and Abubakar’s children will have this condition?

Show all work including a Punnett square below.

Kyra is a carrier for Fragile X disorder and is unaffected by albinism. Kyra’s mother has unpigmented skin, eyes and hair. Kyra marries Randy who is an albino, and does not have Fragile X disorder.

a. What is Kyra’s genotype?

b. What is Randy’s genotype?

c. What are the possible genotypes and phenotypes of their children? Use a Punnett square.

d. What is the chance of having a child with Fragile X disorder that is also an albino?

e. What would be the gender of this child?