Question

2. In humans, fingerprint ridge count follows a polygenic inheritance pattern. The minimum number of ridges is 80 in males and 70 in females. Each active allele will produce an additional 12 ridges in males and 9 ridges in females. Active alleles are represented by uppercase letters.

• What is the fingerprint ridge count in a male having the following genotype? o AABBccDd? __________ aabBccdd ____________

• What is the fingerprint ridge count in a female having the following genotype? o AABBCCDD?_________ aabbCcDD?____________

• What is the least number of active alleles a child born of the following couple could inherit? Father: AABBccDD Mother: aabbCCdd • What is the probability that a child would have a genotype with the least number of active alleles?

3. The gene for sickle cell anemia is recessive. Suppose a man and woman are both carriers.

• What is the probability that they will have a child with normal blood cells?

• What is the probability that they will have a child who has sickle cell anemia?

• What is the probability that they will have a child who is a carrier?

4. A woman with type A- blood marries a man with type B+ blood. Is it possible for this couple to have a child with:

• Type O blood?

• Type AB blood?

• Type A blood?

• Type B blood?

5. Hair texture in humans is presumed to follow an incomplete dominance inheritance pattern whereby genotype CC results in curly hair, cc results in straight hair, and Cc results in wavy hair. Suppose a man with straight hair has children with a woman who has wavy hair. What is the probability that a child born to this couple will have:

• Straight hair?

• Wavy hair?

• Curly hair?

6. The production of coat color in Labrador retrievers involves two sets of genes where one set is epistatic to the other. Suppose the gene for black (B) fur is dominant to brown (b) fur, and the production of coat pigment requires at least one copy of the dominant gene, E. Lack of coat pigment (ee) results in a yellow lab. A cross is made between two dogs with genotype, BbEe. Which color dog would be most rare?

*Need all questions answered*

Answer 1

2.

Minimum ridges in males = 80

Minimum ridges in females = 70

An active allele will produce an additional 12 ridges in males and 9 ridges in females

a) Males :

Fingerprint ridge count in a male with genotype AABBccDd =140

There are 5 active alleles (represented by capital letters) in the genotype. Therefore,

Fingerprint ridge count = 80 (minimum ridges in a male) + (12 * 5) =140

Fingerprint ridge count in a male with genotype aaBbccdd = 92

80 + 12 (1 active allele) = 92

b) Females

Fingerprint ridge count in a female with genotype AABBCCDD = 142

There are 8 active alleles in the genotype (maximum rigid count). therefore,

Fingerridge count = 70 + (9 * 8) = 142

Fingerprint ridge count in a female with genotype aabbCcDD = 70 + (9 * 3) = 97

c)

parent cross :AABBccDD (Father) * aabbCCdd (Mother)

Gametes of father = ABcD

Gametes of mother = abCd

Offspring = AaBbCcDd ( 4 active alleles)

the least number of active alleles the child will inherit = 4

The probability that the child will have genotype AaBbCcDd = 100%

If the child is male, then minimum ridge count = 80 + ( 12 * 4) = 128

If the child is female, then finger ridge count = 70 + ( 9 * 4) = 106

3)

Sickle cell anemia is an autosomal recessive trait

Let S be the dominant allele for normal hemoglobin

Let s be the recessive cell allele for sickle cell trait

Parent cross: Ss (carrier father) * Ss (carrier mother)

Gametes of father: S, s

Gametes of the mother: S, s

Offspring: SS (normal), Ss (carrier), Ss (carrier), ss (affected)

genotypic ratio: 1(SS): 2 (Ss): 1(ss)

Phenotypic ratio: 1(normal): 2(carriers): 1(sickle cell disease)

a)the probability that child will have normal blood cells: 25% = 1/4

b) the probability that the child will have sickle cell anemia: 25% = 1/4

c) the probability that the child will be a carrier: 50% = 2/4

4.

Genotype of A blood type = AA or AO

Genotype of B blood type = BB or BO

a) Yes, the couple will have a child with O blood type if the genotype of the man is BO and the genotype of the woman is AO

b) Yes, the couple can  have a child with AB blood type whatever be the genotype of the father (BB or BO) and mother (AA or AO)

c) Yes, the couple will have a child with A blood type, if the father has BO genotype

d) Yes, the couple will have a child with B blood type, if the mother has AO genotype

5)

Parent cross: cc (straight hair) * Cc (Wavy hair)

Gametes of male parent: c

gametes of female parent: C, c

Offspring: Cc (Wavy), cc (straight)

Genotypic ratio : 1(Cc) : 1 (cc)

Phenotypic ratio : 1 (wavy) : 1 (straight)

the probability that the child will have straight hair = 50% = 1/2

the probability that the child will have wave hair = 50% = 1/2

The probability that the child will have curly hair (CC) = 0%

6)

The given problem is an example of recessive epistasis. In recessive epistasis, the homozygous recessive alleles of one gene mask the effect of dominant and recessive alleles of another gene. Here, the homozygous recessive alleles (ee) of gene E mask the effect of dominant allele B and recessive allele b of gene B

Genotype of black color: BBEE or BBEe or BbEE or BbEe

Genotype of brown: bbEE or bbEe

Genotype of yellow color: BBee, Bbee, bbee

parent cross: BbEe (Black) * BbEe (black)

Gametes of male parent: BE, Be, bE, be

Gametes of female parent: BE, Be, bE, be

Offspring:

BBEE (Black), BBEe (black), BbEE (black), BbEe (black),

BBEe (black), BBee (yellow), BbEe (black), Bbee (yellow),

BbEE (black), BbEe (black), bbEE (brown), bbEe (brown),

BbEe (black), Bbee (yellow), bbEe (brown), bbee (yellow)

Phenotypic ratio: 9(black) : 4 (yellow) : 3 (brown)

Among offspring, the rare color would be brown (3/16)