1. Monohybrid Cross a. Let’s assume that the character for eye color is controll

Question

1. Monohybrid Cross

a. Let’s assume that the character for eye color is controlled by 1 gene in humans.

Brown eyes are dominant over blue eyes. A brown eyed man marries a blue eyed

woman and they have 6 children: 4 are brown eyed and 2 are blue eyed.

i. What are the genotypes of each parent? How do you know?

ii. What gametes will the parents produce?

iii. Draw a Punnett square that illustrates this marriage and give the ratios of each

possible phenotype.

b. One of the sons with brown eyes from the marriage above marries a woman with blue

eyes and all of their 3 children also have blue eyes.

i. What is the genotype of this son? How do you know? What kind of cross is

this?

ii. What are the possible phenotypes for the children of this marriage and what

ratios will they have?

c. Let’s assume that straight hair is controlled by a single gene in humans and is

dominant over curly hair. A woman who has straight hair marries a man with curly

hair. The woman’s mother has curly hair and her father has straight hair.

i. What is the genotype of the woman?

ii. What is the probability that her and her man will have a child with curly hair?

Explanation / Answer

a] i - Father is Bb and mother bb. B for Brown and b for Blue. Since mother has recessive color, she is homozygous recessive i.e. bb. Since the children have blue color too, and one allele they got from each parent. Thus father is in heterozygous state i.e. Bb.

ii - Father will produce B and b, mother will produce only b.

iii - Punnett square-

Ratio of phenotypes is 1:1.

b] i - Son has genotype Bb. This is because again children are having blue eyes, that means both parents have recessive allele b. This is a kind of test cross as the girl is showing recessive trait.

ii - Same ratio as above, 1:1. Children will be both Brown and Blue eyed.

c] S = straight , s = curly

i - Woman's genotype is Ss. This is because offspring gets one allele from each parent. Since her mother is showing homozygous recessive trait i.e. ss. She will get one recessive allele and thus have heterozygous genotype Ss - straight hair.

ii - They will have both curly and straight hair children with equal probability i.e. 50%.

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