Question

If a heterogeneous woman with traits for dark hair and freckles has a bay with a man with the same traits what is the chance their child would be born no freckles and light haired

Answer 1

The correct answer is
`(1)/(16)` or zero, depending on the fathers genotype.

Step-by-step explanation:

• According to the question,

1. dark hair phenotype is dominant over light hair.
2. freckled phenotype is dominant over non-freckled phenotype.

• Let for hairs, dominant allele be D and the recessive allele be d.
• Let for freckles, dominant allele be F and the recessive allele be f.
• According to the question, the mother is heterozygous for both the traits, that is, she has the genotype, DdFf and phenotype, dark hair and freckles.
• But the father is said to have the same traits but not mentioned to be homozygous (DDFF) or heterozygous (DdFf).
• If he is homozygous, then he would produce only DF gametes.
• The mother produces the following gametes : DF, Df, dF, df.
• On crossing homozygous father (DDFF) with heterozygous mother (DdFf), all the offspring will have dark hair with freckles, so probability of obtaining a child with light hair and no freckles is zero.
• If both the parents are heterozygous, DdFf,
• Crossing them,

DF Df dF df

DF DDFF DDFf DdFF DdFf

(D/F) (D/F) (D/F) (D/F)

Df DDFf DDff DdFf Ddff

(D/F) (D/NF) (D/F) (D/NF)

dF DdFF DdFf ddFF ddFf

(D/F) (D/F) (L/F) (L/F)

df DdFf Ddff ddFf ddff

(D/F) (D/NF) (L/F) (L/NF)

• In the above cross, in the brackets (),

1. D means dark hair.
2. F means freckled.
3. L means light hair.
4. NF means non-freckled.

• The phenotypes observed are :

1. Dark hair with freckled - 9.
2. Dark hair with non-freckled - 3.
3. Light hair with freckled -3.
4. Light hair with non-freckled - 1.

• Probability that a child born would be light haired with no freckles (ddff) is
  `(1)/(16)`.