4. Consider the autosomal sex-influenced trait of pattern baldness (B), which is
ID: 93223 • Letter: 4
Question
4. Consider the autosomal sex-influenced trait of pattern baldness (B), which is dominant in men and recessive in women. A heterozygous bald (B'B) man marries a heterozygous non-bald (B B) woman. A. What is the probability among sons that the son will be bald? B. What is the probability among daughters that a daughter will be bald? C. Suppose a female, who has a bald father that is heterozygous for the baldness allele and a mother who does not carry the allele for baldness, marries a bald man with genotype B'B. The couple has a son. What is the probability that the son will be bald as an adult?Explanation / Answer
Answer A. The male in this question is B’B (heterozygous), married a woman which is heterozygous (B’B).
B’
B
B’
B’B’
B’B
Y
B’Y
BY
The probability that a son will be born with pattern baldness is ½. As baldness is sex linked trait carried by X chromosome (represented here by B’). In male, another chromosome can be mentioned as Y (B). From punnet square, it is clear that 50% of male children will have male-pattern baldness.
Answer B. The probability that a daughter will have baldness is ½ i.e. 50%. In female, B’B’ genotype will be the female (tendency for hair loss).
Answer C. Form question what we know based on the information provided:
Relation
Genotype
Father
B’B (B’Y)
Mother
BB
Husband
B’B (B’Y)
B
B (Mother)
B’
BB’
B’B
Y
Father
BY
BY
The probable genotype of daughter is B’B and is married to man having genotype B’B (B’Y).
B’
B (Mother)
B’
B’B’
B’B
Y
B’Y
BY
The probability that a son will be born with pattern baldness is ½ i.e. 50%.
B’
B
B’
B’B’
B’B
Y
B’Y
BY
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