Question at position 23 Duchenne muscular dystrophy (DMD) is caused by rare X-linked recessive allele. It results in progressive muscular wasting and usually leads to death before age 20. An affected person is one with a severe form of DMD caused by hemizygosity or homozygosity for the disease allele. What is the probability that the son of a woman whose brother is affected will be affected? 1/41/21/803/41/187/8

Question

BlackTom AI · Accepted Answer

The scenario involves an X-linked recessive disorder (Duchenne muscular dystrophy). An affected male has genotype X^d Y. The sister of an affected male has a mother who must be a carrier (X^D X^d) for the affected son to exist in the family, because affected males come from mothers who pass the mutant X to their sons. For a daughter from a carrier mother, there is a 50% chance she will be a carrier herself (X^D X^d) and a 50% chance she will be non-carrier (X^D X^D).

If the woman in question is a carrier, each of her sons has a 50% chance of being affected (X^d from carrier mother times Y from father). Therefore the probability that the son of this woman will be affected is the product of the probability that she is a carrier (50%) and the probability that a carrier mother passes the mutant allele to a son (50%), giving 0.5 × 0.5 = 0.25, i.e., 1/4.

Now, evaluating each option:
- Option 1: 1/4 — This matches the calculated probability (carrier status × affected-son probability). It is correct based on Mendelian inheritance for X-linked recessive traits.
- Option 2: 1/2 — This would be correct if we knew the woman is definitely a carrier and were calculating only the chance a son is affected given carrier status, but that ignores the initial 50% chance she is a carrier. It overestimates by assuming carrier status with certainty.
- Option 3: 1/8 — This would correspond to 25% × 1/2 or 50% × 1/4 in some miscomputed logic; it wrongly combines probabilities and does not reflect the two-stage process (carrier probability and then affected-son probability).
- Option 4: 0 — Suggests impossibility; this is incorrect since there is a real risk for the son to be affected if the mother is a carrier.
- Option 5: 3/4 — This would imply a very high probability, inconsistent with X-linked recessive inheritance and the two-gene (carrier) considerations involved.
- Option 6: 1/18 — A nonsensical fraction in this genetic context, not arising from standard Mendelian probabilities for this scenario.
- Option 7: 7/8 — An even larger fraction that does not fit the two-step inheritance calculation for an X-linked recessive condition in this family context.

In summary, the correct reasoning leads to a final probability of 1/4 for the son to be affected, while all other listed options misstate the risk due to overlooking the necessary carrier probability of the mother or applying incorrect inheritance logic.

查看解析

登录即可查看完整答案

类似问题

If a female chimpanzee that is heterozygous for red-green color blindness (a recessive X-linked mutation) is crossed with a wild-type male, what predicted proportion of the progeny will be females that display the mutant phenotype?

Which of the following children will inherit an X-linked recessive allele from a woman who is heterozygous for color blindness (assuming the man is not a carrier: XNY)?

Question at position 44 A man will pass on an allele of an X-linked gene to:all of his daughtershalf of his sonsall of his sonshalf of his daughtersall of his children

In cats, black fur color is determined by an X-linked allele; the other allele at this locus determines orange color. The heterozygote is tortoiseshell. What kinds of offspring would you expect from the cross of a black female and an orange male?

With X-linked traits, who is more likely to be a carrier, and who is more likely to express them?

更多留学生实用工具

考试浏览器助手

风格化写作助手

论文查重助手

文献引用助手

课堂转译助手

课堂笔记助手

Quiz搜索助手

学校历年真题

智能刷题助手

智能匹配练习

希望你的学习变得更简单