SAT Subject Biology Practice Question 388: Answer and Explanation

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Question: 388

18. A man who is color blind marries a normal woman, and together they have a daughter who is not color blind. If the daughter marries a man with normal vision, what is the probability of their firstborn child being a son who is color blind?

A. 0%
B. 25%
C. 50%
D. 75%
E. 100%

Correct Answer: B

Explanation:

B This is a typical genetics probability question with a twist. Make sure you read the questions carefully! You should know that color blindness is caused by an X-linked recessive allele. The only way the color blind man can produce a daughter is if he passes on his X chromosome; therefore, the daughter, even though normal, must be a carrier of color blindness. The daughter marries a normal man, meaning that he does not carry the allele for color blindness on his X chromosome. For them to produce a son, he must pass on his Y chromosome (so it doesn't really matter what his X is like, anyway) and the probability of him passing on his Y is 50%. The final probability lies with the daughter and whether she passes on her normal X or her color blind X to her son; this is also a 50% probability. It's very tempting to choose choice C; however, here's the twist: The question asks for the probability that BOTH events (male child AND color blind) occur TOGETHER. The easiest way to solve this is to a the Punnett square.

The four possibilities in offspring are XXc (carrier daughter), XX (normal daughter), Xc Y (color blind son), and XY (normal son). The probability of producing a color blind son is 1 out of 4, or 25%.

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