| A) |
0.0001 |
| B) |
0.0005 |
| C) |
0.01 |
| D) |
0.02 |
| E) |
0.025 |
| Answer:[A]0.0001 |
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| **Color blindness is an X-linked recessive trait. A male is hemizygous |
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| for the X chromosome, and thus has only one copy of each trait. The frequency of an X-linked |
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| recessive in males is thus equal to the frequency of the allele in the population. From this, we |
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| know that q = 0.01 and p = 0.99. A female has two copies of each gene on the X chromosome, so |
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| the equation for Hardy-Weinberg equilibrium is the same as for the autosomal traits. In this |
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| case, a homozygous recessive female would occur at a frequency of q2 or 0.0001. |
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| Choice B, 0.0005, is incorrect. If you remembered that color blindness was more frequent in |
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| males, but did not know how to use the equations to get the true estimate, you might have |
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| guessed this answer. |
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| Choice C, 0.01, makes the assumption that the trait is autosomal, and so the frequencies of |
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| affected males and affected females are equal. |
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| Choice D, 0.02, assumes that q = 0.01, and then calculates the frequency of carrier females |
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| (2pq).** |
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