Suppose that extra fingers and toes are caused by a recessive trait, but it appears in only 60% of homozygous recessive individuals. Two heterozygotes conceive a child. What is the probability that this child will have extra fingers and toes? 0.10 0.25 0.33 0.15 0.05

0.15 = 15%

Step-by-step explanation:

Let's call the recessive allele x, and the dominant allele X.

The homozygotes have the pair xx, and the heterozygotes have the pair Xx.

If two heterozygotes conceive a child, there are four configurations that the child can have:

xx, xX, Xx, XX

the only configuration that allows the child the possibility to have extra fingers and toes is the xx, as this aspect is a recessive trait, so the probability of the child being xx is 25% (1 option between 4 options, all of them with equal probabilities).

If the child is xx, there is a chance of 60% of the child having extra fingers and toes. So, the final probability is:

25% * 60* = 0.25 * 0.6 = 0.15 = 15%

answer: answer

step-by-step explanation:

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