A slightly more confusing situation arises in determining the probability that a

Question

A slightly more confusing situation arises in determining the probability that an unaffected child of two known-heterozygous parents is itself heterozygous. For example, in the pedigree shown, we know with certainty that the parents of individual 10 must be heterozygous. The question, then, is "What is the probability that one of their phenotypically normal children is heterozygous? We know that in the mating ac x the expected ratio of offspring is icc2a1cc, that is, among the expected 9 10 I 12 13 14 5 16 17 FIGURE 2.1 Human pedigree showing four generations, Circles represent females; squares represent males.

Explanation / Answer

Answer:

d) 10x14:

As given the probability of individual 10 for heterozygous carrier is 1/2

Lets assume Aa for individual 10

The parent of individual 14 , male parent is heterozygoys Carrier Aa and female is normal AA, hence probability of individual as carrier would be 1/2.

So cross between Aa and Aa

The probability of albino would be

1/4 x 1/2 x 1/2 = 1/16

e) The probability of individual 3 for heterozygous carrier would be 1/2

The probability of individual 9 for carrier is 1/2 same as that of individual 10

If individual 9 with Aa marries normal AA Individual. The probability of carrier in 17 would be 1/2

So cross between Aa and Aa

The probability of albino would be

1/4 x 1/2 x 1/2 = 1/16

f) 3x15

The probability for Aa in individual 3 is 1/2 and that in individual 15 is 1/2

So cross between Aa and Aa

The probability of albino would be

1/4 x 1/2 x 1/2 = 1/16

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