Answer to Question #66783 in Genetics for Siddharth
Smooth (S) is dominant over wrinkled (s). What fraction of the offspring in the
following cross would be homozygous recessive for all gene pairs in the cross:
YyTtss x YyttSs?
First, we must find the probability of this genotype, which is a product of probabilities for each locus.
To find the probability for the locus, we must find the product of probailities for each parent to give the recessive allele.
For the first locus it is the following:
each parent can give either Y or y allele. So, the probability for each parent to give y allele is 1/2.
Thus the probability for the first locus to be yy is: 1/2*1/2 = 1/4.
For the second locus:
the first parent can give either T or t, while the second can give only t. Thus, the probability to give allele t for the first parent is 1/2,
while for the second one it is 1.
Thus, the probability for the second locus to be tt is: 1/2*1 = 1/2
For the third locus:
the first parent can give only s, while the second can give either S or s. Thus, the probability to give allele s for the first parent is 1, while for the second one it is 1/2.
Thus, the probability for the third locus to be ss is: 1*1/2 = 1/2
The probability of genotype is: 1/4*1/2*1/2 = 1/16.
Thus, the proportion of offspring, which will be homozygous recessive for all gene pairs is : 100%*1/16 = 6.25%, or 0.0625.
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