Assume that there is complete dominance and complete penetrance at each locus an
ID: 271186 • Letter: A
Question
Assume that there is complete dominance and complete penetrance at each locus and that epistasis does not occur. (These are the same conditions we used in class). Write your answers in numeric form (not in words).
Assume that there is complete dominance and complete penetrance at each locus and that epistasis does not occur. (These are the same conditions we used in class). Write your answers in numeric form (not in words) Referring to the cross DdFFGghhRrTT x DDffGgHhRRTt: How many different gametes does the DdFFGghhRrTT parent produce? 8 How many different genotypes are found among the progeny of this cross? 27 How many different phenotypes are found among the progeny of this cross? 8 What is the probability that the first offspring from this cross will show the dominant phenotype for al loci? (Type in as a decimal and round to 6 digits.) What is the probability that the first offspring from this cross will be heterozygous for all loci? (Type in as a decimal and round to 6 decimal digits.) What is the probability that the first offspring from this cross will be a son with genotype DDFfGghhRrTt? (Type in as a decimal and round to 6 decimal digits)Explanation / Answer
We have the following two genotypes being crossed: DdFFGghhRrTT × DDffGgHhRRTt
Dd × DD = 1
FF × ff = 1
Gg × Gg = 3/4= .75
hh × Hh = 1/2 = 0.50
Rr × RR = 1
TT × Tt =1
So, to find the probability of all these events occurring simultaneously, we follow the probability multiplication rule.
Multiplying all the probabilities we get,
1 × 1 × 0.75 × 0.50 × 1 × 1 = 0.375000
Dd × DD = 0.50
FF × ff = 1
Gg × Gg = 0.50
hh × Hh = 0.50
Rr × RR = 0.50
TT × Tt =0.50
And again, to find the probability of all the loci having heterozygous genotype, we apply the multiplication rule of probability and get,
0.50 × 1 × 0.50 × 0.50 × 0.50 × 0.50 = 0.031250
The probability of occurrence of :
DD is 0.50
Ff is 1
Gg is 0.50
hh is 0.50
Rr is 0.50
Tt is 0.50
For all these genotypes to occur simultaneously, we apply the multiplication rule
0.50 × 1 ×0.50 × 0.50 × 0.50 × 0.50 = 0.031250
Then, we know that the probability of the offspring being a son or a daughter is 1/2 = 0.50
For the offspring to be a son and have the given genotype, the probability is =0.50 × 0.031250 = 0.015625
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