Problem: In fruit flies, the gene for normal wings is dominant over the gene for stubby wings. A heterozygous male fly is mated with a stubby winged female, and 200 offspring were counted about two weeks later.
a. According to the laws of probability and inheritance, what would the expected ratio be of normal to stubby winged flies?
b. What is the genotype ratio for the above cross? What is the phenotype ratio?
c. What are the chances of obtaining a fly with stubby wings?
d. If the first two flies had stubby wings what are the chances of the second fly having stubby wings?
Another Genetics question problem. Please help?
a) 50% normal (N) and 50% stubby (S)
b) genotype: 50% SN, 50% SS and 0% NN, since the stubby winged female is homozygouos for stubby, there are no normal winged offspring that are not heterozygous, so the phenotype ratio is also 50% stubby 50% normal
c) 50%
d) 50%, as the probaility for each fly is 50% independent of how many of what type have been counted before
Reply:When dealing with simple Mendelian ratios,
the odds are 50-50 when you mate a heterozygous individual to a mate that is recessive.
These odds however, are accurate only over a large number of individuals from such a mating. It would be possible for the progeny of any single mating to be all one or the other in phenotype or anywhere inbetween.
sex distribution is the same way as to ratios, you can have a litter of pups for example, that is all males, or all females, but over the course of several litters the total should average out to be about 50-50.
Reply:This can be answered using a Punnett square.
The ratio will be 50-50.
The genotype ratio (Ww x ww) is Ww, Ww, ww, ww.
The chance of obtaining a stubby-winged fly is 50%.
If the first two flies had stubby wings, the chances of the second fly having stubby wings is 100%, because you just stated the first TWO had stubby wings. That would include the second fly and the first fly.
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