1.) In tomato plants, spines ( s ) are recessive to no spines ( S ), and white f
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Question
1.) In tomato plants, spines (s) are recessive to no spines (S), and white fruit color (w) is recessive to red fruit color (W). The genes for spines and fruit color are located on the same chromosome; findings from mapping experiments indicate that they are 13.2 m.u. apart. A tomato plant having spines and white fruit color is crossed with a plant that is homozygous for no spines and red fruit color. The F1 are crossed with plants that have spines and white fruit color. Which of the following phenotypes and proportions are expected in the progeny of this cross? Check all that apply.
a) Red, no spines = 43.4%; white, spines = 43.4%
b) White, no spines = 43.4%; red, spines = 43.4%
c) Red, no spines = 6.6%; red, spines = 6.6%
d) White, no spines = 6.6%; white, spines = 6.6%
e) White, no spines = 6.6%; red, spines = 6.6%
f) Red, no spines = 43.4%; red, spines = 43.4%
g) Red, no spines = 6.6%; white, spines = 6.6%
h) White, no spines = 43.4%; white, spines = 43.4%
2.) In Drosophila, the map positions of genes are given in map units numbering from one end of a chromosome to the other. The X chromosome of Drosophila is 66 m.u. long. The X-linked gene for body color-with two alleles, y +for gray body and y for yellow body-resides at one end of the chromosome at map position 0.0. A nearby locus controlling bristle form, with f +for normal bristles and f for forked bristles is located at map position 1.5. A third X-linked gene for eye color, with alleles w +for red eye and w for white eye, is located at map position 56.7. Each gene resides on the X chromosome, and at each locus the wild-type allele is dominant over the mutant allele. Do you expect any of these gene pair(s) to assort independently?
a) Yes, y and w are each expected to assort independently of f
b) Yes, all the genes are expected to assort independently
c) Yes, y and f are each expected to assort independently of w
d) No, there are no genes expected to assort independently
Explanation / Answer
Question No 1
Answer : (a) and (e)
Genotype of tomato plant having spines and white fruit color = sw / sw
Genotype of tomato plant homozygous for no spine and red fruit color = SW / SW
Now perform genetic cross between them and we get
sw / sw x SW / SW
Gametes sw SW
F1 Progeny
Genotype = sw / SW
Phenotype = No spine Red fruit color tomatoes
Genotype of tomato plant having spines and white fruit = sw / sw
Performing a cross with F1 progeny
sw / SW x sw / sw
Gametes formed by the heterozygous parent ( sw / SW) = sw, SW, sW, Sw
Gametes formed by other parent (sw / sw) = sw
F1 Progeny
Genotypes = sw / sw ; SW / sw ; sW / sw ; Sw / sw
Phentopes = spine white color fruit ; no spine red color fruit (non-recombinant progeny)
spine red color fruit ; No spine white color fruit (recombinant progeny)
A recombinant frequency (RF) of 1% is equivalent to 1 m.u. But this equivalence is only a good approximation for small percentages; the largest percentage of recombinants cannot exceed 50%, which would be the situation where the two genes are at the extreme opposite ends of the same chromosomes. In this situation, any crossover events would result in an exchange of genes, but only an odd number of crossover events (a 50-50 chance between even and odd number of crossover events) would result in a recombinant product of meiotic crossover.
The recombination frequency tells us that 13.2% of the gametes produced by the heterozygous parent will be recombinants. Because there are two types of recombinant gametes, each should arise with a frequency of 13.2%/2 = 6.6%. All the other gametes will be nonrecombinants; so they should arise with a frequency of 100% - 13.2% = 86.8%.Because there are two types of nonrecombinant gametes, each should arise with a frequency of 86.8%/2 = 43.4%.
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