GENETICS QUESTION - PLS HELP On the planet Trellus, in the Rayban region, the mo
ID: 176990 • Letter: G
Question
GENETICS QUESTION - PLS HELP
On the planet Trellus, in the Rayban region, the mountain lion species Totalis Frigus has certain physical attributes determined by three loci:
Gene Alferiore (Allele A = mutant; tremendous body strength; allele a = standard body strength).
Gene Bingenti (Allele B = mutant; tremendous speed; allele b = standard speed).
Gene Cakratuv (Allele C = mutant; exceptional long range vision; allele c = standard vision).
Note each capital letter is dominant to each lower case letter. Also, we can have homozygous dominant phenotypes (i.e, no heterozygous lethal).
In a mating of a true breeding Totalis Frigus lion with tremendous strength, tremendous speed, and exceptional long range vision, with a true breeding Totalis Frigus lion with standard attributes for all, is it possible to offspring counts as in the following table?
Note: Litters among Totalis Frigus lions are very large, typically.
A.
Yes, if Gene Alferiore is in the middle and:
1. Recombination fraction between Alferiore and Bingenti is close to 0.40;
2. Recombination fraction between Alferiore and Cakratuv is close to 0.40.
B. No, it is not possible.
C.
Yes, Gene Bingenti is in the middle:
1. Recombination fraction between Bingenti and Alferiore is near 0.40;
2. Recombination fraction between Bingenti and Cakratuv is near 0.40.
D.
Yes, if Gene Alferiore is in the middle and:
1. Recombination fraction between Bingenti and Alferiore is close to 0.40;
2. Recombination fraction between Alferiore and Cakratuv is close to 0.50.
E.
Yes, if Gene Bingenti is in the middle and:
1. Recombination fraction between Alferiore and Bingenti is close to 0.0;
2. Recombination fraction between Bingenti and Cakratuv is close to 0.0.
Explanation / Answer
C). Yes, Gene Bingenti is in the middle:
1. Recombination fraction between Bingenti and Alferiore is near 0.40;
2. Recombination fraction between Bingenti and Cakratuv is near 0.40.
Haplotypes
Counts
abc
193 --à Parental
ABC
159 --à Parental
AbC
58 à Double crossover
aBc
64 à Double crossover
ABc
111 à Single crossover between B and C
abC
113 à Single crossover between B and C
Abc
144 à Single crossover between A and B
aBC
138 à Single crossover between A and B
from the double crossover and single crossover gametes, we can say that the gene order is ABC.
Now, distance between A and B
= number of gametes produced by single crossover of A and B + number of double crossover gametes*100/ total number of gametes.
= (144+138+58+64)*100/ 980 = 2000/200 = 41.22 map units.
Now, distance between B and C
= number of gametes produced by single crossover of B and C + number of double crossover gametes*100/ total number of gametes.
= (111+113+58+64)*100/ 980 = 35.3 map units.
Haplotypes
Counts
abc
193 --à Parental
ABC
159 --à Parental
AbC
58 à Double crossover
aBc
64 à Double crossover
ABc
111 à Single crossover between B and C
abC
113 à Single crossover between B and C
Abc
144 à Single crossover between A and B
aBC
138 à Single crossover between A and B
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