AT&T; 12:59 PM extra credit.pdf Name 1. Two linked genes, (A) and (), are separa

Question

AT&T; 12:59 PM extra credit.pdf Name 1. Two linked genes, (A) and (), are separated by 18 m.u. A man with genotype Aa Bb marries a woman who is aa bb. The man's father was A4 BB. What is the probability that their first two children will both be ablab? A) 0168 B) 0008I C) 0032 D) 0062 E) 013 2. A series of two-point crosses were carried out among twelve loci (a, b, e,d. e.f.g.h, i.j. k I), producing the following recombination frequencies. Using these recombination freqguencies, map the twche loci, showing their linkage groups, the onder of loci in each linkage group, and the distamces between the loci of 50 50 14 50 50 50 5 50 50 850 5050 16 50 50150 | 50 | 14 | 50 | 50 | 50 | 50 50 50 50 11550 19 50 50 50 50 Page I Open With Print

Explanation / Answer

Answer
Linkage : when two genes lies at smaller distance and remain together while gamete formation then these genes are called linkage. Linkage group represents multiple together on a chromosome.
As. Given in the question gene A and B are linked genes.
In this cross. Male(AaBb) X. Female (aabb)
Gametes A B X a b   
a————b. a———-b
F1. AaBb. aabb
Probability of parental and recessive progeny is 50% for parental and 50% of recessive.
But the distance between A and B genes is 18 m.u i.e. 18 cM which is equal to 18% recombinase frequency also.
Then the observe probability for the Child to be recessive is 50 – 9= 41%
= 41/100   
So probability of two recessive child would be = 41/100 *41/100
=. 1681/10000
= 0.168 answer

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