A Geneticist is mapping genes in a raccoon. She takes raccoon females heterozygo
ID: 223290 • Letter: A
Question
A Geneticist is mapping genes in a raccoon. She takes raccoon females heterozygous for eahc of 3 autosomal recessive mutations (red fur [a], no rings around eyes [b], and absence of striped tail [c]), and testcrosses them to males showing all three mutant phenotypes producing 1000 progeny of the following phenotype:
red hair, no ring, no stripe: 432
Wild type: 429
red hair, no ring: 37
Red hair, no stripe: 35
no ring: 34
no stripe: 33
a) show the arrangement/order of alleles in the heterozygous trihybrid female parent
b) draw the genetic map that explains the obained data
c) show whether or not interference is occuring
Explanation / Answer
Answer:-
Phenotype
Number
+ + +
429 ( Wild Type )
b a c
432
b a +
37
+ a c
35
b + +
34
+ + c
33
Total
1000
The most frequent reciprocal pairs of phenotypic classes which arise from non-crossover gametes tell us the genotype of the original parents. Here (+ + +) and (b a c) are parental classes which have 460 and 450 members respectively for a total of 861 (429+432).
A single cross over between b-a produces b++ and +ac which have 34 and 35 members respectively for a total of (34+35)=69. Therefore, the frequency of S.C.O between b-a is (69/1000)*100 = 6.9%
Similarly, other single crossing over between a-c produces ba+ and ++c phenotypic classes which have 37 and 33 members repectively for a total of (37+33) = 70. Therefore the frequency of S.C.O between a-c is (70/1000)*100 = 0.7%
a) Arrangement/order of alleles in the heterozygous trihybrid female parent is - b a c
Where b = no rings around eyes, a = Red fur, c = absence of striped tail.
b) Genetic map is –
b-----------------------6.9mu-----------------------------a-------0.7mu-------c
c) There is no inference is occurring because here D.C.O has not occurred.
Phenotype
Number
+ + +
429 ( Wild Type )
b a c
432
b a +
37
+ a c
35
b + +
34
+ + c
33
Total
1000
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