You testcross a fly that is heterozygous for three genes, A, B, and C. The wild

Question

You testcross a fly that is heterozygous for three genes, A, B, and C. The wild type allele is designated with a "+" and the mutant allele is designated with a lower case letter. Below are the results of your test cross:

Phenotype Observed
-------------------------------------
abc 300
-------------------------------------
+++ 300
-------------------------------------
ab+ 50
-------------------------------------
++c 50
-------------------------------------
a++ 125
-------------------------------------
+bc 125
-------------------------------------
a+c 25
-------------------------------------
+b+ 25
------------------------------------
1000
We've determined that x2 value/ total progeny value is 740

a) How do we draw best map for these genes...pls, include map units

b) Is there any interference? how much?

Explanation / Answer

Recombination frequencies:

            a-b Recombinants:

125+25; 125+25=

RF = 200/1000 = 30%

-----------------------

            b-c Recombinants

50+25; 50+25= RF = 100/1000 = 10%

-------------------------------------------

            a-c Recombinants: double crossover s

125+50; 125+50

RF = 300/1000 = 30%

RF = 0.30 + [2 x (25 + 25)/1000] = 0.3+ 0.10 = 0.4, or 40%.

---------------------------------------------------

Map

a------------10 –a.m.u ---------b------------------30 a.m.u----------------c

a--------------------------------------------------40 a.m.u----------------c

Interference between chiasma in regions I and II:

Observed number of double crossovers = 25 + 25 = 50

Expected number of double crossovers = 0.1 x 0.3 x 1000 = 30

I = 1 – (50/30) = 1 – 1.6 = 0.6, or 60% interference.

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