Given genes A, B, and C with phenotypes: Gene A Phenotype: Triceratops leg curva
ID: 55069 • Letter: G
Question
Given genes A, B, and C with phenotypes:
Gene A Phenotype: Triceratops leg curvature;
Gene B Phenotype: Farfeggnuggen wing vein;
Gene C Phenotype: Muppeteritus antennae.
And given the following table of phenotype counts (where "+" refers to wild-type phenotype):
ABC = 236
AB+ = 6
A+C = 26
A++ = 124
+BC = 121
+B+ = 32
++C = 4
+++ = 314
Which gene is in the middle?
A. There is no middle gene. A is linked to B, but not to C.
B. B
C. A
D. C
E. There is no middle gene. A is on a different chromosomes than B and
Based on my working below, the
100
Recombination frequency = 32 AC is 13
I am stuck here on how to find the middle gene. Since AC is 13, AB is 2.72 and BC is 0.41 which is where I am getting stuck.
Can some one help me understand how to solve this.
ABC 326 34.20776 AB+ 26 2.728227 A+C 124 13.01154 A++ 6 0.629591 + BC 4 0.419727 "+B+" 121 12.69675 "++C" 32 3.357817 "+++" 314 32.94858 953100
Explanation / Answer
From the given data, the following can be seen:
ABC and +++ are parental genotypes since they are the most frequent ones
AB+ and ++C are double recombinants since they are the least frequent genotypes
From the double recombinants and the parental genotypes, it can be clearly seen that gene C is flipped, and hence it is in the middle
To explain in detail:
If gene A is the middle gene, then double recombinants should be +BC and A++
If gene B is in the middle, then double recombinants should be A+C and +B+
Both of the above said recombinants are numerous in number and are not double recombinants. Hence, genes A and B are not in the middle
Gene C is the middle gene. The double recombinants are A+B and +C+. The frequency of A+B is 6 and that of +C+ is 4, as given in the data. These are the least frequent genotypes and hence, represent double recombinants.
Therefore, the gene that is present in the middle is gene ‘C’. Correct answer is option (D).
If recombination frequency is required, they are calculated as follows:
Recombination between genes A and C = (121+124+6+4)/863 = 0.295
Recombination between genes C and B = (26+32+6+4)/863 = 0.079
Double crossovers = (6+4)/863 = 0.0116
Expected DCO = 0.295 × 0.079 = 0.19
The table with correct gene orders is as follows:
ACB
236
A+B
6
A C +
26
A++
124
+C B
121
++B
32
+C+
4
+++
314
Total
863
ACB
236
A+B
6
A C +
26
A++
124
+C B
121
++B
32
+C+
4
+++
314
Total
863
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