Numbers 37 through 40. Consider the following: A female heterozygous for three g
ID: 95207 • Letter: N
Question
Numbers 37 through 40. Consider the following: A female heterozygous for three genes A, B, C (AaBbCc) is mated to a male that is homozygous recessive for all three gene loci. They produce 1,000 offspring with the following genotypes: AaBbCc 304 aabbcc 310 aaBbCc 142 Aabbcc 145 aabbcc 40 AaBbCc 39 aabbCc 9 AaBbcc 11 Work area: _____ The mother's original chromosomes were in: A. coupling B. repulsion _____ The correct order for these gene loci is: A. C-B-A B. C-A-B C. B-C-A _____ The appropriate distances between the gene pairs is A. AC 28.7/BC 7.9 B. AC 9.9/BC 30.7 C. AB 9.9/CB 30.7 D. CA 30.7/CB 9.9 _____ Finally: please determine if there is or is not interference A. there is not interference in this cross. B. there is interference at a level of approximately 10% C. there is interference at a level of approximately 33% D. there is interference at a level of approximately 50% E. there is interference at a level of approximately 67%Explanation / Answer
37. The mother’s original chromosomes were in heterozygous state. The statement mentioned above, a female heterozygous for 3 genes A, B and C.
The repulsion state of chromosomes refers where the mother’s chromosomes have one dominant and one recessive allele from the two genes. You can see here the mother chromosome is AaBbCc, so this is a form of repulsion. This phase is also known as Trans phase.
The right answer is B. Repulsion.
38. and 39. Answers we will be calculating together.
As you can see there is triple cross between the A and B and C genes and their alleles.
Now the question has mentioned about the single and double cross overs, that we need to take.
The cross over rate is high when the genes are far away and the rate is low when the genes are nearer.
Here the crossover we need to check for the gene s between A---C and from C----B.
Let us consider the formula through which we can calculate it.
SCO + DCO / Total number of offspring *100 = No. of map units between genes.
(SCO – single cross over, DCO- double cross over).
Calculating for A-C, here the SCO is aabbcc – 40, AaBbCc – 39, together = 40 + 39 = 79. For DCO, aabbcc – 9, AaBbcc – 11, = 11 +9 = 20.
Formula for A-C calculation is SCO + DCO / Total number of offspring *100 = No. of map units between genes.
79 + 20/ 1000 *100 = 9.9 mapunits.
Same for the B-C calculation, here SCO is aaBbCc- 142, Aabbcc – 145, total = 142 + 145 = 287, and DCO = same as the previous one, 11 + 9 = 20.
So, SCO + DCO /1000 * 100 = 287 + 20 /1000 *100 = 30.7
38. Answer will be B. C-A-B.
39. Answer will be B. AC 9.9 / BC 30.7
40. Finally the level of interference calculation also known as coefficient of coincidence (c.o.c.). C.O.C is calculated when there is a 3 point cross with double recombinants. If the individual genes, (here it is A-C and B-C), if known, then the C.O.C between the individual cross regions can be calculated from the rate of double recombination.
Formula is, C.O.C = actual frequency of double recombinant/ expected frequency of double recombinant. Then the calculation of Interference is = 1 - C.O.C
Here the double recombinant is 20, for A-C the single recombinant is 99, and between B-C is 307,
Now, (99/1000) * (307/1000) = 0.099 * 0.307 = .030 or 30 per 1000.
The C.O.C is 20 / 30 = 0.66
The Interference is 1- 0.66 = approximately 33 %, the right answer is option C.
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