Gene mapping question.

Genetics-Gene Mapping?

• Hi, I'm just trying to learn gene mapping and I'm a bit confused as to why the answer this question is D. I just don't get it, and I'd really appreciate it if someone could help work it through so I'd understand how they got that answer to be wrong and the other one's to be right. Esmé has her Bio199 project planned. She will determine the map distance between the tubby and vestigial genes in Drosophila. tb+ normal, long body tb tubby, short body vg+ big wings vg small wings She began by crossing true-breeding, homozygous parents and picking an F1 female from this cross to testcross. Esmé generated the following table of the progeny from the testcross: Observed phenotype number normal, long body and big wings 283 normal, long body and small wings 1294 tubby, short body and big wings 1418 tubby, short body and small wings 241 Which of the following statements is not true about this experiment? A. The recessive alleles of the F1 heterozygous female were in trans. B. The F1 heterozygous female genotype was tb+ vg/tb vg+. C. The frequency of recombination was 16%. D. vg is 19 map units from tb. E. All of the statements are true. Thanks in advance.

• Answer:

  The two offspring classes that have the most offspring in them: normal, long body and small wings 1294 tubby, short body and big wings 1418 are the non-recombinant classes, which means they represent the actual chromosomes in the heterozygous parent. So, you know that the heterozygous parent had the normal, long body allele (tb+) on the same chromosome as the small wings allele (vg). So, we know answer B. is true. Also, this formation (as opposed to having both the dominant alleles on the same chromosome) is called the trans orientation (as opposed to cis). So we also know that answer A. is true. The two other phenotype classes of offspring: normal, long body and big wings 283 tubby, short body and small wings 241 are the recombinant classes and the product of crossing over. To find the recombination rate, you take the total number of recombinant offspring and divide it by the total number of offspring overall: (241 + 283)/3236 = 0.16 or 16% recombination. Therefore, we now know that answer C. is true. In gene mapping, % recombination = map units. So, the vg and tb loci are 16 map units apart, not 19. So, answer D. is not true.