Gene Mapping Question

greenpak07 wrote:Hi guys,
I'm new here, I found this site while trying to look for help with bio questions, and it helped me.
I'm just trying to learn gene mapping and I'm a bit confused as to why the answer this question is D. I don't understand how they determined which gene is in the middle,I just don't get it, 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. For someone who's just learned this, it's really hard to work backwards, so I'd appreciate any type of help.

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.

First, you have to understand what was done. In general, F0 generation for the cross is usually true breeding (homozygous) mutants for a single trait. Thus, F0 parents in this case are tb+tb+vgvg (normal,small wings) and tbtbvg+vg+ (tubby, big wings). Cross between the two will result in all heterozygous tbtb+vgvg+ (normal, big wings).

Now, test cross means a backcross to the homozygous recessive individual:

tbtb+vgvg+ x tbtbvgvg or normal, big wings x tubby, small wings

If you do this cross, you will find out that you would expect to have

1 normal, big wings: 1 tubby, short wings: 1 tubby, big wings 1 normal, small wings
1 tb+tbvg+vg: 1 tbtbvgvg: 1tbtbvg+vg: 1 tb+tbvgvg
IF the two traits are unlinked!

However, observed data gives you the ratio of
283:241:1418:1294 or
1.17normal, big wings: 1 tubby, small wings: 5.88 tubby, big wings: 5.37 normal, small wings

Meaning that traits are linked (on the same chromosome close by) and that heterozygous parent had tubby, bid wings traits on 1 chromosome and normal, small wing trait on the sister chromosome – genotype tbvg+/tb+vg – the original combination from parental F0 cross.

For the recombination frequency, you need to divide number of recombinants by total number of individuals in the cross: (283+241)/( 283+241+1418+1294)=0.16 or 16% which the same as saying that two genes are 16 m.u. apart since 1 m.u. means 1% recombination frequency.

So, the only wrong statement is D.

Good luck,

Alla.

P.S. I would appreciate if you register and post further question here.

Hi Alla,
Thanks again for the help. So i guess since this was not a 3 point cross we did not have to find out which gene was in the middle by comparing the phenotypes of the parentals to the recombinants?

thanks so much!

That is correct and you are welcome!