9)Two recessive genes in Drosophilia (b and vg) produce black body and vestigial

Question

9)Two recessive genes in Drosophilia (b and vg) produce black body and vestigial wings respectively. When wildtype flies are testcrossed the Fl all have the dominant genes on one chromosome and the recessive genes on the homologous chromosome. Testcrossing the female Fl produced 1930 wildtype 412 black 1888 black and vestigial 370 vestigial (a) Calculate the distance between b and vg (b) Another recessive gene cn lies between the loci of b and vg producing cinnabar eye color. When wildtype flies are testcrossed the F1 are all trihybrid. Testcrossing the FI females produced 72 black, cinnabar 664 wildtype 68 vestigial 4 black, vestigial 652 black, cinnabar, vestigial 70 black 61 cinnabar, vestigial 8 cinnabar (d) do the b-vg distances coincide between the two (c) Calculate the map distances experiments? Explain

Explanation / Answer

Test Cross are the crosses in which one parent is always recessive for the given genes.

In the given scenario we are talking about two pheotypes expressed when genes are recessive in nature i.e. Black(bb) and Vestigial (vgvg)

Wild type flies will have genotype that is dominant i.e BBVGVG which is not black and not vestigial. When they are test crossed:

We will first make the gametes:

For wild type flies gamestes will be: BVG

For recessive flies it will be : bvg

So the F1 progeny will be BbVGvg i.e. not black and not vestigial.

Further when F1 females (BbVGvg) are test crossed (bbvgvg) : we can have gametes as

F1 female gametes can be: BVG, Bvg, bVG and bvg, while gametes for recessive female will be bvg. So drawing a punnet square we wil get progeny with following genotype:

So the recombinants that we get are not black and vestigial flies (412) and Black and not vestigial flies (370)

Recombination frequency (RF) = No. of recombinants/total no. of progeny * 100

RF = (412 + 370)/ 4600 * 100% = 17%

Therefore, the two genes coding for black colour and vestigial wings lie on the same chromosome since RF< 50%.

b.) As explained in the above answer, when we have a third gene the first cross leads to a trihybrid with genotype :

Lets refer cinnabar eyes with only cc and vestigial wings with vv

So the F1 progeny is: BbCcVv

And its gamets are: BCV, BcV, Bcv, bCV, bcV, bcv, bCv and BCv.

And for recessive phenotype the gametes are bcv

Drawing a punnet square we have:

To calculate RF for three genes we choose two genes first, lets take black and cinnbar first, for calculation of RF between b and c, lets check recombinants for these two genes:

BbccVv: 8

Bbccvv: 61

bbCcVv: 70

bbCcvv: 4

RF (between b and c) = 143/1599 * 100% = 8.9%

For RF (between c and v):

BbccVv: 8

BbCcvv: 68

bbCcvv: 4

bbccVv: 72

RF (between c and v) = 152/1599* 100% = 9.5%

RF (between b and v) = 271/1599 *100% = 16.9%

Therfore map distance between b and cn is 8.9 map unit, between cn and vg is 9.5 map unit and between b and vg is 16.9 map unit.

c) Yes map distance between b and vg loci coincide in both the expirements suggesting cn lies in between b and vg loci.

bvg Phenotype No. of flies BVG BbVgvg Wild type 1930 Bvg Bbvgvg Not black and vestigial 412 bVG bbVGvg Black and not vestigial 370 bvg bbvgvg Black and vestigial 1888 Total flies 4600

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