Question

w m W m wmlwm White eye Miniature wing wim/YO Red eye Full wing w* m* CO W m CO Wm wmlwm Red eye Full wing wm/YO White eye Miniature wing Phenotypes and genotypes of the F, Females Males W* m* Number observed w * m * 791 W m w mlwm Red eye Full wing w m/Y Red eye Full wing w m um 750 W m wm/wm White eye Miniature wing wm/Y White eye Miniature wing w m W* m W m Wtm/wm Red eye Miniature wing w m/Y Red eye Miniature wing W m * mo 455 CO W m wm/wm White eye Full wing wm /Y White eye Full wing

The results for the F2 progeny are shown for a F1 cross using the two X-linked markers:

w⁺ and m⁺.

As shown in the figure, the F1 cross is between the following two parents:

- a phenotypically wild-type (red eye, full wing), heterozygous female: w⁺m⁺/wm

- a double mutant male (white eye, miniature wing): wm/Y

Answer the following questions:

1) In the F1 cross, can you infer the genotype for the phenotype in the female? Why or why not?

2) Can you infer the genotype for the phenotype in the male? Why or why not?

3) ) In the F2 progeny, can you infer the genotype for the phenotype in the female? Why or why not?

3) Recombination in meiosis can modify the linkage between markers in the gametes. WHERE on the X chromosome can a recombination event occur to have such an effect for the w⁺ and m⁺ loci?

4) Can recombination modify linkage between the w⁺ and m⁺ loci in the male? Why or why not?

5) Can recombination modify linkage between the w⁺ and m⁺ loci in the female? Why or why not?

6) What is/are the parental gamete(s) for the male?

7) What is/are the parental gamete(s) for the female?

8) What is/are the recombinant gamete(s) for the male?

9) What is/are the recombinant gamete(s) for the female?

10) What is/are the parental phenotypic type(s) you predict in the F2 (progeny of the F1 cross)?

11) What is/are the recombinant type(s) you predict in the F2 (progeny of the F1 cross)?

12) Now that you can distinguish parental vs recombinant types, you can calculate the recombination between w+ and m+:

write the fraction and calculate the recombination frequency:

number of recombinant types       

_______________________ = ___________ =   

Total number of progeny

Now express the recombination frequency in centimorgans:

13) what would be the expected numbers for each phenotypic class in the F2, if the genes were UNLINKED? HINT: think of what would happen if w+ were on an autosome, with the same F1 parental cross.

Answer 1

1. as the cross done b/w the Red eye full wing which is dominant trait which is present in male , and the white eye miniature wing in the female which results into the RED eye full wing female in the F1 progeny as they get the red eye dominant gene from male during cross since they follow the X-linked inheritance so genotype is RED EYE FULL WING .

2. same as we discussed above regarding the female .the Y chromosome is imp. for the male trait , which during cross came with the white eye miniature wing which is the trait present in the parental female . The reason of such shuffling is due to the linkage. the dominant w+m+/Y is absent in the male game of

3. as the crossing taken place b/w the red eye full wing female and the white eyed miniature winged male , which upon cross gave us the following female , having genotypes

w+m+/wm –red eye full wing; wm/ wm-white eye miniature wing = they are like parental ones .

w+m/wm- red eye miniature wing ; wm+ /wm – white eye miniature wing = they are recombinant one’s having linkage associated with them among genes .

4. yes the recombination modifies the linkage b/w the w+ and m+ loci in male this linkage causes the parental genes to become recombinant one which is red eye miniature wing (w+m/Y )and white eye full wing( wm+ /Y). which in absence of the linkage gives the w+m+/Y –red eye full wing; wm/ Y-white eye miniature wing gametes .

5. yes the recombination modifies the linkage b/w the w+ and m+ loci in female this linkage causes the parental genes to become recombinant one linkage b/w red eye white eyes they came together and crossed among themselves making red eye miniature wing (w+m/wm)and white eye full winged (wm+/wm). Which otherwise become red eye full wing ( w+m+/wm )and white eye miniature wing e(wm/wm).

6. w+m+/Y–red eye full wing; wm/ Y -white eye miniature wing = they are like parental gametes for male

7. w+m+/wm –red eye full wing; wm/ wm-white eye miniature wing = they are parental gametes for females

8. red eye miniature wing (w+m/Y)and white eye full winged (wm+/Y). Are recombinant gametes for the male

9. red eye miniature wing (w+m/wm)and white eye full winged (wm+/wm) re recombinant gametes in female .

10. w+m+/wm red eye full wing        and wm/Y white eye miniature wing is predicted in the F2 progeny

11. red eye miniature wing (w+m/wm) and white eye full wing (wm+/wm) is predicted recombinant in the progeny

12. recombination frequency =no. of recombinant gametes /total no of gametes

=(w+m/wm) + (wm+/wm) in both male and female = (445+455 =900)

So , 900/2441=0.36 cM or 36 %

13. expected no of gametes for each phenotype is

red eye full wing=   9/16 *2441= 1373

white eye miniature wing =3/16 *2441= 457

red eye miniature wing =3/16 *2441= 457

white eye full wing =1/16 *2441= 152

since it is two point cross we use here 9:3:3:1 ratio 9/16 *2441