Exam 4 Review & Notes

B recombina8on between one gene and its centromere c a

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: w A is in the middle. In the three point cross AaMmTt by aammh, the following phenotypes among the progeny were obtained: AMT- 25 amt- 25 AmT- 425 aMt- 425 AMt- 275 amT- 275 Amt- 4,275 aMT- 4,275 9. For the ques8on above, the distance between A and M in map units is: a) 6 b) 9 c) 14 d) 94 e) None of the above In the three point cross AaMmTt by aammh, the following phenotypes among the progeny were obtained: AMT- 25 amt- 25 AmT- 425 aMt- 425 recombinant AMt- 275 amT- 275 Amt- 4,275 aMT- 4,275 9. For the ques8on above, the distance between A and M in map units is: a) 6 b) 9 c) 14 d) 94 e) None of the above Given the info above, we can say A and M are linked and then use the recombinant classes to find the recombina8on frequency. RF = recombinants/total x 100% RF = 50+550/10,000 x 100% = 6% = 6 m.u. 10. For the cross DeF/dEf by def/def, the percent of the progeny with the DEF phenotype will be: 10. For the cross DeF/dEf by def/def, the percent of the progeny with the DEF phenotype will be: D e F d e f d E f d e f Given the info above, and that fact that e is in the middle, we can tell that the DEF phenotype is a double recombinant. No mater how this recombines and separates you can only ever get def D E F d e f The probability of the double recombina8on occurring is the probability of one occurring D/E (.10) 8mes the probability of the other occurring E/F (.20) 8mes one half. (1/2)(0.10)(0.20) = 0.01 or 1% 11. Using the map in the ques8on above, for the cross BH/bh by bh/bh, the percent of the progeny with the bh phenotype will be: a) 25 b) 32.5 c) 50 d) 65 e) None of the above 11. Using the map in the ques8on above, for the cross BH/bh by bh/bh, the percent of the progeny with the bh phenotype will be: a) 25 b) 32.5 c) 50 d) 65 e) None of the above Because these two genes are more than 50 map units apart, you are equally likely to produce any of the progeny. B H b h b h b h No mater how this recombines and separates you can only ever get b/h B H B h b H b h b h b h b h b h Therefore there is a ¼ chance the progeny would have the bh phenotype, or 25% 12. Two yeast...
View Full Document

Ask a homework question - tutors are online