Exam 4 Review & Notes

B recombina8on between one gene and its centromere c a

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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...
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