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ww.macmillanhighered.com launchpad pierceessentials3e 8102522#/launchpad item/MODULE bs 6534D6AF 3296 4D25 9638 F6DA905 ningCurve: Chapter 5 100/100 pts 06 ? Back to Study Plan Score: 355/300 Quest In a three-point cross of ABC/abc x abc/abc, the following genotypes and number of progeny are observed (total number of progeny 1100): Genotypes Number of Progeny ABC/abc 482 abc/abc 454 Abc/abc 39 aBC/abc 35 ABc/abc 38 abClabc 41 Abc abc aBcabc5 What is the map distance between the a and b genes? Need help on this question? E Read the ebook page on this topic Get a hint Show answer no points) no penalty) fewer points)Explanation / Answer
Answer:
Distance between the genes a & b = 7.82 cM
Explanation:
Hint: Always recombinant genotypes are smaller than the non-recombinant genotypes.
Hence, the parental (non-recombinant) genotype is ABC/abc
1).
If single crossover occurs between A&B
Normal combination: AB/ab
After crossover: Ab/aB
Ab progeny= 39+7=46
aB progeny = 35+5=40
Total this progeny = 86
The recombination frequency between A&B = (number of recombinants/Total progeny) 100
RF = (86/1100)100 = 7.82%
2).
If single crossover occurs between B&C..
Normal combination: BC/bc
After crossover: Bc/bC
Bc progeny= 38+5=43
bC progeny = 7+41 = 48
Total this progeny = 91
The recombination frequency between B&C = (number of recombinants/Total progeny) 100
RF = (91/1100)100 = 8.27%
3).
If single crossover occurs between A&C..
Normal combination: AC/ac
After crossover: Ac/aC
Ac progeny= 39+38=77
aC progeny = 35+41=76
Total this progeny = 153
The recombination frequency between A&C = (number of recombinants/Total progeny) 100
RF = (153/1100)100 = 13.91%
Recombination frequency (%) = Distance between the genes (cM)
a----------7.82cM--------b-----------8.27cM--------------c
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