You are mapping genes A and B, on a chromosome and you obtained 450 NON-recombinant offspring out of a total of 500. How far apart are these genes? \text{Recombination frequency} = \frac{\text{recombinant progeny}}{\text{total progeny}} 1\% \text{ recombination} = 1 \text{ map unit (m.u.)} 1 \text{ map unit} = 1 \text{ centimorgan (cM)}
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Step 1: Calculate the number of recombinant offspring: 500 total offspring - 450 non-recombinant offspring = 50 recombinant offspring. Show more…
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Sri K.
Map the above loci along a chromosome. The frequency of crossing over is proportional the distance separating two genes on a chromosome. The greater the distance, the more likely for crossing over to recombine those alleles. Scientists use a simple calculation to estimate the distance separating two genes on a chromosome. The percent of offspring showing recombinant phenotypes equals the distance between the two chromosomes. The unit of this distance is centimorgans (cM) or map units (MU). a. Calculate the separation distance of the A and B genes using the data on the frequency of offspring phenotypes above. [Number of recombinant phenotypes__________/total offspring__________] X 100 = __________ 5. Another gene in flies that is on the same chromosome as the A and B genes is body. The gray allele is dominant over yellow. A breeding study similar to the one above determined that recombination rates between the G locus and the B locus were 3% and the recombination rate between the A locus and the G locus were 28%. Using this information, draw a map of the chromosome indicating the relative positions of the A, B, and G loci.
1. During a dihybrid cross involving two linked genes, 23 percent of the resulting gametes showed a recombinant genotype. These two linked genes are _____ map units apart. A. 77 B.100 C.23 D. 27 2. Genes A and B are 20 map units apart. Consider the following cross: AB/ab x ab/ab. What is the probability of getting offspring with the genotype AB/ab? A. 0.6 B. 0.4 C. 0.2 D. 0.1
Madhur L.
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