10. A cross is made between two E. coli strains: (Hfr arg+?bio+?leu+) x (F? arg?
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10. A cross is made between two E. coli strains: (Hfr arg+?bio+?leu+) x (F? arg??bio??leu?). Interrupted-mating studies show that arg+ enters the recipient last, so arg+ recombinants are selected on a medium containing biotin and leucine only. These recombinants are tested for the presence of bio+ and leu+. The following numbers of individuals are found for each genotype: arg+ bio+ leu+ 320 arg+ bio- leu+ 0 arg+ bio+ leu- 8 arg+bio- leu- 48 A: what is the gene order B:what are the map distances in recombination percentagesExplanation / Answer
Step 1: Determine the parental genotypes. The most abundant genotypes are the partenal types. These genotypes are v cv+ ct+ and v+ cv ct. What is different from our first three-point cross is that one parent did not contain all of the dominant alleles and the other all of the recessive alleles. Step 2: Determine the gene order To determine the gene order, we need the parental genotypes as well as the double crossover geneotypes As we mentioned above, the least frequent genotypes are the double-crossover geneotypes. These geneotypes are v cv+ ct and v+ cv ct+. From this information we can determine the order by asking the question: In the double-crossover genotypes, which parental allele is not associated with the two parental alleles it was associated with in the original parental cross. From the first double crossover, v cv+ ct, the ct allele is associated with the v and cv+ alleles, two alleles it was not associated with in the original cross. Therefore, ct is in the middle, and the gene order is v ct cv. Step 3: Determing the linkage distances. v - ct distance caluculation. This distance is derived as follows: 100*((89+94+3+5)/1448) = 13.2 cM ct - cv distance calculation. This distance is derived as follows: 100*((45+40+3+5)/1448) = 6.4 cM Step 4. Draw the map. Three-point crosses also allows one to measure interference (I) among crossover events within a given region of a chromosome. Specifically, the amount of double crossover gives an indication if interference occurs. The concept is that given specific recombination rates in two adjacent chromosomal intervals, the rate of double-crossovers in this region should be equal to the product of the single crossovers. In the v ct cv example described above, the recombination frequency was 0.132 between genes v and ct, and the recombination frequency between ct andcv was 0.064. Therefore, we would expect 0.84% [100*(0.132 x 0.64)] double recombinants. With a sample size of 1448, this would amount to 12 double recombinants. We actually only detected 8. To measure interference, we first calculate the coefficient of coincidence (c.o.c.) which is the ratio of observed to expected double recombinants. Interference is then calculated as 1 - c.o.c. One linkage map unit (LMU) is 1% recombination. Thus, the linkage map distance between two genes is the percentage recombination between those genes. In this case, we have a total of 300 recombinant offspring, out of 2000 total offspring. Map distance is calculated as (# Recombinants)/(Total offspring) X 100. So our map distance is (300/2000)x100, or 15 LMU. Linkage map units do not correspond to any fixed length of chromosome. Frequency of crossover (and thus of recombination) can be affected by location (crossing over is repressed close to the centromere) and proximity to another corssover. Also, since scoring of recombinant offspring ignores what else might be going on between the two genes in question, the number of recombination events is always underestimated because double crossovers are missed. A distance of 50 LMU corresponds to independence. Note that, if two gene loci are far enough apart on a large chromosome, the crossover rate between them may be so high that they will map as if they are not linked.
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