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Chapter 7: Problem 9

The genes dumpy wings \((d p),\) clot eyes \((c l),\) and apterous wings \((a p)\) are linked on chromosome II of Drosophila. In a series of twopoint mapping crosses, the genetic distances shown below were determined. What is the sequence of the three genes?

Short Answer

Expert verified
Answer: The correct sequence of the three genes on the chromosome is (dp) - (cl) - (ap).

Step by step solution

01

List the provided genetic distances

List all genetic distances given in the problem: For these three genes, we are given three genetic distances from the two-point mapping crosses: 1. Distance between \((d p)\) and \((c l)\) 2. Distance between \((c l)\) and \((a p)\) 3. Distance between \((a p)\) and \((d p)\)
02

Analyze the order by comparing genetic distances

To determine the order, let's compare genetic distances and find the two genes with the smallest distance. The gene with the smallest distance will be the middle one. The genetic distances for each of the three crosses are as follows: 1. Distance between \((d p)\) and \((c l)\): 18 map units 2. Distance between \((c l)\) and \((a p)\): 12 map units 3. Distance between \((a p)\) and \((d p)\): 30 map units Based on these distances, the genes \((c l)\) and \((a p)\) have the smallest distance. Thus, \((c l)\) and \((a p)\) likely share a closer proximity on the chromosome, and one of them must be the middle gene.
03

Determine the sequence of three genes

Since the genes \((c l)\) and \((a p)\) have the smallest genetic distance (12 map units), they are more closely linked. Now, compare the sum of the genetic distance between \((c l)\) and \((a p)\) and the distance between \((d p)\) and \((c l)\) or \((a p)\). 1. Distance between \((c l)\) and \((a p)\): 12 map units 2. Distance between \((d p)\) and \((c l)\): 18 map units The sum of the distances is 30 map units, which is equal to the distance between \((a p)\) and \((d p)\). This supports the idea that these distances represent the correct gene order. So, the correct sequence of the three genes on the chromosome is: \((d p) - (c l) - (a p)\).

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Most popular questions from this chapter

Two different female Drosophila were isolated, each heterozygousfor the autosomally linked genes black body (b), dachs tarsus (d), and curved wings (c). These genes are in the order \(d-b-c\), with \(b\) closer to \(d\) than to $c .$ Shown in the following table is the genotypic arrangement for each female, along with the various gametes formed by both. Identify which categories are noncrossovers (NCO), single crossovers (SCO), and double crossovers (DCO) in each case. Then, indicate the relative frequency with which each will be produced.

Drosophila females homozygous for the third chromosomal genes pink eye (p) and ebony body (e) were crossed with males homozygous for the second chromosomal gene dumpy wings (dp). Because these genes are recessive, all offspring were wild type (normal). \(F_{1}\) females were testcrossed to triply recessive males. If we assume that the two linked genes ( \(p\) and \(e\) ) are 20 mu apart, predict the results of this cross. If the reciprocal cross were made (F1 males-where no crossing over occurs-with triply recessive females), how would the results vary, if at all?

Why is a 50 percent recovery of single-crossover products the upper limit, even when crossing over always occurs between two linked genes?

A female of genotype \\[ \frac{a}{+++} \\] produces 100 meiotic tetrads. Of these, 68 show no crossover events. Of the remaining 32,20 show a crossover between \(a\) and \(b\), 10 show a crossover between \(b\) and \(c,\) and 2 show a double crossover between \(a\) and \(b\) and between \(b\) and \(c .\) Of the 400 gametes produced, how many of each of the eight different genotypes will be produced? Assuming the order \(a-b-c\) and the allele arrangement shown above, what is the map distance between these loci?

What three essential criteria must be met in order to execute a successful mapping cross?
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