Create a genetic mapping homework problem. In the problem you create, the studen

ID: 101774 • Letter: C

Question

Create a genetic mapping homework problem. In the problem you create, the student is given the results of a testcross involving two of the genes on the map attached and is asked to determine the recombination frequency. In other words, pick two linked genes from the map of the Drosophila X chromosome at the right and make up data for a test cross so that when the student calculates the recombination frequency they get the same distance as given on this map.

How to use the genetic map The diagram to the right is a genetic map of the Drosophila X chromosome. It shows the positions of genes along the chromosome and the units are centimorgans (100 x ). To find the distance between two genes, subtract their map positions. For example if gene p was at position 65.5 and gene n was at position 59, then the distance between them is 65.5-59 = 6.5 cM. That means that you should expect to see a recombination frequency = 0.065

On this map, if the mutant allele is recessive to wild type, it’s symbol is made with lowercase letters. If the mutant allele is dominant to wild type, the gene symbol is made with capital letters. The only gene on this map where the mutant is dominant to wild type is Bar eyes.

Note that these genes are all X-linked. That will affect how you do the cross and interpret the data. In a mapping testcross, the F1 generation must be heterozygous for both genes, but only females have the possibility of being heterozygous because they have two X chromosomes.

1 (X) yellow body, y scute bristles, sc 0.0 I . 3.0 5.5 7.5 white eyes, w facet eyes, fa echinus eyes, ec ruby eyes, rb crossveinless wings, cv 13.7 20.0 21.0 cut wings, ct singed bristles, sn 27.5 27.7 tan body,it lozenge eyes, lz 33.0ver vermilion eyes, v miniature wings, m 36.1 sable body, s garnet eyes, g 43.0 44.0 51.5 scalloped wings, sd forked bristles, f 56.7 57.0 9.5 62.5 66.0 68.1 Bar eyes, B fused veins, fu carnation eyes, car bobbed hairs, bb little fly, If

Explanation / Answer

We can make the question as following.-

Take any three genes; here we take yellow body (y), crossveinless mutant wings (cv), and cut wings (ct). These are recessive mutations. A cross has made between true black body (y+), wild type wings (cv+) and uncut wings (ct+). The F1 progeny has true configuration. After that a test cross has made between F1 to recessive true breeding parent. The F2 cross show following 1000 progenies with their phenotypes-

Phenotypes

Number of progenies

black body wild type and uncut wings parental

600

black body wild type cut wings SCO

black body crossveinless and uncut wings DCO

black body crossveinless and cut wings SCO

yellow body wild type and uncut wings SCO

yellow body crossveinless and uncut wings SCO

yellow body wild type and cut wings DCO

yellow body crossveinless and cut wings parental

206

Find the linkage map from above data.

Note- the number of progenies are determined by the map distance formula. That helps to create questions according the distance of original genes.