Drosophila females heterozygous for each of three recessive...

Drosophila females heterozygous for each of three recessive autosome mutations with independent phenotypic effects (thread antennae [th], hairy body [ $h]$, and scarlet eyes [ $s t$ ]) were testcrossed to males showing all three mutant phenotypes. The 1000 progeny of this testeross were $$ \begin{array}{lr} \text { thread, hairy, scarlet } & 432 \\ \text { wild type } & 429 \\ \text { thread, hairy } & 37 \\ \text { thread, scarlet } & 35 \\ \text { hairy } & 34 \\ \text { scarlet } & 33 \end{array} $$ a. Show the arrangement of alleles on the relevant chromosomes in the triply heterozygous females. b. Draw the best genetic map that explains these data. c. Calculate any relevant interference values.

Drosophila females heterozygous for each of three recessive autosome mutations with independent phenotypic effects (thread antennae [th], hairy body [ $h]$, and scarlet eyes [ $s t$ ]) were testcrossed to males showing all three mutant phenotypes. The 1000 progeny of this testeross were
$$
\begin{array}{lr}
\text { thread, hairy, scarlet } & 432 \\
\text { wild type } & 429 \\
\text { thread, hairy } & 37 \\
\text { thread, scarlet } & 35 \\
\text { hairy } & 34 \\
\text { scarlet } & 33
\end{array}
$$
a. Show the arrangement of alleles on the relevant chromosomes in the triply heterozygous females.
b. Draw the best genetic map that explains these data.
c. Calculate any relevant interference values.

Key Concepts

Double Crossover Events and Interference

Double crossover events occur when two separate crossover events happen between three linked loci, leading to new combinations of alleles. Interference is the phenomenon that one crossover event can reduce the probability of an adjacent second crossover. Calculating interference values can help understand the non-random distribution of crossover events along the chromosome.

Gene Linkage and Allelic Arrangement

Gene linkage refers to the physical proximity of genes on the same chromosome, which causes them to be inherited together more often than expected by independent assortment. In cases involving heterozygotes, analyzing the parental and recombinant offspring can reveal whether the alleles are arranged in a coupling (cis) configuration, where all dominant or all recessive alleles are on one chromosome, or in repulsion (trans), which is key to interpreting testcross data.

Testcross

A testcross involves mating an individual with unknown genotype (typically heterozygous for the traits of interest) with a homozygous recessive individual. This cross reveals the gamete types produced by the heterozygote because the recessive homozygote does not mask any recombinant phenotype, thereby allowing direct inference of the genetic arrangement of alleles.

Recombination Frequency and Genetic Mapping

Recombination frequency is calculated by dividing the number of recombinant offspring by the total offspring, which gives an estimate of the genetic distance between genes measured in centimorgans (cM). This concept is crucial for constructing genetic maps and helps to determine the order and relative spacing of genes on a chromosome.

Transcript

00:01 Drosophila female heterozygous for each of the three recessive autosomal mutations with independent phenotypic effects, thred, th, harry, h, scarlet, i, st were test crossed to male showing all three mutant phenotypes.

00:20 The 1 ,000 progeny of this test cross were shown on a board.

00:24 So a showed a range of alleles on a relevant chromosome in the triplet heterozygous female.

00:31 So if we look at our distribution of this phenotype, you can see there are two missing phenotypes.

00:38 Harry, scarlet, and thred is missing.

00:42 So this tells us the two most common type of progeny are called parental.

00:49 The parental progeny came from the non -crossover gametes.

00:54 So this means there are three different genes that are linked.

00:57 The wild type has plus, plus, and plus.

01:01 The thred, harry, scarlet has three mutations, th, h, and st.

01:13 And if there's no crossover between two of the genes, we call this parental.

01:18 And this usually has the most common.

01:20 So in this case, 432 and 429, they're both the most common progeny.

01:27 So we have already decided they are parental.

01:30 Now there are again, as i just mentioned, the harry, scarlet, and thred, these two phenotypes are missing.

01:37 So this tells us that this pair is the least common.

01:47 So therefore, usually in a progeny, the least common are double crossover dco.

01:54 So you have three genes.

01:55 There are two crossover events between two of the genes.

01:59 So due to the rareness of the double crossover event, usually you have almost have the least common progeny.

02:08 In this case, you do not have any double crossover.

02:11 So this is the least common.

02:13 And the rest of others, four are single crossover between two of the genes, the first two and the second and third.

02:19 So there is crossover, but it's less common than parental, but it's more than the double crossover.

02:29 So now let's compare the double crossover with parental.

02:33 So i have already written down the genotype of the parental.

02:36 Later i will compare that with the double crossover.

02:45 So the double crossover has h, st, and plus, and the thred is plus, plus, th.

03:03 Now we compare the genotype.

03:06 So you can see that if compare herriot's scarlet with the thred herriot's scarlet, h and st, they are both parental.

03:16 Hth, sorry, h and st, both are parental.

03:23 But in the parent, you have a th, but you have a plus here.

03:28 So this means the thred gene, th gene is a recombinant.

03:34 The other two, parental.

03:37 The same, if you compare thred with wild type, h and st gene, they're both parental, but th gene again, is a recombinant.

03:49 So after we compare the dco with parental p, we figure out that th gene must be in the middle.

04:01 So what the arrangement looks like is st, th, and h.

04:08 So this is parental arrangement.

04:12 Now due to double crossover, you have a cross between st and th, and then another cross between th and h.

04:22 So you end up having st plus an h.

04:28 This will give you the herriot and scarlet, and also you will have plus, th, and plus.

04:35 This will give you thred.

04:37 So now we figure out that th is in the middle.

04:41 So answer the first question, th gene in the middle.

04:49 So we have the order of these three genes, as i just mentioned, st, th, and h.

05:06 B, draw the best genetic map to explain these data.

05:12 So you have to calculate distance between st and th and th and h.

05:17 And this information comes from the single crossover.

05:20 So let's take a look at a single crossover.

05:24 Th, h, plus th, and plus, and st...

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