Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

In Drosophila, a female fly is heterozygous for three mutations, Bar eyes \((B),\) miniature wings \((m),\) and ebony body \((e)\) Note that Bar is a dominant mutation. The fly is crossed to a male with normal eyes, miniature wings, and ebony body. The results of the cross are as follows. Interpret the results of this cross. If you conclude that linkage is involved between any of the genes, determine the map distance(s) between them.

Short Answer

Expert verified
Question: Determine if there is any linkage between the genes involved in the mutations and calculate the map distance(s) between the linked genes based on the cross between a female fly heterozygous for Bar eyes (B), miniature wings (m), and ebony body (e) and a male fly with normal eyes, miniature wings, and ebony body. Answer: To determine the presence of linkage and calculate map distance, perform the following steps: 1. Determine the genotypes of the parent flies (female: BbMmEe, male: bbmmee). 2. Identify the possible gametes produced by each parent fly. 3. Perform the cross between the two parent flies and note the resulting offspring genotypes. 4. Investigate the likelihood of linkage between the genes by comparing the observed offspring ratio to the expected 1:1:1:1 ratio for independent assortment. 5. Calculate the recombination frequency between the linked genes using the provided offspring data. 6. Use the recombination frequency to determine the map distance between the linked genes. 7. Interpret the results and draw conclusions on the presence of linkage and the map distance between the genes. Note that to perform steps 5 and 6, you will need the specific data on the offspring counts from the cross.

Step by step solution

01

Identifying the genotypes of the parent flies

The female fly is heterozygous for three mutations: Bar eyes (B), miniature wings (m), and ebony body (e), so her genotype can be represented as BbMmEe. The male fly has normal eyes, miniature wings, and ebony body, so his genotype is bbmmee.
02

Determining possible gametes

Each parent fly can produce different gametes as a result of independent assortment during meiosis. For the female, the possible gametes are BME, BM, Be, BmE, Bm, bMe, be, bME (note that the last gamete is impossible, as there is no B dominant allele). For the male, the only possible gamete is bme.
03

Performing the cross and noting the resulting offspring

We need to perform the cross between the two parent flies and note the genotypes of the resulting offspring. We cross BM, Be, Bm, and bMe from the female fly with bme from the male fly. The resulting genotypes are: BbMmEe, Bbe, BbmE, and bbmMe.
04

Investigating the likelihood of linkage

If the genes are assorting independently, we should see a 1:1:1:1 ratio among the four possible offspring genotypes. However, if there is evidence of any other ratio, this may suggest that linkage is involved between some of the genes. We have to analyze the data we received from the cross to determine any linkage.
05

Calculating recombination frequency

The first step to calculate map distances between two loci (assuming linkage) is to determine the recombination frequency. The recombination frequency can be calculated using the formula: Recombination frequency = (Number of recombinant offspring) / (Total number of offspring) We have to determine the count of each type of offspring in the given data and apply the formula.
06

Using the recombination frequency to determine map distance

The recombination frequency can be used to determine map distance between linked genes, as 1% recombination frequency corresponds to 1 map unit (or 1 centimorgan). We can calculate map distance using the recombination frequency obtained in step 5.
07

Interpreting the results and drawing conclusions

Based on the calculated map distances between genes and their corresponding recombination frequencies, we can determine if there is any linkage between the genes involved in the mutations. If a linkage is present, then we have a clear map distance between them. If not, then the genes are assorting independently.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Drosophila genetics
Drosophila, commonly known as fruit flies, have long been a cornerstone of genetic research. With a small number of chromosomes, just four pairs, they are an ideal model organism. Researchers can easily observe inheritance patterns across generations. In the exercise above, we look at Drosophila genetics to understand genetic linkage. Specifically, the inheritance of mutations affecting bar eyes, miniature wings, and ebony bodies. Studying flies with these traits enables geneticists to explore how these mutations are passed on. A female fly with heterozygous traits can produce a variety of gametes, mixing and matching these mutations. By observing the offspring, scientists can determine whether genes are linked or assort independently. Understanding the basic genetics of Drosophila is a stepping stone to exploring more complex genetic phenomena.
recombination frequency
Recombination frequency is key in determining how often genes are linked or inherited together. This important concept helps in understanding the physical distance between genes on a chromosome. During meiosis, homologous chromosomes can exchange pieces of genetic material. This process is called recombination. If genes are far apart, such exchanges happen frequently, leading to a higher recombination frequency. However, if genes are close together, recombination is less frequent. In the exercise we're discussing, calculating recombination frequency helps us infer the genetic distance. By examining how many offspring show recombinant phenotypes versus parental types, the recombination frequency is calculated. This frequency gives insights into whether genes are linked, and if so, how closely they are situated on the chromosome.
gene mapping
Gene mapping is a fascinating method of determining the location of genes on a chromosome. Once recombination frequencies are known, they can be converted into map units, also known as centimorgans (cM). A 1% recombination frequency corresponds to 1 map unit, helping geneticists create a genetic map that highlights gene positions. If two genes are close, they will have a lower recombination frequency and will map closer together. Conversely, genes with higher recombination frequencies map further apart. In our Drosophila example, mapping involves calculating these frequencies from offspring data to see how the bar eyes, miniature wings, and ebony body genes are situated in relation to each other. Gene maps provide a visual representation of gene positions, offering insight into genetic linkages and the likelihood of inheritance patterns.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The gene controlling the Xg blood group alleles \(\left(X g^{+} \text {and } X g^{-}\right)\) and the gene controlling a newly described form of inherited recessive muscle weakness called episodic muscle weakness \((E M W X)\) (Ryan et al., 1999 ) are closely linked on the X chromosome in humans at position \(\mathrm{Xp} 22.3\) (the tip of the short arm \() .\) A male with EMWX who is \(\mathrm{Xg}^{-}\) marries a woman who is \(\mathrm{Xg}^{+}\), and they have eight daughters and one son, all of whom are normal for muscle function, the male being \(\mathrm{Xg}^{+}\) and all the daughters being heterozygous at both the \(E M W X\) and \(X g\) loci. Following is a table that lists three of the daughters with the phenotypes of their husbands and children. (a) Create a pedigree that represents all data stated above and in the following table. (b) For each of the offspring, indicate whether or not a crossover was required to produce the phenotypes that are given.

List some of the differences between a linkage map obtained by analyzing crossovers and a physical map obtained by sequencing the DNA.

Why are double-crossover events expected less frequently than single-crossover events?

In a certain plant, fruit is either red or yellow, and fruit shape is either oval or long. Red and oval are the dominant traits. Two plants, both heterozygous for these traits, were testcrossed, with the following results. Determine the location of the genes relative to one another and the genotypes of the two parental plants.

Phenotypically wild \(\mathrm{F}_{1}\) female Drosophila, whose mothers had light eyes \((l t)\) and fathers had straw \((s t w)\) bristles, produced the following offspring when crossed with homozygous lightstraw males: $$\begin{array}{lc} \text { Phenotype } & \text { Number } \\ \text { light-straw } & 22 \\ \text { wild } & 18 \\ \text { light } & 990 \\ \text { straw } & 970 \\ \text { Total } &{2000} \end{array}$$ Compute the map distance between the light and straw loci.

See all solutions

Recommended explanations on Biology Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free