A wild-type fly (heterozygous for gray body and normal wings) is mated with a black fly with vestigial wings. The offspring have the following phenotypic distribution: wild type, 778; black vestigial; 785; black normal, 158; gray vestigial, 162. What is the recombination frequency between these genes for the body color and wing size? Is this consistent with the results of the experiment in Figure 15.9?

Short Answer

Expert verified

17% of recombination exists between the genes for the body color and wing size.

Yes, the recombination frequency is consistent with the experiment results, as the recombination frequency obtained in the experiment was also 17%.

Step by step solution

01

Recombination frequency

The measure of recombinant offspring obtained in a genetic cross concerning the number of offspring produced is called recombination frequency. The recombination frequency between gene pairs helps to calculate the distance between the genes on a chromosome.

The following formula calculates the recombination frequency between gene pairs:\({\rm{Recombination frequency}}\,{\rm{ = }}\frac{{{\rm{Number of recombinants}}}}{{{\rm{Total number of offsprings}}}}\,{\rm{ \times 100}}\)

02

Recombination frequency between genes for the body color and wing size

Four types of offspring were obtained from the cross. These were wild type, black vestigial, black normal, and gray vestigial. Wild type and black vestigial are wild type offspring, while black normal and gray vestigial are recombinants obtained from the cross.

The total number of offspring produced in the cross is \({\rm{778 + 785 + 158 + 162 = 1883}}\).

The total number of recombinants is \({\rm{162 + 158 = 320}}\).

The recombination frequency is calculated by substituting the values into the following equation:

\({\rm{Recombination frequency}}\,{\rm{ = }}\frac{{320}}{{1883}}\,{\rm{ \times 100}}\)

\(\begin{aligned}{l}{\rm{Recombination frequency}}\, = \,16.99\,\% \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \approx 17\% \end{aligned}\)

Thus, the recombination frequency between the genes for the body color and wing size is approximately 17%.

03

Explanation for consistency with the experiment result

In the testcross performed in the experiment, between the body color and wing size, the recombination frequency obtained was 17%. It is consistent with the result of the cross here, as the recombination frequency obtained is also 17%.

The consistency is expected because the same genes were considered in both the experiment and the cross. The distance between the genes does not change from one cross to the other.

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Most popular questions from this chapter

Two genes of a flower, one controlling blue (B) versus white (b) petals and the other controlling round (R) versus oval (r) stamens, are linked and are 10 map units apart. You cross a homozygous blue oval plant with a homozygous white round plant. The resulting F1 progeny are crossed with homozygous white oval plants, and 1,000 offspring plants are obtained. How many plants of each of the four phenotypes do you expect?

The goodness of fit is measured by\({\chi ^{^2}}\). This statistic measures the amounts by which the observed values differ from their respective predictions to indicate how closely the two sets of values match. The formula for calculating this value is

\({\chi ^{}} = \sum \frac{{{{\left( {o - e} \right)}^2}}}{e}\)

Where o=observed and e= expected. Calculate the\({\chi ^{^2}}\)value for the data using the table below. Fill out the table, carrying out the operations indicated in the top row. Then add up the entries in the last column to find the\({\chi ^{^2}}\)value.

Testcross Offspring

Expected

(e)

Observed

(o)

Deviation

(o-e)

(o-e)2

(o-e)2/e

(A-B-)

220

(aaB-)

210

(A-bb)

231

(aabb)

239

\({\chi ^2}\) =sum

A white-eyed Drosophila is mated with a red-eyed (wild-type) male, the reciprocal cross of the one shown in Fig 15.4. What phenotypes and genotypes do you predict for the offspring from this cross?

Assume that genes A and B are on the same chromosome and are 50 map units apart. An animal heterozygous at both loci is crossed with one that is homozygous recessive at both loci. What percentage of the offspring will show recombinant phenotypes resulting from crossover? Without knowing these genes are on the same chromosome, how would you interpret the results of this cross?

Using the information from problem 4, scientists do a further testcross using a heterozygote for height and nose morphology. The offspring are tall upturned snout, 40; dwarf upturned snout, 9; dwarf downturned snout, 42; tall downturned snout, 9. Calculate the recombination frequency from these data, and then use your answer from problem 4 to determine the correct order of the three linked genes.

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