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Chapter 15: Q15.3-2CC (page 306)

For each type of offspring of the test-cross in Figure 15.9, explain the relationship between its phenotype and the alleles contributed by the female parent. (It will be useful to draw out the chromosomes of each fly and follow alleles throughout the cross.)

Short Answer

Expert verified

The female parent contributes the alleles, and the male parent contributes recessive alleles. The contribution made by the female parent determines the offspring's phenotype.

Step by step solution

01

Description of offspring

Offspring is the young one of the organism. In sexual reproduction, the fusion of male and female gametes results in the formation of gametes. The offspring produced from sexual reproduction tends to possess the character of parents.

02

Description of the dominant and recessive allele

An allele is a variant present in the gene. The recessive allele is the allele that masks its effect. The other type of allele is the dominant alleles that get expressed in the offspring.

03

Allelic expression in the phenotype

The phenotype of the test crossing produced offspring such as wild type (gray normal), black vestigial, gray vestigial, and black normal. Out of these offspring, only female characters are more prominently seen in these offspring.

The male characters are recessive alleles, so the expression is not prominent. The alleles that are found in the eggs are exposed to the phenotype of the organism.

Hence, the female characters are predominantly seen in the test cross.

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Most popular questions from this chapter

A wild-type fruit fly (heterozygous for the gray body color and red eyes) is mated with a black fruit fly with purple eyes. The offspring are wild-type, 721; black purple, 751; gray purple, 49; black red, 45. What is the recombination frequency between these genes for the body color and eye color? Using information for problem 3, what fruit flies (genotypes and phenotypes) would you mate to determine the order of the body color, wing size, and eye color genes on the chromosome?

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The results in the data table are from a simulated F1 dihybrid test cross. The hypothesis that the two genes are unlinked predicts that the offspring phenotypic ratio will be 1:1:1:1. Using this ratio, calculate the expected number of each phenotype out of the 900 total offspring, and enter the values in that data table.

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

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