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Chapter 15: Q1ITD (page 294)

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.

Short Answer

Expert verified

F1test hybrid is obtained from the crossing made between the cosmos plants. The expected number of each phenotype, out of the 900 offspring, is 225.

Step by step solution

01

Genes and their inheritance

Gene is the basic genetic material found in every organism. There are two types of alleles present in it such as linked alleles and unlinked alleles. Linked alleles are present in the chromosomes at a closer distance within a gene, and unlinked alleles are found in the different chromosomes.

02

Calculation of the expected from the given conditions

The expected number of unlinked genes is calculated as follows:

Total offspring\( = 900\)

Offspring phenotypic ratio\( = 1:1:1:1\)

Since the ratio equals the four different offspring, the total value divided by four will give the final expected number.

Expected number\( = \frac{{total{\rm{ }}number}}{{no.{\rm{ }}of{\rm{ }}offsprings}} = \frac{{900}}{4} = 225\)

The expected number of each offspring is 225.

03

Tabulation with the expected number results

Offspring from test cross of AaBb(F1)* aabb

Purple stem/short petals(A-B-)

Green stem/short petals (aaB-)

Purple stem/long petals(A-bb)

Green stem/ long petals(aabb)

Expected ratio if the genes are unlinked

1

1

1

1

Expected number of offspring (of 900)

225

225

225

225

The observed number of offspring (of 900)

220

210

231

239

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Most popular questions from this chapter

Consider what you learned about dominant and recessive alleles in Concept 14.1. If a disorder were caused by a dominant X-linked allele, how would the inheritance pattern differ from what we see for recessive X-linked disorders?

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?

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

Review the description of meiosis (see Figure 13.8) and Mendel鈥檚 laws of segregation and independent assortment (see Concept 14.1). What is the physical basis for each of Mendel鈥檚 laws?

Gene A, B, and C are located on the same chromosome. Test crosses show that the recombination frequency between A and B is 28% and that between A and C is 12%. Can you determine the linear order of these genes?
See all solutions

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