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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.

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

Butterflies have an X-Y sex determination system that is different from that of flies or humans. Female butterflies may be either XY or X0, while butterflies with two or more X chromosomes are males. This photograph shows a tiger swallowtail gynandromorphy, which is half male (left side) and half female (right side). Given that the first division of the zygote divides the embryo into the future right and left halves of the butterfly, propose a hypothesis that explains how nondisjunction during the first mitosis might have produced this unusual looking butterfly.

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.

A man with hemophilia (a recessive, sex-linked condition) has a daughter without the condition. She marries a man who does not have hemophilia. What is the probability that their daughter will have hemophilia? Their son? If they have four sons, what is the probability that all will be affected?

The ABO blood type locus has been mapped on chromosome 9. A father with type AB blood and a mother who has type O blood have a child with trisomy nine and type A blood. Using this information, can you tell in which parent the non-disjunction occurred? Explain your answer. (See Figures 14.11 and 15.13).

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?

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