Animation Activity: X-Linked Inheritance

Chapter 4. Animation Activity: X-Linked Inheritance

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Question 1 of 10

Animation Assessment

Question

The results of which of the following crosses allowed T. H. Morgan to conclude that the white-eye trait in Drosophila was not autosomal recessive?

crosses between true-breeding red-eyed females and white-eyed males

crosses between true-breeding red-eyed females and red-eyed males

crosses between true-breeding white-eyed females and white-eyed males

crosses between true-breeding white-eyed females and red-eyed males

None of the above crosses would give a different result than what would be expected if the white-eye trait was autosomal recessive.

Correct. Crosses between true breeding white-eyed females and red-eyed males would result in 50% red-eyed females and 50% white-eyed males because the white eye trait is linked to the X-chromosome and males only inherit their X chromosome from their mother. If the white eye trait were autosomal recessive, this cross would have yielded all red-eyed F1 progeny. All of the other crosses described would yield identical results whether or not the white-eyed trait was X-linked.

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Incorrect. Crosses between true breeding white-eyed females and red-eyed males would result in 50% red-eyed females and 50% white-eyed males because the white eye trait is linked to the X-chromosome and males only inherit their X chromosome from their mother. If the white eye trait were autosomal recessive, this cross would have yielded all red-eyed F1 progeny. All of the other crosses described would yield identical results whether or not the white-eyed trait was X-linked.

Question

Calvin Bridges determined that the rare white-eyed males produced from the mating of true-breeding red-eyed females with white-eyed males resulted from a nondisjunction of X chromosomes in the female parent during meiosis. If such a mating produced 1234 red-eyed progeny and three white-eyed males, what is the observed rate of nondisjunction of the X chromosomes?

0.97%

0.49%

0.24%

0.12%

This is impossible to calculate with the given information.

Correct. Each nondisjunction of X chromosomes in the red-eyed female will generate gametes with two X chromosomes and with no X chromosomes with equal frequency. The white-eyed male will generate gametes with an X chromosome, bearing the white trait, or a Y chromosome, again with equal frequency. The union of these gametes generates four potential types of offspring. If an egg with two X chromosomes is fertilized by a sperm with one X chromosome, the result would be an X⁺X⁺X^w metafemale fly that is nonviable. If an egg bearing two X chromosomes is fertilized by a sperm bearing a Y chromosome, the result would be an X⁺X⁺Y female that will be phenotypically indistinguishable from the red-eyed females resulting from gametes without any nondisjunction. If an egg with no X chromosomes is fertilized with an X chromosome bearing sperm, the result will be an X X^wmale with white eyes. If an egg with no X chromosomes is fertilized with a Y-bearing sperm, the zygote will die because survival requires at least one X chromosome. So if there are three white-eyed males and 1234 red-eyed offspring, we can assume that there were (3×4) 12 zygotes resulting from nondisjunction gametes. Of these, we would have expected only six to survive. Therefore the rate of nondisjunction is 12/(1234 + 9) = 0.0097. The total number of offspring was 1243 to account for the six zygotes that would not have survived and the three white offspring which led to the discovery of the nondisjunction.

You have one more chance to answer the question correctly. Try again.

Incorrect. Each nondisjunction of X chromosomes in the red-eyed female will generate gametes with two X chromosomes and with no X chromosomes with equal frequency. The white-eyed male will generate gametes with an X chromosome, bearing the white trait, or a Y chromosome, again with equal frequency. The union of these gametes generates four potential types of offspring. If an egg with two X chromosomes is fertilized by a sperm with one X chromosome, the result would be an X⁺X⁺X^w metafemale fly that is nonviable. If an egg bearing two X chromosomes is fertilized by a sperm bearing a Y chromosome, the result would be an X⁺X⁺Y female that will be phenotypically indistinguishable from the red-eyed females resulting from gametes without any nondisjunction. If an egg with no X chromosomes is fertilized with an X chromosome bearing sperm, the result will be an X X^wmale with white eyes. If an egg with no X chromosomes is fertilized with a Y-bearing sperm, the zygote will die because survival requires at least one X chromosome. So if there are three white-eyed males and 1234 red-eyed offspring, we can assume that there were (3×4) 12 zygotes resulting from nondisjunction gametes. Of these, we would have expected only six to survive. Therefore the rate of nondisjunction is 12/(1234 + 9) = 0.0097. The total number of offspring was 1243 to account for the six zygotes that would not have survived and the three white offspring which led to the discovery of the nondisjunction.

Question

If nondisjunction of X-chromosomes occurs at a rate of 2% in a certain population of Drosophila, approximately how many white-eyed males would you expect to see from 1000 fertilized eggs resulting from breeding a true breeding red-eyed female with a white- eyed male?

This is impossible to calculate with the given information.

Correct. Each nondisjunction of X chromosomes in the red-eyed female will generate gametes with two X chromosomes and with no X chromosomes with equal frequency. The white-eyed male will generate gametes with an X chromosome, bearing the white trait, or a Y chromosome, again with equal frequency. The union of these gametes generates four potential types of offspring. If an egg with two X chromosomes is fertilized by a sperm with one X chromosome, the result would be an X⁺X⁺X^w metafemale fly that is nonviable. If an egg bearing two X chromosomes is fertilized by a sperm bearing a Y chromosome, the result would be an X⁺X⁺Y female that will be phenotypically indistinguishable from the red-eyed females resulting from gametes without any nondisjunction. If an egg with no X chromosomes is fertilized with an X chromosome bearing sperm, the result will be an X^w male with white eyes. If an egg with no X chromosomes is fertilized with a Y-bearing sperm, the zygote will die because survival requires at least one X chromosome. Therefore, if the nondisjunction rate is 2%, this means that 20 of the 1000 offspring were the result of nondisjunction-bearing gametes, of which 1/4 would be white-eyed males.

You have one more chance to answer the question correctly. Try again.

Incorrect. Each nondisjunction of X chromosomes in the red-eyed female will generate gametes with two X chromosomes and with no X chromosomes with equal frequency. The white-eyed male will generate gametes with an X chromosome, bearing the white trait, or a Y chromosome, again with equal frequency. The union of these gametes generates four potential types of offspring. If an egg with two X chromosomes is fertilized by a sperm with one X chromosome, the result would be an X⁺X⁺X^w metafemale fly that is nonviable. If an egg bearing two X chromosomes is fertilized by a sperm bearing a Y chromosome, the result would be an X⁺X⁺Y female that will be phenotypically indistinguishable from the red-eyed females resulting from gametes without any nondisjunction. If an egg with no X chromosomes is fertilized with an X chromosome bearing sperm, the result will be an X^w male with white eyes. If an egg with no X chromosomes is fertilized with a Y-bearing sperm, the zygote will die because survival requires at least one X chromosome. Therefore, if the nondisjunction rate is 2%, this means that 20 of the 1000 offspring were the result of nondisjunction-bearing gametes, of which 1/4 would be white-eyed males.