Genetics Biology: Journal Questions For Sex Linked Traits PDF

Preview

Summary

assignment ...

Description

Journal Questions for Sex Linked Traits 1. In a mating between a red-eyed male fruit fly and a red-eyed

heterozygous female, what percentage of the female offspring is expected to be carriers? How did you determine the percentage? 50% of the females are expected to be carriers. This was determined by completing a genetic cross using a punnett square to determine the genotypes of the offspring. A red-eyed male (XRY) was crossed with a red-eyed female carrier (XRXr). The genotypes that resulted were 1 XRXR, 1 XRXr, 1 XrY. One out of two, or 50% of the females are carriers. 2. In a mating between a red-eyed male fruit fly and a white-eyed female fruit fly, what percentage of the male offspring will have white eyes? Describe how you determined the percentage. 100 percent of the male offspring will have white eyes. This is determined by crossing either a red eyed male fruit fly with a white eyed female fruit fly in the Drosophila Laboratory and observing the results or by performing a genetic cross using a punnett square and crossing a red eyed male (XRY) and a white eyed female (XrXr) to determine that the genotypes were: 1 XRXr, 1 XrY, one out of one, which is 100% of the males are white eyed. 3. Hemophilia, a blood disorder in humans, results from a sex-linked recessive allele. Suppose that a daughter of a mother without the allele and a father with the allele marries a man with hemophilia. What is the probability that the daughter's children will develop the disease? Describe how you determined the probability. There is a 50:50 chance (50%) that her offspring will have the disease. This was determined by performing two punnett squares. One with a female with the allele (XHXH) was crossed with a male with an allele (XhY). The second one was a female carrier (XHXh) is crossed with a male with hemophilia (XhY). 4. Colorblindness results from a sex-linked recessive allele. Determine the genotypes of the offspring that result from a cross between a colorblind male and a homozygous female who has normal vision. Describe how you determined the genotypes of the offspring. All females will have normal vision, but they will be carriers of the allele. All mates will have normal vision. This was determined by completing a punnett square and crossing a color blind male (XbY) and a homozygous normal female (XBXB). The genotypes were 1 XBXb : 1 XBY. 5. Explain why sex-linked traits appear more often in males than in females. This is because males only have one X-chromosome, (most sexlinked diseases are found in the X chromosome) and therefore rely on all information provided by this chromosome. This means that it only takes on malfunctioning sex-linked gene for the male to express the condition for they don't, unlike females, have an extra copy. 6. In humans, hemophilia is a sex-linked recessive trait. It is located on the X chromosome. Remember that the human female genotype is XX and the male genotype is XY. Suppose that a daughter of a mother without the allele and a father with the allele marries a man with

hemophilia. What is the probability that the daughter's children will develop the disease? Describe how you determined the probability The probability of the daughter having a child with the disease is 50%. There is a 25% chance she will have a child that is a carrier, and there is a 25% chance she will have a child that does not have the allele. 7. Colorblindness also results from a sex-linked recessive allele on the X chromosome in humans. Determine the genotypes of the offspring that result from a cross between a color-blind male and a homozygous female who has normal vision. Describe how you determined the genotypes of the offspring. A colorblind male can be represented as (XcY) and a homozygous female with normal vision can be represented as (XCXC). Therefore, your cross is as follows (Xc Y) x (XC XC). Based on this crossing, you can conclude that: 100% females will have the genotype (XCXc) and 100% males will have the genotype (XCY). The phenotypes would be 100% females will be carriers, and will have normal vision because the allele for normal vision dominates the recessive allele for colorblindness. 8. Based on the traits explained in questions 6 and 7, explain why sexlinked traits in humans appear more often in males than in females. Males

only have one X chromosome, so one copy of an X-linked trait is all that is needed for the male to have that trait.

Trial #

Eye Color of Fruit Flies Parental Parental Offspring Genotypes Phenotypes Genotypes

1

XRXR XRY

2

XRXR XrY XRXr XRY

Red Eye Red Eye 100% red eye Red Eye White Eye Red Eye Red Eye

XRXr XrY XrXr XRY XrXr XrY

Red Eye White Eye White Eye Red Eye White eye White eye

3 4 5 6

1 XRXR: 1 XRy

100% red eyed

1 XRXr: 1XRY

100% Red Eye 3 Red Eyed: 1 White Eyed

1 XRXR: 1XRXr: 1 XRY: 1 XrY 1XRXr: 1XrXr: 1XRY: 1XrY 1 XRXr: 1XrY 1 XrXr: 1XrY

Key Male Female White Eye Red Eye

Offspring Phenotypes

XY XX r R

1 Red eye: 1 White eye 1 Red Eyed: 1 White eyed 100% White Eyed...