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Chapter 4: Problem 10

A husband and wife have normal vision, although both of their fathers are red- green color-blind, inherited as an X-linked recessive condition. What is the probability that their first child will be (a) a normal son, (b) a normal daughter, (c) a color-blind son, (d) a color-blind daughter?

Short Answer

Expert verified
Answer: The probabilities for their first child are as follows: - Normal son: 25% - Normal daughter: 25% - Color-blind son: 25% - Color-blind daughter: 25%

Step by step solution

01

1. Identify the genotypes of the parents

The problem states that both the husband and the wife have normal vision, but both of their fathers are red-green color-blind. Since the color-blindness gene is X-linked recessive, we can determine their genotypes: - The husband (the male) has XY chromosomes, and since he is not color-blind, his genotype must be X^NY (N stands for normal, and his Y chromosome does not carry the colorblind gene). - The wife (the female) has XX chromosomes. As both her father and her paternal grandfather were colorblind, she must have inherited an X^N chromosome from her mother, but since she is not colorblind, her genotype must be X^NX^R (R stands for color-blind).
02

2. Create a Punnett square

Now, we will create a Punnett square to determine the genotypes of their potential offspring. With one letter from each parent, the square will look like this: X^N (from father) Y (from father) X^N (from mother) X^NX^N X^NY X^R (from mother) X^NX^R X^RY
03

3. Calculate the probabilities

Using the Punnett square, we can now calculate the probability for each type of child: (a) For a normal son (X^NY): The probability is 1 out of 4 squares, so the probability is 1/4 or 25%. (b) For a normal daughter (X^NX^N): The probability is 1 out of 4 squares, so the probability is 1/4 or 25%. (c) For a color-blind son (X^RY): The probability is 1 out of 4 squares, so the probability is 1/4 or 25%. (d) For a color-blind daughter (X^NX^R): The probability is 1 out of 4 squares, so the probability is 1/4 or 25%. Final probabilities for their first child: - Normal son: 25% - Normal daughter: 25% - Color-blind son: 25% - Color-blind daughter: 25%

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Most popular questions from this chapter

Pigment in the mouse is produced only when the \(C\) allele is present. Individuals of the \(c c\) genotype have no color. If color is present, it may be determined by the \(A\) and \(a\) alleles. AA or Aa results in agouti color, whereas aa results in black coats. (a) What \(F_{1}\) and \(F_{2}\) genotypic and phenotypic ratios are obtained from a cross between \(A A C C\) and aace mice? (b) In the three crosses shown here between agouti females whose genotypes were unknown and males of the aacc genotype, what are the genotypes of the female parents for each of the following phenotypic ratios? (1) 8 agouti (2) 9 agouti (3) 4 agouti 8 colorless 10 black \(\quad 5\) black 10 colorless

In foxes, two alleles of a single gene, \(P\) and \(p,\) may result in lethality \((P P),\) platinum coat \((P p),\) or silver coat \((p p) .\) What ratio is obtained when platinum foxes are interbred? Is the \(P\) allele behaving dominantly or recessively in causing (a) lethality; platinum coat color?

As in the plants of Problem \(6,\) color may be red, white, or pink and flower shape may be personate or peloric. Determine the \(\mathrm{P}_{1}\) and \(\mathrm{F}_{1}\) genotypes for the following crosses: (a) red, peloric \(\times\) white, personate \(\mathrm{F}_{1}:\) all pink, personate (b) red, personate \(\times\) white, peloric \(\mathrm{F}_{1}:\) all pink, personate (c) pink, personate \(\times\) red, peloric $F_{1:}\left\\{\begin{array}{l}1 / 4 \text { red, personate } \\ 1 / 4 \text { red, peloric } \\ 1 / 4 \text { pink, personate } \\ 1 / 4 \text { pink, peloric }\end{array}\right.$ (d) pink, personate \(x\) white, peloric $\mathbf{F}_{1:}\left\\{\begin{array}{l}1 / 4 \text { white, personate } \\ 1 / 4 \text { white, peloric } \\ 1 / 4 \text { pink, personate } \\ 1 / 4 \text { pink, peloric }\end{array}\right.$ (e) What phenotypic ratios woud result from crossing the \(\mathrm{F}_{1}\) of (a) to the \(F_{1}\) of \((b) ?\)

In humans, the ABO blood type is under the control of autosomal multiple alleles. Red-green color blindness is a recessive X-linked trait. If two parents who are both type A and have normal vision produce a son who is color- blind and type \(0,\) what is the probability that their next child will be a female who has normal vision and is type \(0 ?\)

In four o'clock plants, many flower colors are observed. In a cross involving two true-breeding strains, one crimson and the other white, all of the \(P_{1}\) generation were rose color. In the \(F_{2}\), four new phenotypes appeared along with the \(P_{1}\) and \(F_{1}\) parental colors. The following ratio was obtaincd: \(1 / 16\) erimson \(2 / 16\) orange \(1 / 16\) yellow \(2 / 16\) magenta \(4 / 16\) rose \(2 / 16\) pale yellow \(4 / 16\) white Propose an explanation for the inheritance of these flower colors.
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