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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 \(\operatorname{son},(d)\) a color-blind daughter?

Short Answer

Expert verified
Answer: The probabilities are as follows: (a) normal son: 1/4, (b) normal daughter: 1/4, (c) color-blind son: 1/4, and (d) color-blind daughter: 0.

Step by step solution

01

Understanding the given information

Since both husbands and wives have normal vision, it means they are both carriers of the colorblind gene (Xc) inherited from their fathers, who were colorblind. Their genotypes are XcX for the wife and XcY for the husband.
02

Creating the Punnett Square

We will create a Punnett square for the possible genotypes of their children. In the rows, we will place the wife's alleles (Xc and X) and in the columns, we will place the husband's alleles (Xc and Y): Xc Y -------------- Xc | XcXc | XcY -------------- X | XX | XY
03

Identifying the outcomes

We can now identify the possible outcomes for their children based on the Punnett square: - XcXc (color-blind daughter) - XX (normal daughter) - XcY (color-blind son) - XY (normal son)
04

Calculating the probabilities

To find the probability of each outcome, we will calculate the ratio of the desired outcome among the four possibilities. (a) The probability of having a normal son (XY) is 1/4 (b) The probability of having a normal daughter (XX) is 1/4 (c) The probability of having a color-blind son (XcY) is 1/4 (d) The probability of having a color-blind daughter (XcXc) is 0/4 or 0, as it's not possible for them to have a color-blind daughter since the mother has a normal X chromosome. In conclusion, the husband and wife have probabilities of having: (a) a normal son: 1/4 (b) a normal daughter: 1/4 (c) a color-blind son: 1/4 (d) a color-blind daughter: 0

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Most popular questions from this chapter

Pigment in the mouse is produced only when the \(C\) allele is pres- ent. Individuals of the ce 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 \(\mathrm{F}_{1}\) and \(\mathrm{F}_{2}\) 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

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