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Ch. 4 - Extensions of Mendelian Genetics
Klug - Concepts of Genetics 12th Edition
Klug12th EditionConcepts of Genetics ISBN: 9780135564776Not the one you use?Change textbook
All textbooksKlug 12th EditionCh. 4 - Extensions of Mendelian GeneticsProblem 24
Chapter 4, Problem 24

In humans, the ABO blood type is under the control of autosomal multiple alleles. 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 is type O, what is the probability that their next child will be a female who has normal vision and is type O?

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Determine the genotypes of the parents for the ABO blood group. Since both parents are type A but have a child with type O, each parent must be heterozygous with genotype \(I^A i\) (where \(I^A\) is the allele for type A and \(i\) is the allele for type O).
Analyze the inheritance of the ABO blood group for the next child. Use a Punnett square to find the probability that the child will inherit the \(i\) allele from both parents, resulting in blood type O (\(ii\) genotype).
Determine the genotypes of the parents for the X-linked color blindness trait. Both parents have normal vision, but they have a color-blind son. Since color blindness is recessive and X-linked, the mother must be a carrier with genotype \(X^N X^c\) (where \(X^N\) is the normal allele and \(X^c\) is the color-blind allele), and the father must have normal vision with genotype \(X^N Y\).
Calculate the probability that the next child will be female and have normal vision. The child must inherit an \(X^N\) from the mother and an \(X^N\) from the father to have normal vision and be female (\(X^N X^N\)).
Combine the probabilities from the ABO blood group and the X-linked trait, and multiply by the probability that the child is female (which is 1/2), to find the overall probability that the next child will be a female with normal vision and blood type O.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Multiple Alleles and ABO Blood Group Inheritance

The ABO blood group is determined by three alleles (IA, IB, and i) located on an autosome. IA and IB are codominant, while i is recessive. Blood type A can be genotype IAIA or IAi, and blood type O is ii. Understanding these allele interactions is essential to predict offspring blood types.
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Haploid Genetics

X-linked Recessive Inheritance and Color Blindness

Color blindness is caused by a recessive allele on the X chromosome. Males (XY) express the trait if they inherit the affected X, while females (XX) must inherit two copies to be affected. Carrier females have one normal and one affected allele and show normal vision but can pass the trait to offspring.
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Purpose of X Inactivation

Probability Calculation in Genetic Crosses

Calculating the probability of offspring traits involves combining independent probabilities of genotype and sex. This requires determining parental genotypes, using Punnett squares for autosomal and sex-linked traits, and multiplying probabilities for combined traits like blood type, vision, and sex.
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Related Practice
Textbook Question

In a unique species of plants, flowers may be yellow, blue, red, or mauve. All colors may be true breeding. If plants with blue flowers are crossed with red-flowered plants, all F₁ plants have yellow flowers. When these produced an F₂ generation, the following ratio was observed:

9/16 yellow: 3/16 blue: 3/16 red: 1/16 mauve

In still another cross using true-breeding parents, yellow-flowered plants are crossed with mauve-flowered plants. Again, all F₁ plants had yellow flowers, and the F₂ showed a 9:3:3:1 ratio, as just shown. Determine the F₁ and F₂ results of a cross between true-breeding red and true-breeding mauve-flowered plants.

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Textbook Question

Five human matings (1–5), identified by both maternal and paternal phenotypes for ABO and MN blood-group antigen status, are shown on the left side of the following table:

Each mating resulted in one of the five offspring shown in the right-hand column (a–e). Match each offspring with one correct set of parents, using each parental set only once. Is there more than one set of correct answers?

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Textbook Question

A husband and wife have normal vision, although both of their fathers are red–green color-blind, an inherited 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?

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Textbook Question

In Drosophila, an X-linked recessive mutation, scalloped (sd), causes irregular wing margins. Diagram the F₁ and F₂ results if (a) a scalloped female is crossed with a normal male; (b) a scalloped male is crossed with a normal female. Compare these results with those that would be obtained if the scalloped gene were autosomal.

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Textbook Question

Another recessive mutation in Drosophila, ebony (e), is on an autosome (chromosome 3) and causes darkening of the body compared with wild-type flies. What phenotypic F₁ and F₂ male and female ratios will result if a scalloped-winged female with normal body color is crossed with a normal-winged ebony male?

Work out this problem by both the Punnett square method and the forked-line method.

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Textbook Question

In Drosophila, the X-linked recessive mutation vermilion (v) causes bright red eyes, in contrast to the brick-red eyes of wild type. A separate autosomal recessive mutation, suppressor of vermilion (su-v), causes flies homozygous or hemizygous for v to have wild-type eyes. In the absence of vermilion alleles, su-v has no effect on eye color. Determine the F₁ and F₂ phenotypic ratios from a cross between a female with wild-type alleles at the vermilion locus, but who is homozygous for su-v, with a vermilion male who has wild-type alleles at the su-v locus.

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