Skip to main content
Ch. 14 - Mendel and the Gene Idea
Chapter 14, Problem 6

Hemochromatosis is an inherited disease caused by a recessive allele. If a woman and her husband, who are both carriers, have three children, what is the probability of each of the following? a. All three children are of normal phenotype. b. One or more of the three children have the disease. c. All three children have the disease. d. At least one child is phenotypically normal.

Verified step by step guidance
1
Step 1: Understand the inheritance pattern. Hemochromatosis is caused by a recessive allele, meaning a child must inherit two recessive alleles (one from each parent) to express the disease. Each parent being a carrier has one recessive allele (h) and one dominant allele (H), so their genotype is Hh.
Step 2: Calculate the probability of each child's genotype. Using a Punnett square, the possible genotypes for each child are: HH (25%), Hh (50%), and hh (25%).
Step 3: Address part (a) - Calculate the probability that all three children are of normal phenotype (do not have the disease). This occurs if none of the children have the hh genotype. The probability of a child being either HH or Hh (normal) is 75%. Therefore, the probability that all three children are normal is 0.75 * 0.75 * 0.75 = 0.421875 or 42.19%.
Step 4: Address part (b) - Calculate the probability that one or more of the three children have the disease (hh genotype). First, find the probability that none of the children have the disease (all are HH or Hh) which is 0.421875 as calculated in step 3. The probability that at least one child has the disease is 1 - 0.421875 = 0.578125 or 57.81%.
Step 5: Address parts (c) and (d) - Calculate the probability that all three children have the disease (all are hh). The probability of one child being hh is 25%, so for three children it is 0.25 * 0.25 * 0.25 = 0.015625 or 1.56%. For part (d), the probability that at least one child is phenotypically normal is the complement of all children having the disease, which is 1 - 0.015625 = 0.984375 or 98.44%.

Verified Solution

Video duration:
1m
This video solution was recommended by our tutors as helpful for the problem above.
Was this helpful?

Key Concepts

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

Genetics and Alleles

Genetics is the study of heredity and variation in organisms. Alleles are different forms of a gene that can exist at a specific locus on a chromosome. In the case of hemochromatosis, the disease is caused by a recessive allele, meaning that an individual must inherit two copies of this allele (one from each parent) to express the disease phenotype.
Recommended video:
02:56
Genetics and Allele Frequencies Example 1

Punnett Squares

A Punnett square is a diagram used to predict the genotypes of offspring from parental crosses. It helps visualize the probability of inheriting specific traits based on the alleles contributed by each parent. For hemochromatosis, a Punnett square can illustrate the potential combinations of alleles from two carriers, aiding in calculating the probabilities of their children's phenotypes.
Recommended video:
Guided course
01:37
Punnett Squares

Probability in Genetics

Probability in genetics refers to the likelihood of certain genetic outcomes occurring in offspring. It is calculated based on the ratios of possible genotypes and phenotypes derived from parental alleles. Understanding basic probability principles allows one to determine the chances of various scenarios, such as the likelihood of children being affected by a recessive condition like hemochromatosis.
Recommended video:
Guided course
03:38
Punnett Square Probability
Related Practice
Textbook Question

A man with type A blood marries a woman with type B blood. Their child has type O blood. What are the genotypes of these three individuals? What genotypes, and in what frequencies, would you expect in future offspring from this marriage?

1824
views
Textbook Question

DRAW IT Two pea plants heterozygous for the characters of pod color and pod shape are crossed. Draw a Punnett square to determine the phenotypic ratios of the offspring.


2038
views
Textbook Question

Flower position, stem length, and seed shape are three characters that Mendel studied. Each is controlled by an independently assorting gene and has dominant and recessive expression as indicated in Table 14.1. If a plant that is heterozygous for all three characters is allowed to self-fertilize, what proportion of the offspring would you expect to be each of the following? (Note: Use the rules of probability instead of a huge Punnett square.) a. homozygous for the three dominant traits b. homozygous for the three recessive traits c. heterozygous for all three characters d. homozygous for axial and tall, heterozygous for seed shape

1534
views
Textbook Question

The genotype of F1 individuals in a tetrahybrid cross is AaBbCcDd. Assuming independent assortment of these four genes, what are the probabilities that F2 offspring will have the following genotypes? a. aabbccdd b. AaBbCcDd c. AABBCCDD d. AaBBccDd e. AaBBCCdd

3505
views
Textbook Question

What is the probability that each of the following pairs of parents will produce the indicated offspring? (Assume independent assortment of all gene pairs.) a. AABBCC×aabbcc→AaBbCc b. AABbCc×AaBbCc→AAbbCC c. AaBbCc×AaBbCc→AaBbCc d. aaBbCC×AABbcc→AaBbCc

2314
views
Textbook Question

Karen and Steve each have a sibling with sickle-cell disease. Neither Karen nor Steve nor any of their parents have the disease, and none of them have been tested to see if they carry the sickle-cell allele. Based on this incomplete information, calculate the probability that if this couple has a child, the child will have sickle-cell disease.

1935
views