Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A man with hemophilia (a recessive, sex-linked condition) has a daughter without the condition. She marries a man who does not have hemophilia. What is the probability that their daughter will have hemophilia? Their son? If they have four sons, what is the probability that all will be affected?

Short Answer

Expert verified

The probability that the daughter will be hemophilic is zero because neither of her parents is affected.

The son's probability of hemophilic is half (1/2) as they might receive a defective allele from their carrier mother.

The probability that all four will be hemophilic is 1/16 because each son has ยฝ probability of being hemophilic.

Step by step solution

01

Explanation for the probability that the daughter will be hemophilic

Hemophilia is an X-linked recessive disease. This meansa male with a hemophilic X chromosome will acquire the disease because males have one X and one Y chromosome.

Females with two hemophilic X chromosomes will acquire the disease as they have two X chromosomes. However, with one hemophilic X chromosome, they will be a carrier for the disease.

Here, the man is hemophilic, so he would transmit his hemophilic X chromosome to his daughter. Thus, her daughter must be the carrier for the disease as she is not hemophilic.

When the carrier woman marries a normal male, their daughter's chance to be hemophilic is zero.This is because the daughter must receive a hemophilic X chromosome from both the parent to be hemophilic.

However, this is not possible because his father does not have the defective allele for the trait.

02

Explanation for the probability that the son will be hemophilic

The males receive their X chromosome from their mother.Here, the mother being a carrier, can transmit either the normal X chromosome or hemophilic X chromosome to the son.

This means there is a 50 percent chance that the son might inherit the hemophilic X chromosome. Thus, the probability that the son is hemophilic is ยฝ because he has half chance of inheriting the hemophilic X chromosome from the mother.

03

Probability that all four sons will be hemophilic

The probability that the son would be is ยฝ. There are four sons, so the likelihood that all four sons will be hemophilic can be calculated as 12ร—12ร—12ร—12ร—12=116.

Therefore, the probability that all the sons will be hemophilic is 1/16.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A wild-type fruit fly (heterozygous for the gray body color and red eyes) is mated with a black fruit fly with purple eyes. The offspring are wild-type, 721; black purple, 751; gray purple, 49; black red, 45. What is the recombination frequency between these genes for the body color and eye color? Using information for problem 3, what fruit flies (genotypes and phenotypes) would you mate to determine the order of the body color, wing size, and eye color genes on the chromosome?

Review the description of meiosis (see Figure 13.8) and Mendelโ€™s laws of segregation and independent assortment (see Concept 14.1). What is the physical basis for each of Mendelโ€™s laws?

Reciprocal cross between two primrose varieties, A and B, produced the following results: Afemaleร—Bmaleโ†’offspringwith all green (non-variegated) leaves;Bfemaleร—Amaleโ†’offspring with patterned (variegated) leaves. Explain these results

The goodness of fit is measured by\({\chi ^{^2}}\). This statistic measures the amounts by which the observed values differ from their respective predictions to indicate how closely the two sets of values match. The formula for calculating this value is

\({\chi ^{}} = \sum \frac{{{{\left( {o - e} \right)}^2}}}{e}\)

Where o=observed and e= expected. Calculate the\({\chi ^{^2}}\)value for the data using the table below. Fill out the table, carrying out the operations indicated in the top row. Then add up the entries in the last column to find the\({\chi ^{^2}}\)value.

Testcross Offspring

Expected

(e)

Observed

(o)

Deviation

(o-e)

(o-e)2

(o-e)2/e

(A-B-)

220

(aaB-)

210

(A-bb)

231

(aabb)

239

\({\chi ^2}\) =sum

Consider what you learned about dominant and recessive alleles in Concept 14.1. If a disorder were caused by a dominant X-linked allele, how would the inheritance pattern differ from what we see for recessive X-linked disorders?

See all solutions

Recommended explanations on Biology Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free