/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 13 Albinism, lack of pigmentation i... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 13

Albinism, lack of pigmentation in humans, results from an autosomal recessive gene (a). Two parents with normal pigmentation have an albino child. (a) What is the probability that their next child will be albino? (b) What is the probability that their next child will be an albino girl? (c) What is the probability that their next three children will be albino?

Short Answer

Expert verified
Answer: The probability of their next child being an albino girl is 12.5%.

Step by step solution

01

(Identify genotypes)

Since both parents have normal pigmentation but have produced an albino child, we can determine that both of them are heterozygous carriers of the albinism gene. This means their genotypes are Aa x Aa, where "A" represents the dominant allele for normal pigmentation and "a" is the recessive allele for albinism.
02

(Create a Punnett square)

To determine the probability of different genotypes for the next child, we can use a Punnett square to list all possible combinations of alleles from both parents. The Punnett square is as follows: | A | a ----------- A | AA | Aa ----------- a | Aa | aa
03

(a) Calculate the probability of an albino child)

The probability of an albino child is determined by the probability of inheriting the recessive "aa" genotype. According to the Punnett square, there is one "aa" genotype out of four possible genotypes, so the probability of their next child being albino is 1/4 or 25%.
04

(b) Calculate the probability of an albino girl)

To determine the probability of the next child being an albino girl, we first need to consider the gender probability. The probability of having a girl is 1/2 or 50%. Therefore, the probability of an albino girl is the product of the probabilities of being a girl and being albino: (1/2) * (1/4) = 1/8 or 12.5%
05

(c) Calculate the probability of three consecutive albino children)

To determine the probability of their next three children all being albino, we need to calculate the probability of three consecutive albino children. This probability can be calculated by taking the probability of one albino child and raising it to the power of 3 (because there are three independent events): (1/4) * (1/4) * (1/4) = (1/4)^3 = 1/64 or 1.56%

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 91影视!

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

In a study of black guinea pigs and white guinea pigs, 100 black animals were crossed with 100 white animals, and each cross was carried to an \(\mathrm{F}_{2}\) generation. In 94 of the crosses, all the \(\mathrm{F}_{1}\) offspring were black and an \(\mathrm{F}_{2}\) ratio of 3 black: 1 white was obtained. In the other 6 cases, half of the \(\mathrm{F}_{1}\) animals were black and the other half were white. Why? Predict the results of crossing the black and white \(\mathrm{F}_{1}\) guinea pigs from the 6 exceptional cases.

Albinism in humans is inherited as a simple recessive trait. For the following families, determine the genotypes of the parents and offspring. (When two alternative genotypes are possible, list both.) (a) Two normal parents have five children, four normal and one albino. (b) A normal male and an albino female have six children, all normal. (c) A normal male and an albino female have six children, three normal and three albino. (d) Construct a pedigree of the families in (b) and (c). Assume that one of the normal children in (b) and one of the albino children in (c) become the parents of eight children. Add these children to the pedigree, predicting their phenotypes (normal or albino).

Dentinogenesis imperfecta is a rare, autosomal, dominantly inherited disease of the teeth that occurs in about one in 8000 people (Witkop 1957 ). The teeth are somewhat brown in color, and the crowns wear down rapidly. Assume that a male with dentinogenesis imperfecta and no family history of the disease marries a woman with normal teeth. What is the probability that (a) their first child will have dentinogenesis imperfecta? (b) their first two children will have dentinogenesis imperfecta? (c) their first child will be a girl with dentinogenesis imperfecta?

The autosomal (not X-linked) gene for brachydactyly, short fingers, is dominant to normal finger length. Assume that a female with brachydactyly in the heterozygous condition is married to a man with normal fingers. What is the probability that (a) their first child will have brachydactyly? (b) their first two children will have brachydactyly? (c) their first child will be a brachydactylous girl?

Mendel crossed peas having round seeds and yellow cotyledons (seed leaves) with peas having wrinkled seeds and green cotyledons. All the \(F_{1}\) plants had round seeds with yellow cotyledons. Diagram this cross through the \(\mathrm{F}_{2}\) generation, using both the Punnett square and forked-line, or branch diagram, methods.
See all solutions

Recommended explanations on Biology Textbooks

Biology Experiments

Read Explanation

Responding to Change

Read Explanation

Communicable Diseases

Read Explanation

Microbiology

Read Explanation

Control of Gene Expression

Read Explanation

Heredity

Read Explanation
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.