/*! 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 22 Alkaptonuria is a metabolic diso... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 22

Alkaptonuria is a metabolic disonder in which affected persons produce black urine. Alkaptonuria results from an allele \((a)\) that is recessive to the allele for normal metabolism \((A) .\) Sally has normal metabolism, but her brother has alkaptonuria. Sally's father has alkaptonuria and her mother has normal metabolism. a. Give the genotypes of Sally, her mother, her father, and her brother. b. If Sally's parents have another child, what is the probability that this child will have alkaptonuria? c. If Sally marries a man with alkaptonuria, what is the probability that their first child will have alkaptonuria?

Short Answer

Expert verified
a. Sally: Aa, Mother: Aa, Father: aa, Brother: aa. b. Probability = 50%. c. Probability = 50%.

Step by step solution

01

Define Genotypes

In alkaptonuria, the disease allele is recessive \( (a) \), while the normal metabolism allele is dominant \( (A) \). An individual has normal metabolism if they have at least one dominant allele (AA or Aa), and they have alkaptonuria if they have two recessive alleles (aa).
02

Determine Sally's Genotype

Sally's father has alkaptonuria, so his genotype must be \( aa \). Her mother has normal metabolism, so her genotype can be \( AA \) or \( Aa \). Since Sally has normal metabolism, she must have at least one dominant allele. Thus, her genotype must be \( Aa \) because she received a recessive allele \( (a) \) from her father.
03

Determine Sally's Family Genotypes

Sally's father's genotype is \( aa \). Sally's mother also must be \( Aa \) because Sally's brother has alkaptonuria, meaning he must be \( aa \) and therefore received \( a \) from both parents.
04

Calculate Probability for Sally's Parents' Next Child

If Sally's parents (Aa and aa) have another child, the possible genotypes for the child are found using a Punnett square: \[ \begin{array}{c|c|c} & A & a \ \hline a & Aa & aa \ \hline a & Aa & aa \ \end{array} \] The probability the child will have alkaptonuria \( (aa) \) is \( \frac{2}{4} = \frac{1}{2} \) or 50\%.
05

Calculate Probability for Sally's Child

If Sally (Aa) marries a man with alkaptonuria (aa), the possible genotypes for their children are: \[ \begin{array}{c|c|c} & A & a \ \hline a & Aa & aa \ \hline a & Aa & aa \ \end{array} \] The probability their first child will have alkaptonuria \( (aa) \) is \( \frac{2}{4} = \frac{1}{2} \) or 50\%.

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!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Alkaptonuria
Alkaptonuria is a rare genetic metabolic disorder characterized by the accumulation of homogentisic acid, which leads to the discoloration of urine, turning it black when exposed to air. This happens because the body lacks the enzyme homogentisic acid oxidase, which is necessary to break down this compound.

People with alkaptonuria often have other symptoms, such as darkened skin and connective tissue, as well as joint and cartilage issues later in life. This disorder is inherited in a specific manner, which leads us to the understanding of how genes function in relation to this condition.
  • The disease occurs due to the presence of two recessive alleles.
  • Individuals with only one recessive allele do not exhibit symptoms, as they have a normal dominant allele.
Genotype
A genotype is essentially a collection of genes or alleles in an individual's DNA that determine specific traits and characteristics. In the case of alkaptonuria, the relevant genes come in two variants: the normal metabolism allele (A) and the alkaptonuria allele (a).

The genotype affects whether an individual will exhibit symptoms of alkaptonuria or remain healthy:
  • Individuals with genotype AA possess two normal alleles and thus have normal metabolism.
  • Genotype Aa, known as heterozygous, still results in normal metabolism, because the dominant allele (A) masks the effect of the recessive allele (a).
  • Only genotype aa will result in alkaptonuria, as there are no dominant alleles to counter the recessive ones.
Understanding genotypes helps in predicting the probability of inheriting disorders like alkaptonuria.
Recessive Allele
In genetics, alleles are different forms of a given gene. A recessive allele, like the one responsible for alkaptonuria (represented as 'a'), needs to be present in two copies (homozygous aa) for its effects to manifest in an individual.

Recessive alleles are often overshadowed by dominant alleles, which makes the disorder only appear when two recessive alleles are inherited:
  • Having one recessive allele (Aa) results in a carrier status, where the individual doesn't show symptoms but can pass the allele to offspring.
  • Two recessive alleles (aa) mean that no dominant allele will suppress the disorder, resulting in the condition being expressed.
The pattern of inheritance for recessive alleles explains why genetic disorders like alkaptonuria often skip generations or affect children of unaffected parents.

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 watermelons, bitter fruit \((B)\) is dominant over sweet fruit \((b),\) and yellow spots \((S)\) are dominant over no spots \((s) .\) The genes for these two characteristics assort independently. A homozygous plant that has bitter fruit and yellow spots is crossed with a homozygous plant that has sweet fruit and no spots. The \(F_{1}\) are intercrossed to produce the \(\mathrm{F}_{2}\) a. What are the phenotypic ratios in the \(\mathrm{F}_{2} ?\) b. If an \(F_{1}\) plant is backcrossed with the bitter, yellow-spotted parent, what phenotypes and proportions are expected in the offspring? c. If an \(F_{1}\) plant is backcrossed with the sweet, nonspotted parent, what phenotypes and proportions are expected in the offspring?

What is the chromosome theory of heredity? Why was it important?

Phenylketonuria (PKU) is a disease that results from a recessive gene. Two normal parents produce a child with PKU. a. What is the probability that a sperm from the father will contain the PKU allele? b. What is the probability that an egg from the mother will contain the PKU allele? c. What is the probability that their next child will have PKU? d. What is the probability that their next child will be heterozygous for the PKU gene?

Hairlessness in A merican rat terriers is recessive to the presence of hair. Suppose that you have a rat terrier with hair. How can you determine whether this dog is homozygous or heterozygous for the hairy trait?

What is the principle of seg regation? Why is it important?
See all solutions

Recommended explanations on Biology Textbooks

Ecological Levels

Read Explanation

Reproduction

Read Explanation

Astrobiological Science

Read Explanation

Responding to Change

Read Explanation

Biology Experiments

Read Explanation

Cells

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.