/*! 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 19 The hemoglobin B gene ( \(H b\) ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 18: Problem 19

The hemoglobin B gene ( \(H b\) ) has a common allele \((A)\) of an \(\operatorname{SNP}(r s 334)\) that encodes the \(H b^{A}\) form of (adult) hemoglobin and a rare allele ( \(T\) ) that encodes the sickling form of hemoglobin, \(H b^{S}\). Among 57 members of the Yoruba tribe of Nigeria, 44 were \(A / A\) and 13 were \(A / T .\) No \(T / T\) individuals were observed. Use the \(\chi^{2}\) test to determine whether these observed genotypic frequencies fit Hardy-Weinberg expectations.

Short Answer

Expert verified
The genotypic frequencies fit Hardy-Weinberg expectations (\(\chi^2 = 0.906\), \(p > 0.05\)).

Step by step solution

01

Determine Observed Genotype Frequencies

The observed genotypes in the sample are as follows: 44 individuals are homozygous A/A, and 13 are heterozygous A/T, with no individuals being homozygous T/T.
02

Determine the Total Number of Alleles

Calculate the total number of alleles by multiplying the number of individuals by 2 since each individual carries two alleles. \[ 57 \times 2 = 114 \text{ alleles} \]
03

Calculate Allele Frequencies

Calculate the frequencies of alleles A and T using the observed genotype counts.- Frequency of A allele: - From A/A individuals: \[ 44 \times 2 = 88 \text{ A alleles} \] - From A/T individuals: \[ 13 \times 1 = 13 \text{ A alleles} \] - Total A alleles: \[ 88 + 13 = 101 \] - Frequency of A allele: \[ \frac{101}{114} \approx 0.886 \]- Frequency of T allele: - Total T alleles, from A/T individuals: \[ 13 \times 1 = 13 \] - Frequency of T allele: \[ \frac{13}{114} \approx 0.114 \]
04

Calculate Expected Genotype Frequencies under Hardy-Weinberg Equilibrium

Use the Hardy-Weinberg principle which states that the frequencies of genotypes are:- Expected frequency of A/A: \[ (p^2) = (0.886)^2 \approx 0.785 \]- Expected frequency of A/T: \[ 2pq = 2 \times 0.886 \times 0.114 \approx 0.202 \]- Expected frequency of T/T: \[ (q^2) = (0.114)^2 \approx 0.013 \]Convert these frequencies to expected numbers by multiplying by the total number of individuals:- Expected number of A/A: \[ 57 \times 0.785 \approx 44.7 \]- Expected number of A/T: \[ 57 \times 0.202 \approx 11.5 \]- Expected number of T/T: \[ 57 \times 0.013 \approx 0.7 \]
05

Perform Chi-Square Test

Calculate the chi-square statistic using the formula: \[\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}\]where \(O_i\) is the observed count and \(E_i\) is the expected count:- For A/A: \[ \frac{(44 - 44.7)^2}{44.7} \approx 0.011 \] - For A/T: \[ \frac{(13 - 11.5)^2}{11.5} \approx 0.195 \]- For T/T: \[ \frac{(0 - 0.7)^2}{0.7} \approx 0.7 \]Total chi-square value: \[\chi^2 \approx 0.011 + 0.195 + 0.7 \approx 0.906\]
06

Compare with Critical Value

Determine the degrees of freedom, \(df = \text{number of genotypes} - \text{alleles} = 3 - 2 = 1 \). At \(\alpha = 0.05\) and \(df = 1\), the critical value is 3.841 from chi-square distribution tables.Since the calculated \(\chi^2\) value of 0.906 is less than 3.841, we fail to reject the null hypothesis.

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.

Chi-Square Test
The Chi-Square Test is a statistical method used to determine if there is a significant difference between the expected frequencies and the observed frequencies in one or more categories. In genetics, it is commonly used to test the fit of observed genotype frequencies to those expected under the Hardy-Weinberg Equilibrium.

The formula for calculating the chi-square statistic is:
  • \[ \chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}\]
where:
  • \(O_i\) represents the observed frequency of a genotype.
  • \(E_i\) represents the expected frequency of a genotype under Hardy-Weinberg.
The chi-square value calculated is then compared to a critical value from the chi-square distribution table based on the degrees of freedom and chosen significance level (typically \(\alpha = 0.05\)).

A high chi-square value suggests that the observed frequencies significantly differ from expectations, indicating other evolutionary influences on allele frequencies or genetic drift. A low chi-square value leads to a failure to reject the null hypothesis, suggesting the population is in Hardy-Weinberg Equilibrium.
Allele Frequencies
Allele frequencies measure how common an allele is in a population. It is calculated by dividing the number of copies of an allele by the total number of alleles for a gene in a population. Allele frequencies help us understand the genetic structure of a population and predict future genotypic makeup under certain conditions.

To calculate allele frequencies:
  • Count the total number of individuals and multiply by 2 since each carries two alleles.
  • For allele \(A\), count all \(A/A\) individuals twice and \(A/T\) individuals once. Add these to get the total \(A\) alleles. Then divide by the total number of alleles.
  • Do the same for allele \(T\), considering its presence in \(A/T\) individuals.
In the given population:
  • Total \(A\) alleles count as 101, leading to an \(A\) frequency of \(0.886\).
  • Total \(T\) alleles count as 13, resulting in a \(T\) frequency of \(0.114\).
Studying allele frequencies helps in predicting Hardy-Weinberg genotypes frequencies and determining genetic stability of a population.
Genotype Frequencies
Genotype frequencies refer to the proportions of different genotypes - combinations of alleles - within a population. Calculating these frequencies is vital in understanding the genetic diversity and predicting evolutionary trends.

In Hardy-Weinberg Equilibrium, genotype frequencies are derived from allele frequencies with no evolutionary influences like selection, mutation, or migration. The formulae are:
  • \(p^2\): frequency of homozygous dominant genotypes (e.g., \(A/A\))
  • \(2pq\): frequency of heterozygous genotypes (e.g., \(A/T\))
  • \(q^2\): frequency of homozygous recessive genotypes (e.g., \(T/T\))
where \(p\) and \(q\) are the frequencies of alleles \(A\) and \(T\) respectively.

In the given population:
  • The expected frequency of \(A/A\) is \(0.785\).
  • The expected frequency of \(A/T\) is \(0.202\).
  • The expected frequency of \(T/T\) is \(0.013\).
These expected frequencies are then used to calculate the numbers of individuals expected to have these genotypes, facilitating the chi-square analysis necessary for evaluating genetic equilibrium.

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 randomly mating laboratory population of Drosophila, 4 percent of the flies have black bodies (encoded by the autosomal recessive \(b\) ), and 96 percent have brown bodies (the wild type, encoded by \(B\) ). If this population is assumed to be in Hardy-Weinberg equilibrium, what are the allele frequencies of \(B\) and \(b\) and the genotypic frequencies of \(B / B\) and \(B / b ?\)

Color blindness in humans is caused by an X-linked recessive allele. Ten percent of the males of a large and randomly mating population are color- blind. A representative group of 1000 people from this population migrates to a South Pacific island, where there are already 1000 inhabitants and where 30 percent of the males are colorblind. Assuming that Hardy-Weinberg equilibrium applies throughout (in the two original populations before the migration and in the mixed population immediately after the migration), what fraction of males and females can be expected to be color-blind in the generation immediately after the arrival of the migrants?

If the recessive allele for an X-linked recessive disease in humans has a frequency of 0.02 in the population, what proportion of individuals in the population will have the disease? Assume that the population is 50: 50 male : female.

A group of 50 men and 50 women establish a colony on a remote island. After 50 generations of random mating, how frequent would a recessive trait be if it were at a frequency of \(1 / 500\) back on the mainland? The population remains the same size over the 50 generations, and the trait has no effect on fitness.

In a population of mice, there are two alleles of the \(A\) locus \(\left(A_{1} \text { and } A_{2}\right) .\) Tests showed that, in this population, there are 384 mice of genotype \(A_{1} / A_{1}, 210\) of \(A_{1} / A_{2}\) and 260 of \(A_{2} / A_{2}\). What are the frequencies of the two alleles in the population?
See all solutions

Recommended explanations on Biology Textbooks

Biological Organisms

Read Explanation

Genetic Information

Read Explanation

Cellular Energetics

Read Explanation

Astrobiological Science

Read Explanation

Responding to Change

Read Explanation

Control of Gene Expression

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.