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

Next, use the Hardy-Weinberg equation (p2 + 2pq + q2 = 1) to calculate the day seven expected frequencies of genotypes CGCG, CGCY, and CYCY for a population in Hardy-Weinberg equilibrium.

Short Answer

Expert verified

The expected frequencies of the genotypes \({C^G}{C^G}\),\({C^G}{C^Y}\), and \({C^Y}{C^Y}\)calculated from the day seven data are 0.23, 0.50, and 0.26, respectively.

Step by step solution

01

Hardy-Weinberg equilibrium

The mathematical equation for Hardy-Weinberg equilibrium is-

\({p^2} + 2pq + {q^2} = 1\), where \({p^2}\) is the genotype frequency for the homozygous dominant character, \({q^2}\) is the genotype frequency for the homozygous recessive character, and \(2pq\) is the genotype frequency for heterozygous individuals.

And \(p + q = 1\) , where \(p\)is the frequency of the dominant allele, and\(q\)is the frequency of the recessive allele.

A Hardy-Weinberg population is large where random mating occurs and is devoid of natural selection, mutation, and migration influences. In short, the population is not undergoing evolution.

02

Expected genotype frequency

Expected genotype frequency for homozygous dominant genotype and homozygous recessive genotype are the squares of dominant allele frequency and recessive allele frequency, respectively.

For heterozygous dominant genotype, the expected genotype frequency is two times the frequency of dominant and recessive alleles. It provides a measure of the number of a particular genotype that is predicted after a cross.

03

Expected frequencies of genotypes CGCG, CGCY, and CYCY

From the observed frequencies from Day 7, it is given that:

Number of homozygous dominant or green seedlings (\({C^G}{C^G}\))= 49

Total number of seedlings= 216

The genotypic frequency of \({C^G}{C^G}\) (\({p^2}\)) is:

\({p^2} = \frac{{Number{\rm{ }}of{\rm{ }}homozygous{\rm{ }}dominant{\rm{ }}or{\rm{ }}green{\rm{ }}seedlings{\rm{ }}\left( {{C^G}{C^G}} \right)}}{{Total{\rm{ }}number{\rm{ }}of{\rm{ }}seedlings}}\)


The allele frequency for\({C^G}\)allele (\(p\)) is:

\(p = \sqrt {\frac{{Number{\rm{ }}of{\rm{ }}homozygous{\rm{ }}dominant{\rm{ }}or{\rm{ }}green{\rm{ }}seedlings{\rm{ }}\left( {{C^G}{C^G}} \right)}}{{Total{\rm{ }}number{\rm{ }}of{\rm{ }}seedlings}}} \)

\(\begin{aligned}{c}p &= \sqrt {\frac{{49}}{{216}}} \\ &= 0.476\\ \simeq 0.48\end{aligned}\)

The expected genotypic frequencyof \({C^G}{C^G}\) (\({p^2}\)) is:

\(\begin{aligned}{l}{p^2} &= 0.48 \times 0.48\\{p^2} &= 0.23\end{aligned}\)

From the observed frequencies from Day 7, it is given that:

Number of homozygous dominant or yellow seedlings (\({C^Y}{C^Y}\))= 56

Total number of seedlings= 216

The genotypic frequency of \({C^Y}{C^Y}\) (\({q^2}\)) is:

\({q^2} = \frac{{Number{\rm{ }}of{\rm{ }}homozygous{\rm{ }}recessive{\rm{ }}or{\rm{ }}yellow{\rm{ }}seedlings{\rm{ }}\left( {{C^Y}{C^Y}} \right)}}{{Total{\rm{ }}number{\rm{ }}of{\rm{ }}seedlings}}\)

The allele frequency for\({C^Y}\)allele (\(q\)) is:

\(q = \sqrt {\frac{{Number{\rm{ }}of{\rm{ }}homozygous{\rm{ }}recessive{\rm{ }}or{\rm{ }}yellow{\rm{ }}seedlings{\rm{ }}\left( {{C^Y}{C^Y}} \right)}}{{Total{\rm{ }}number{\rm{ }}of{\rm{ }}seedlings}}} \)

\(\begin{aligned}{c}q &= \sqrt {\frac{{56}}{{216}}} \\ &= 0.509\\ \simeq 0.51\end{aligned}\)

The expected genotypic frequencyof \({C^Y}{C^Y}\) (\({q^2}\)) is:

\(\begin{aligned}{l}{q^2} &= 0.51 \times 0.51\\{q^2} &= 0.26\end{aligned}\)

We have:

\(p = 0.48\)and\(q = 0.51\)

The expected frequency for\({C^G}{C^Y}\)(\(2pq\)) is:

\(\begin{aligned}{l}2pq &= 2 \times 0.48 \times 0.51\\2pq &= 0.489\\2pq &= 0.49\end{aligned}\)

\(2pq = 0.5\)

Thus, the expected frequencies of the genotypes such as\({C^G}{C^G}\),\({C^G}{C^Y}\), and\({C^Y}{C^Y}\)are 0.23, 0.50, and 0.26, respectively.

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

Study anywhere. Anytime. Across all devices.

Sign-up for free