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X Linked Hardy Weinberg Equilibrium Problem

X Linked Hardy Weinberg Equilibrium Problem



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In a given population under Hardy Weinberg equilibrium, 40.0% of men have hemophilia. What is the probability that a random man and random woman will have a daughter with haemophilia?

I think the answer is 16%, but the answer given is 9.6%. According to Hardy-Weinberg principle, p2 + 2pq + q2 = 1. In order to inherit the disease, the mother must either be a carrier of have the disease, which occurs with probability 1-q2 = 0.72

Therefore, the odds of having a child with the disease is (0.84)(0.4). Since it asks for the probability of a girl, the total must be divided by two, so the answer is 0.168

Where am I wrong?


I make it 8%. Here is my reasoning.

The gene is X-linked.

40% mutant males, so freq(mutant allele) = p = 0.4, and freq(wt allele) = q = 0.6

To get a mutant female we have to have a mutant male parent, probability = 0.4 Of these matings one half will produce a female offspring so 0.4*0.5 = 0.2

i.e. 20% of matings derive from a mutant male and produce a female offspring.

Now look at the female mate:

probability(mutant) = p2 = 0.16

probability(carrier) = 2pq = 0.48

probability(wt) = q2 = 0.36

so our 20% of matings that have the potential to produce a mutant female offspring partition as:

mating with a mutant female: 0.2 x 0.16 = 0.032 mutant female offspring

mating with a carrier female: 0.2 x 0.48 = 0.096 of which:

0.048 mutant female offspring

0.048 carrier female offspring

mating with wt female: 0.2 x 0.36 = 0.072 carrier female offspring

(sanity check - 0.032 + 0.096 + 0.072 = 0.2)

Thus the probability of random mating producing a mutant female is 0.032 + 0.048 = 0.08 (8%)

Incidentally there is another way of thinking about this. Note that the H-W frequency of mutant females in the population is 16%. One of the assumptions of H-W is random mating. So the probability of a random mating producing a mutant female = p(female) * p(mutant if female) = 0.5 x 0.16 = 0.08

So - where have I gone wrong?


In a given population, 40% of men have hemophilia - an X-linked recessive disorder. What are the odds that a random woman and a random man from that population will have a daughter with hemophilia? Hemophilia is X-linked and recessive, so the frequency of males having the disease = q. So, q = 0.40. To determine the frequency of the dominant allele in the population, use… p + q = 1 p + 0.4 = 1, p = 0.6 Use these allele frequencies to calculate the genotype frequencies in the females using the Hardy-Weinberg equation: P2 + 2pq + q2 = 1 0.36 + 0.48 + 0.16 = 1 Now use these frequencies in two separate Punnett squares 1) All of the offspring from a homozygous recessive woman and a hemophiliac man will have hemophilia. Thus, (0.16)(0.4) = 0.064. Half of these offspring will be daughters, so 0.064/2 = 0.032

2) Half of the offspring of a cross between a heterozygous woman and a hemophiliac man will have hemophilia. Thus, (0.48)(0.4)(.5) = 0.096. Half of these offspring will be daughters, so 0.096/2 = 0.048.

3) Add the two possibilities together, so 0.032 + 0.048 = 0.08, or 8%


Frequency of allele for Haemophilia (q) = 0.4
Frequency of normal allele (p) = 0.6

Cross between Heterozygote female and hemophiliac male by punnett square:
Probability of hemophiliac daughter = 0.5
P(hemophiliac male) = 0.4
P(heterozygote female) = 2pq = 2 × 0.4 × 0.6 = 0.48
P(hetero female+hemophiliac male+hemophiliac daughter) = 0.5 × 0.4 × 0.48 = 0.096

Hence 9.6%


In a given population, 40% of men have hemophilia - an X-linked recessive disorder. What are the odds that a random woman and a random man from that population will have a daughter with hemophilia? Hemophilia is X-linked and recessive, so the frequency of males having the disease = q. So, q = 0.40. To determine the frequency of the dominant allele in the population, use… p + q = 1 p + 0.4 = 1, p = 0.6 Use these allele frequencies to calculate the genotype frequencies in the females using the Hardy-Weinberg equation: P2 + 2pq + q2 = 1 0.36 + 0.48 + 0.16 = 1 Now use these frequencies in two separate Punnett squares 1) All of the offspring from a homozygous recessive woman and a hemophiliac man will have hemophilia. Thus, (0.16)(0.4) = 0.064. Half of these offspring will be daughters, so 0.064/2 = 0.032

2) Half of the offspring of a cross between a heterozygous woman and a hemophiliac man will have hemophilia. Thus, (0.48)(0.4)(.5) = 0.096. Half of these offspring will be daughters, so 0.096/2 = 0.048.

3) Add the two possibilities together, so 0.032 + 0.048 = 0.08, or 8%


Teaching Hardy-Weinberg Equilibrium using Population-Level Punnett Squares: Facilitating Calculation for Students with Math Anxiety

Hardy-Weinberg (HW) equilibrium and its accompanying equations are widely taught in introductory biology courses, but high math anxiety and low math proficiency have been suggested as two barriers to student success. Population-level Punnett squares have been presented as a potential tool for HW equilibrium, but actual data from classrooms have not yet validated their use. We used a quasi-experimental design to test the effectiveness of Punnett squares over 2 days of instruction in an introductory biology course. After 1 day of instruction, students who used Punnett squares outperformed those who learned the equations. After learning both methods, high math anxiety was predictive of Punnett square use, but only for students who learned equations first. Using Punnett squares also predicted increased calculation proficiency for high-anxiety students. Thus, teaching population Punnett squares as a calculation aid is likely to trigger less math anxiety and help level the playing field for students with high math anxiety. Learning Punnett squares before the equations was predictive of correct derivation of equations for a three-allele system. Thus, regardless of math anxiety, using Punnett squares before learning the equations seems to increase student understanding of equation derivation, enabling them to derive more complex equations on their own.


X Linked Hardy Weinberg Equilibrium Problem - Biology

Approach to Hardy-Weinberg Proportions at sex-linked loci

Consider a sex-linked locus in a species where females are XX (homogametic) and males are XY (heterogametic). Suppose allele frequencies in females and males are initially unequal. (1) In males, he frequency of the allele f( a ) in females of generation nnecessarily determines f( a ) in males in generation n+1. The male f( a ) therefore "chases" the female f( a ) in the preceding generation until they reach approximate equality. (2) In females, because each female in generation n+1 receives an X chromosome for each parent in generation n, the female f( a ) is the mean of the male and female f(a) in the preceding generation. In this example, note that frequencies are within 1% of each other in the seventh generation, even when the initial frequencies are completely divergent.

Note as well that, where each sex makes up half the population, and females contribute two X chromosomes and males one X each, the initial f(a) for a sex-linked locus is 0.6667 and remains constant.

HOMEWORK : Calculate combined genotype frequencies in males & females over the first seven generations.


Biology Final

Scenario I. A particular species of mouse feeds on the seeds of a single species of cherry tree. When the mice eat a seed, they digest it completely. The mice choose seeds of intermediate and large sizes, leaving the very small seeds of the cherry tree uneaten.

Scenario IIa. Small island A contains three separate populations of a single species of cherry tree. The seed size varies between trees. That is, some trees produce seeds that are all in the small size ranges, others produce seeds all in the middle size ranges, and others produce seeds in the large size ranges. A small population of mice is introduced to the island. The mice eat cherries and are the only predators on the cherry trees. When the mice eat a cherry, they completely digest it and the pit or seed inside it. The mice choose medium and large seeds and leave the smallest seeds uneaten.

Scenario IIb. Would your answer for Scenario IIa change given the following information? Explain. As you continue to study the populations of trees, you note that the viability of the seeds varies with size such that the viability of the small seeds is less than that of the middle-sized seeds, which is less than that of the largest seeds.

Scenario IIIa. Small island B contains three separate populations of a single species of cherry tree. Unlike the species on island A, in this species seed size varies within trees. That is, each tree produces seeds that range in size from large to small. A small population of mice is introduced to the island. These are the only predators on the cherry trees. When the mice eat a cherry, they digest it and the pit or seed completely. The mice choose medium and large seeds and leave the smallest seeds uneaten.


Hardy Weinberg Problem Set Mice Answer Key / Solved Hardy Weinberg Problem Set P 2pq Q 1 P 9 Chegg Com

Hardy weinberg equation pogil answer key (1). Collection of hardy weinberg practice problems worksheet with answers. Key for all 4 problems. Watch the short film the making of the fittest: Which of these conditions are never truly met? New p=1/3 and new q=2/3. You can reuse this answer creative commons license. Aa = 0.25, aa = 0.50, and aa = 0.25.

I know that this is 0.2 for the s allele (q in the hardy weinberg equation) and 0.8 for the a allele (p in the hardy weinberg equation). The key to this problem is recalculating p. The mice shown below were collected in a trap. Solving hardy weinberg problems and answers.

28 Evolution Ideas In 2021 Evolution Biology Ap Biology from i.pinimg.com Hardy weinberg problem set key. Data for 1612 individuals are given below: Try setting up a punnett square type arrangement using the 3 genotypes. Below is a data set on wing coloration in the scarlet tiger moth (panaxia dominula). P2 + 2pq + q2 = 1 p + q = 1 p = frequency of the dominant allele in the population q = frequency of the recessive allele in the population. I know that this is 0.2 for the s allele (q in the hardy weinberg equation) and 0.8 for the a allele (p in the hardy weinberg equation). You can reuse this answer creative commons license. Therefore, the number of heterozygous individuals 3. Key ap biology biology 115 at austin college, sherman texas 1. Quizlet is the easiest way to study, practise and master what which of the answer choices reflects a difference in fitness among individuals in a population? He starts with a brief description of a gene pool and shows you how the formula is.

Key ap biology biology 115 at austin college, sherman texas 1.

Whether tackling a problem set or studying for a test, quizlet study sets help you retain key facts about hardy weinberg equilibrium. You can reuse this answer creative commons license. Hardy weinberg problem set key. These would you expect to have poor vision and how many with good vision? The mice shown below were collected in a trap. Below is a data set on wing coloration in the scarlet tiger moth (panaxia dominula). Some basics and approaches to solving problems. He starts with a brief description of a gene pool and shows you how the formula is. Hardy weinberg problem set key. Mice collected from the sonoran desert have two phenotypes, dark (d) and light (d).

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A 2011 study of 93 house mice (mus musculus) from a single barn in texas focused on a single locus with 2 alleles, a & a1.

The genotype frequencies for this locus were found to be Mice collected from the sonoran desert have two phenotypes, dark (d) and light (d). Below is a data set on wing coloration in the scarlet tiger moth (panaxia dominula). Whether tackling a problem set or studying for a test, quizlet study sets help you retain key facts about hardy weinberg equilibrium. Which of these conditions are never truly met? Hardy weinberg equation pogil answer key (1). You can reuse this answer creative commons license. Hardy weinberg problem set key. Hardy weinberg equation pogil answer key (1). The mice shown below were collected in a trap.

This is a little harder to figure out. These would you expect to have poor vision and how many with good vision? One gene pair controls flower height. Answer key hardy weinberg problem set p2 + 2pq + q2 = 1 and p + q = 1 p = frequency of the dominant allele in the population q = frequency of the key ap biology biology 115 at austin college, sherman texas 1. Solving hardy weinberg problems and answers. A 2011 study of 93 house mice (mus musculus) from a single barn in texas focused on a single locus with 2 alleles, a & a1. Hardy weinberg equation pogil answer key (1).

Hardy Weinberg Problem Set Mice Answer Key How To Do Allele Frequency Problems from i2.wp.com Solving hardy weinberg problems and answers. Mice collected from the sonoran desert have two phenotypes, dark (d) and light (d). White coloring is caused by the recessive genotype, aa. The genotype frequencies for this locus were found to be Below is a data set on wing coloration in the scarlet tiger moth (panaxia dominula).

Whether tackling a problem set or studying for a test, quizlet study sets help you retain key facts about hardy weinberg equilibrium.

Solving hardy weinberg problems and answers. The key to this problem is recalculating p. Answer key hardy weinberg problem set p2 + 2pq + q2 = 1 and p + q = 1 p = frequency of the dominant allele in the population q = frequency of the key ap biology biology 115 at austin college, sherman texas 1. Try setting up a punnett square type arrangement using the 3 genotypes. The best answers are voted up and rise to the top. Below is a data set on wing coloration in the scarlet tiger moth (panaxia dominula). A population of ladybird beetles from north carolina was genotyped at a since we had not talked about drift and founder effects prior to the problem set any reasonable answer was given credit. Therefore, the number of heterozygous individuals 3. P2 + 2pq + q2 = 1 p + q = 1 p = frequency of the dominant allele in the population q = frequency of the recessive allele in the population. Which of these conditions are never truly met?

Key ap biology biology 115 at austin college, sherman texas 1.

Quizlet is the easiest way to study, practise and master what which of the answer choices reflects a difference in fitness among individuals in a population?

He starts with a brief description of a gene pool and shows you how the formula is.

Key ap biology biology 115 at austin college, sherman texas 1.

Our most recent study sets focusing on hardy weinberg problems will help you get ahead by allowing you to.

Try setting up a punnett square type arrangement using the 3 genotypes.

Some basics and approaches to solving problems.

Data for 1612 individuals are given below:

This is a little harder to figure out.

One gene pair controls flower height.

Data for 1612 individuals are given below:

Hardy weinberg problem set key.

The genotype frequencies for this locus were found to be

The mice shown below were collected in a trap.

The genotype frequencies for this locus were found to be

Collection of hardy weinberg practice problems worksheet with answers.

Whether tackling a problem set or studying for a test, quizlet study sets help you retain key facts about hardy weinberg equilibrium.

The genotype frequencies for this locus were found to be

Hardy weinberg problem set key.

This is a little harder to figure out.

Whether tackling a problem set or studying for a test, quizlet study sets help you retain key facts about hardy weinberg equilibrium.


Hardy Weinberg Problem Set Answer Key

Hardy Weinberg Problem Set Answer Key. The best answers are voted up and rise to the top. Follow up with other practice problems using human hardy weinberg problem set. I know that this is 0.2 for the s allele (q in the hardy weinberg equation) and 0.8 for the a allele (p in the hardy weinberg equation). This set of 10 questions gives students just enough information to.

Hw equilibrium practice problems key.docx. Input it if you want to receive answer. P2 + 2pq + q2 = 1 p + q = 1 p = frequency of the dominant allele in the population q = frequency of the recessive. Genotypes cgcg, cgcy, and cycy for a population in. Therefore, the number of heterozygous individuals 3.

Assumptions Of Hardy Weinberg Equilibrium Poster By Amoebasisters In 2021 Teaching Biology Biology Teacher Biology Activity from i.pinimg.com The frequency of the aa genotype (q2). Key for all 4 problems. 36%, as given in the problem itself. Is evolution occurring in a soybean population? (a) calculate the percentage of heterozygous individuals in the population. Chemistry 333 protein structure and function fall 2001. Follow up with other practice problems using human hardy weinberg problem set. If the frequency of the aa genotype is 34. This is a little harder to figure out. Hardy weinberg problem set the hardy weinberg equation worksheet answers Hardy weinberg equation pogil answer key (1). Using that 36%, calculate the following:

Use the hardy weinberg equation to determine the allele frequences of traits in a dragon population.

Hardy weinberg problem set the hardy weinberg equation worksheet answers You can reuse this answer creative commons license. 36%, as given in the problem itself. Input it if you want to receive answer. The frequency of the aa genotype (q2). Using that 36%, calculate the following: You have sampled a population in which you know that the percentage of the homozygous recessive genotype (aa) is 36%. Αβγ is an autosomal recessive disorder of man. Genotypes cgcg, cgcy, and cycy for a population in. I know that this is 0.2 for the s allele (q in the hardy weinberg equation) and 0.8 for the a allele (p in the hardy weinberg equation). View answers to practice hardy weinberg problems.pdf from biol 1b at california state university, chico. To solve this problem, determine the charge of each functional group at each ph. Therefore, the number of heterozygous individuals 3. If the frequency of the h allele is 0.45, what percent of the population is homozygous for h (i.e.

The frequency of the aa genotype (q2). This is a little harder to figure out. I know that this is 0.2 for the s allele (q in the hardy weinberg equation) and 0.8 for the a allele (p in the hardy weinberg equation). Try setting up a punnett square type arrangement using the 3 genotypes. Conditions happen to be really good this year for breeding and next year there are 1,245 offspring. You have sampled a population in which you know that the percentage of the homozygous recessive genotype (aa) is 36%. You have sampled a population in which you · students can practice using the hardy weinberg equilibrium equation to determine the allele frequencies in a population. This set of 10 questions gives students just enough information to. You have sampled a population in which you know that the percentage of the the frequency of a is equal to p, so the answer is 40%.

Hardy Weinberg Squirrels from www.biologycorner.com Supposing the matings are random, the frequencies of the matings. 36%, as given in the problem itself. Key ap biology biology 115 at austin college, sherman texas 1. P2 + 2pq + q2 = 1 p + q = 1 p = frequency of the dominant allele in the population q = frequency of the recessive. Is evolution occurring in a soybean population? Do not include the % sign in the box. Which of these conditions are never truly met? Key for all 4 problems. Input it if you want to receive answer. Conditions happen to be really good this year for breeding and next year there are 1,245 offspring.

Some practice questions with answers.

This is a little harder to figure out. To solve this problem, determine the charge of each functional group at each ph. A population of ladybird beetles from north carolina was genotyped at a since we had not talked about drift and founder effects prior to the problem set any reasonable answer was given credit. Below is a data set on wing coloration in the scarlet tiger moth (panaxia dominula). I know that this is 0.2 for the s allele (q in the hardy weinberg equation) and 0.8 for the a allele (p in the hardy weinberg equation). The frequency of the aa genotype (q2). This is a classic data set on wing coloration in the scarlet tiger moth (panaxia dominula). Do not include the % sign in the box. Hardy weinberg equation pogil answer key (1). Use the hardy weinberg equation to determine the allele frequences of traits in a dragon population.

Follow up with other practice problems using human hardy weinberg problem set. P2 + 2pq + q2 = 1 p + q = 1 p = frequency of the dominant allele in the population q = frequency of the recessive. Problem set #1 answer key. Genotypes cgcg, cgcy, and cycy for a population in. Hw equilibrium practice problems key.docx. Therefore, the number of heterozygous individuals (aa) is equal to 2 pq which equals 2 × 0.19 × 0.81 = 0.31 or 31%. Conditions happen to be really good this year for breeding and next year there are 1,245 offspring. View answers to practice hardy weinberg problems.pdf from biol 1b at california state university, chico. You can reuse this answer creative commons license. Is evolution occurring in a soybean population?

Allele Frequency Calculation Practice Problems from www.yumpu.com Therefore, the number of heterozygous individuals (aa) is equal to 2 pq which equals 2 × 0.19 × 0.81 = 0.31 or 31%. Try setting up a punnett square type arrangement using the 3 genotypes. If the frequency of the aa genotype is 34. The frequency of the aa genotype (q2). Problem set #1 answer key. Input it if you want to receive answer. Key for all 4 problems.

You have sampled a population in which you know that the percentage of the the frequency of a is equal to p, so the answer is 40%.

Input it if you want to receive answer. You can reuse this answer creative commons license. Hw equilibrium practice problems key.docx. Use the hardy weinberg equation to determine the allele frequences of traits in a dragon population. Supposing the matings are random, the frequencies of the matings. This is a classic data set on wing coloration in the scarlet tiger moth (panaxia dominula). Hh)?round your answer to the nearest tenth of a percent (i.e. Problem set #1 answer key. To solve this problem, determine the charge of each functional group at each ph. Αβγ is an autosomal recessive disorder of man. The cc is most significant because cc is recessive and the disease form (2 alleles needed) b.

Chemistry 333 protein structure and function fall 2001.

Source: d2j6dbq0eux0bg.cloudfront.net

Chemistry 333 protein structure and function fall 2001.

Key ap biology biology 115 at austin college, sherman texas 1.

Conditions happen to be really good this year for breeding and next year there are 1,245 offspring.

If the frequency of the h allele is 0.45, what percent of the population is homozygous for h (i.e.

Genotypes cgcg, cgcy, and cycy for a population in.

Chemistry 333 protein structure and function fall 2001.

Source: ecdn.teacherspayteachers.com

(a) calculate the percentage of heterozygous individuals in the population.

Use the hardy weinberg equation to determine the allele frequences of traits in a dragon population.

Source: d20ohkaloyme4g.cloudfront.net

This is a little harder to figure out.

Therefore, the number of heterozygous individuals 3.

Input it if you want to receive answer.

I know that this is 0.2 for the s allele (q in the hardy weinberg equation) and 0.8 for the a allele (p in the hardy weinberg equation).

To solve this problem, determine the charge of each functional group at each ph.

Source: ecdn.teacherspayteachers.com

You can reuse this answer creative commons license.

Is evolution occurring in a soybean population?

Hardy weinberg equation pogil answer key (1).

Αβγ is an autosomal recessive disorder of man.

View answers to practice hardy weinberg problems.pdf from biol 1b at california state university, chico.

Source: d20ohkaloyme4g.cloudfront.net

You have sampled a population in which you know that the percentage of the the frequency of a is equal to p, so the answer is 40%.

Hw equilibrium practice problems key.docx.

Some practice questions with answers.

Below is a data set on wing coloration in the scarlet tiger moth (panaxia dominula).


Study Notes on Population Genetics

Study of the frequency of genes and genotypes in a mendelian population is known as population genetics. In other words, it is a branch of genetics which deals with the frequency of genes and genotypes in mendelian populations. Before dealing with population genetics, it is essential to define mendelian population, gene frequency and genotype frequency.

There are two important features of mendelian population, viz:

(2) Equal survival of all genotypes.

In case of random mating, each individual of one sex has equal chance of mating with every individual of opposite sex. In other words, there is no restriction on mating of one individual with other individuals. Such inter-mating populations are also known as panmictic populations.

Random mating populations maintain high level of variability and adaptability. Random mating individuals belong to the same species and same gene pool. The gene pool refers to the sum total of genes in a mendelian population.

Gene frequency refers to the proportion of different alleles of a gene in a random mating population. It is also known as genetic frequency. In other words, the proportion of each type of allele at a particular locus in a random mating population is referred to as gene frequency. The composition of a population is described in terms of gene frequencies.

Estimation of gene frequencies in a population consists of three important steps as given below:

First a random sample of individual is drawn from the random mating population under study. The size of sample should be adequate to represent all the individuals of a population.

After sampling, the individuals are grouped into different classes for a gene and their number is counted.

3. Calculation of gene frequency:

Suppose a random sample of 100 individuals was drawn from a random mating population of four ‘O’ clock plant (Mirabilis jalapa). Out of 100 plants, 30 were with red flower, 40 with pink flower and 30 with white flower.

Now, the allele frequency will be worked out as follows:

(a) In four o’clock plant, a cross between red and white flowered strains produces pink flower in Fi and red, pink and white flowered plants in 1 : 2 : 1 ratio in F2 generation. Thus, plants with red colour are homozygous for dominant allele (RR) and individuals with white flower colour are homozygous for recessive allele (rr).

(b) Each heterozygous individual with pink colour will have dominant (R) and recessive (r) alleles in equal number.

1. Number of R alleles in the Sample (30 individuals)

= 2 (No. of red individuals) + No. of pink individuals

2. Proportion of R alleles in the sample

= Number of RR Alleles/2 (Total plants in a sample)

Similarly, the number of r alleles

Therefore, the frequency of RR and rr alleles is 0.50 each.

It refers to the ratio of different genotypes in a mendelian population. Genotypic frequency is also known as zygotic frequency. The estimation of genotypic frequency for a gene in a population also consists of three important steps mentioned above.

Thus, the genotypic frequency of three types of individuals from the above sample will be calculated as ratio of each individual, class or genotypes to the total individuals in a sample. Thus,

1. Frequency of Red (RR) individuals = 30/100 = 0.30

2. Frequency of Pink (Rr) individuals = 40/100 = 0.40

3. Frequency of White (rr) individuals = 30/100 = 0.30

Hardy-Weinberg Law:

Foundation of population genetics was laid by G.H. Hardy, an English mathematician and W. Weinberg, a German physician in 1908. They independently discovered a principle concerned with the frequency of genes (alleles) in a population. Their principle is commonly known as Hardy-Weinberg Law.

The Hardy-Weinberg Law states that:

1. In a random mating population, the frequency of genes and genotypes remains constant generation after generation, if there is no selection, mutation, migration and random genetic drift.

2. They also developed a mathematical relationship to describe the equilibrium between alleles. According to this relationship, the frequencies of three genotypes for a single locus with two alleles (A and a) are in the ratio of P 2 AA : 2PqAa : q aa. where P and q are the frequencies of allele A and ‘a’ respectively. P + q are always equal to 1 or P = q = 0.50.

P + q = 1 or P= 1 -q and q = 1 – P

Effect of Random Mating:

Random mating results in maintaining the equilibrium of gene frequency in a population. For example, if the frequency of allele A is P and that of ‘a’ is q. If we make a cross between AA and aa, it will produces Aa. If individuals with Aa genotype are allowed to mate randomly, the gene frequency of three genotypes will be in the ratio of P 2 AA + 2PqAa +q 2 aa (Fig. 30.1).

When gene frequencies are in equilibrium, it indicates absence of mutation, selection, migration and genetic drift in a population.

Factors Affecting Gene Frequency:

Hardy-Weinberg principle is based on three main assumptions, viz:

(2) Equal survival of all genotypes, and

(3) Absence of evolutionary forces like selection, mutation, migration and random genetic drift. Non fulfillment of these assumptions will lead to alteration in gene and genotype frequencies in a population.

However, the last assumption is seldom fulfilled. Mutation, migration and genetic drift change gene frequencies in a population. These factors are also known as forces of evolution because they play a key role in natural evolution.

These are briefly discussed below:

Selection refers to a process which favours the survival and reproduction of some individuals in a population. The process of evolution in nature in which the fittest individuals survive and restore wiped out is known as natural selection. Natural selection favours those characters which are advantageous for survival.

The selection by human efforts is known as artificial selection. Such selection favours those plant characters which are useful for mankind like productivity. Before discussing the effect of various types of selection, it is necessary to give brief account of fitness and selection coefficient.

The relative reproductive success of different genotypes of a population in the same environment under natural selection is known as fitness or selective value or adaptive value or selective advantage. It is denoted by W. If the value of W is unity (W = 1), there is 100 per cent survival and if this value is 0 (W = 0), the genotype is completely lethal.

Survival depends on two main factors, viz:

(i) The number of seeds produced by each genotype, and

(ii) The proportion of seeds of each genotype which reaches maturity and produces offspring.

The reproductive rate of different genotypes is estimated in relation to the most fit genotype. If the reproductive rate of the most fit genotype is X] and that of other genotypes is X2 and X3 then Fitness W = X1/X1, X2/X1, X3/X1, etc. The value of W varies between 0 and 1.

Selection Coefficient:

Selection coefficient is a measure of the rate of elimination of different genotypes from a population under natural selection in a particular environment. In other words, it is the measure of the rate of reduction in the adaptive value of a genotype in relation to standard or the most favoured genotype. It is also known as selective disadvantage and is represented by S.

The greater the value of selection coefficient, lesser the survival rate and lesser the value of S greater the survival value. The value of S varies between 0 and 1. If S = 1 there is no survival at all, if S = 0 there is 100 per cent survival.

There is a close relationship between fitness (W) and selection coefficient (S) as given below:

W = 1 – S and S = 1 – W. Thus, selection coefficient is estimated with the help of fitness value or for the estimation of selection coefficient first the value of fitness (W) is estimated. Selection coefficient differs from selection differential in three ways (Table 30.1).

Thus, selection coefficient measures the rate of elimination of different genotypes from a population under natural selection, whereas the selection differential is a measure of difference between the mean phenotypic value of selected plants and mean phenotypic value of parental population under human selection.

Selection may operate at any stage of life (gametic or zygotic) cycle of an individual. Sometimes, selection acts at gametic stage which is referred to as gametic selection. Such selection acts mostly in haploid organisms and in some higher organisms. The tendency of higher organisms to exhibit differential survival rate of gametes is termed as segregation distortion or meiotic drive.

Meiotic drive is generally restricted to either male or female sex in a species. The zygotic selection operates generally in higher organisms. When certain genotypes are favoured by selection, the Hardy-Weinberg equilibrium will be disturbed. In such situation, frequency of some alleles in the population will increase while those of others will decrease.

Zygotic selection may act in three ways, viz:

(i) Against dominant phenotypes,

(ii) Against recessive phenotypes, and

(iii) In favour of heterozygotes.

(i) Selection against Dominant Phenotypes:

When selection acts against dominant phenotypes, it will eliminate both AA and Aa individuals from a population and favour only recessive phenotypes (aa). The elimination process will continue till the entire population is converted into homozygous recessive (aa) phenotypes.

Such selection, leads to fixation of recessive genes and elimination of dominant genes in a population. Since the phenotypes of both homozygous dominant (AA) and heterozygous dominant (Aa) are same, the allele A cannot be protected from elimination even in the heterozygous condition. In such situation, the value of S is 1 for AA and Aa genotypes.

(ii) Selection against Recessive Phenotypes:

Such type of selection leads to elimination of homozygous recessive phenotypes (aa) from a population. Under such type of selection, the value of coefficient of selection (S) is 1 for aa phenotypes. Such selection will lead to increase of AA and Aa genotypes in a population. However, Aa genotypes will continuously produce aa phenotypes due to segregation.

(iii) Selection in Favour of Heterozygotes:

Such type of selection leads to elimination of both dominant and recessive homozygotes (AA and aa). The value of S in such situation is 1 for AA and aa genotypes.

The excess of heterozygotes in a population is an indication of selection in favour of heterozygotes or against both the homozygotes the frequency of homozygotes decreases sharply and the population is dominated by heterozygotes. Such heterozygotes are available in Oenothera.

Genetic Polymorphism:

The regular occurrence of several phenotypes in a genetic population is known as genetic polymorphism. The genetic polymorphism is usually maintained due to superiority of heterozygotes over both homozygotes. When polymorphism is maintained as a result of heterozygote advantage, it is known as balanced polymorphism.

Sometimes it is difficult to identify the polymorphic allelic forms by visual observations. The best way of detecting the polymorphic alleles is the isozyme studies or gel electrophoretic studies. It has been reported that two third of the loci in a population exhibit polymorphism.

Genetic polymorphism increases the adaptive value or buffering capacity of a population by providing increased diversity of genotypes in a population. Thus, genetic polymorphism enhances the adaptability of a population, because heterozygotes are more adaptable than homozygotes.

Mutation refers to a sudden heritable change in the features of an organism. Mutations differ from segregants in terms of their extremely low frequency. Gene mutations are ultimate sources of new alleles and thus of genetic variability.

The new mutation which we observe today would have originated long ago. Mutations lead to alteration of gene frequencies in a population. Alleles change from one form to another by way of mutation. Mutations may occur in both forward and reverse directions, but the frequency of forward mutations is much higher than reverse mutations.

When there is mutation in both the directions the equilibrium condition can be expressed as follows:

The equilibrium is attained very slowly.

Joint Effects of Mutation and Selection:

The rate of change in gene frequency will increase, if mutation and selection are in the same direction. However, if they are in opposite direction which is the usual case, a stable equilibrium may be observed. If a dominant allele arises by mutation at the rate u per generation and is opposed by selection at the rate S, the equilibrium frequency of mutant q will be as follows:

If s equals the selection pressure against the heterozygote and u is the mutation rate from A —>a, then equilibrium value for harmful recessive would be:

Gene flow or migration can also change frequencies of alleles in populations. Migration includes both immigration (in coming) and emigration (outgoing) of alleles in a population. Mass immigration and emigration have tremendous potential in changing allelic frequencies in populations.

Migration generally refers to the movement of individuals into a population from a different populations. Migration may introduce new alleles into the population. These new alleles after mating with the individuals of original population may alter gene and genotype frequencies in a population.

The rate of change in gene frequency, through migration depends on the number of migrants. If the number of migrants is high, the rate of change will be rapid and vice versa. Emigration of some individuals from a population results in decrease in the frequency of alleles migrated to another population.

Random drift or genetic drift refers to random change in gene frequency due to sampling error. Random drift is generally more in case of small sample size. Large sample size provides true representative value of a population or value which is nearer to the population mean.

Therefore, sample size should be adequate to avoid sampling error. Three forces of evolution viz., selection, mutation and migration alter gene and genotype frequency in a particular direction and are called as directional factors. However random genetic drift is a non-directional factor because it does not change the gene frequency in a particular direction.

The direction of change in the gene frequency may differ from generation to generation. In one generation, the change of gene frequency may be in one direction, which may change to opposite direction in the next generation.

Sometimes a new population is established by a single or few individuals in the main population. Such individuals are referred to as founders and effect of such individuals on the gene frequency of a population is known as founder effect. Founder effect is an important factor which sometimes results in the formation of new species.

Significance of Population Genetics:

1. Knowledge of gene and genotype frequency in a population is useful for a plant breeder in the assessment of competitive ability of various genotypes in varietal mixtures. Such studies help in identification of genotypes with high adaptive value.

If such studies are conducted over multiplications, the varietal flexibility or stability can also be assessed in varietal blends. Hardy-Weinberg Law operates in random mating or panmictic species.

2. Study of gene frequency in a population also reveals significance of various factors in natural evolution. In cross pollinated crops, development of composite and synthetic varieties is based on Hardy-Weinberg principle.


Abstract

X-linked red–green color blindness is the most widespread form of vision impairment. The study aimed to determine the prevalence and gene frequencies of red–green color vision impairments among children of six different human populations of Jammu province. A total of 1028 healthy subjects (6–15 years of age) were selected from five Muslim populations and the color vision impairments were determined using the Ishihara's test of color deficiency. The gene frequency was calculated using Hardy–Weinberg equilibrium method. The prevalence of color vision deficiency (CVD) ranged from 5.26% to 11.36% among males and 0.00%–3.03% among females of six different populations. The gender based differences in the frequency of CVD was found to be statistically significant (p < 0.0001), with a higher prevalence among male (7.52%) as compared to female (0.83%) children. We observed high frequency of deutan as compared to protan defects. The incidences of deuteranomaly (5.68%) and deuteranopia (2.27%) were higher among male children of Syed population while the frequencies of protanomaly (1.94%), protanopia (1.28%) and achromacy (2.27%) were the highest among male subjects of Khan, Malik and Syed populations, respectively. The allele and genotype frequencies showed cogent differences among six populations. The population based assessment of CVDs help patients to follow adaptive strategies that could minimize the risks of the disease.


When the Hardy-Weinberg Law Fails

To see what forces lead to evolutionary change, we must examine the circumstances in which the Hardy-Weinberg law may fail to apply. There are five:

  1. mutation
  2. gene flow
  3. genetic drift
  4. nonrandom mating
  5. natural selection

Mutation

The frequency of gene B and its allele b will not remain in Hardy-Weinberg equilibrium if the rate of mutation of B -> b (or vice versa) changes. By itself, this type of mutation probably plays only a minor role in evolution the rates are simply too low. However, gene (and whole genome) duplication - a form of mutation - probably has played a major role in evolution. In any case, evolution absolutely depends on mutations because this is the only way that new alleles are created. After being shuffled in various combinations with the rest of the gene pool, these provide the raw material on which natural selection can act.

Gene Flow

Many species are made up of local populations whose members tend to breed within the group. Each local population can develop a gene pool distinct from that of other local populations. However, members of one population may breed with occasional immigrants from an adjacent population of the same species. This can introduce new genes or alter existing gene frequencies in the residents.

In many plants and some animals, gene flow can occur not only between subpopulations of the same species but also between different (but still related) species. This is called hybridization. If the hybrids later breed with one of the parental types, new genes are passed into the gene pool of that parent population. This process is called introgression. It is simply gene flow between species rather than within them.

Comparison of the genomes of contemporary humans with the genome recovered from Neanderthal remains shows that from 1&ndash3% of our genes were acquired by introgression following mating between members of the two populations tens of thousands of years ago.

Whether within a species or between species, gene flow increases the variability of the gene pool.

Genetic Drift

As we have seen, interbreeding often is limited to the members of local populations. If the population is small, Hardy-Weinberg may be violated. Chance alone may eliminate certain members out of proportion to their numbers in the population. In such cases, the frequency of an allele may begin to drift toward higher or lower values. Ultimately, the allele may represent 100% of the gene pool or, just as likely, disappear from it.

Drift produces evolutionary change, but there is no guarantee that the new population will be more fit than the original one. Evolution by drift is aimless, not adaptive.

Nonrandom Mating

One of the cornerstones of the Hardy-Weinberg equilibrium is that mating in the population must be random. If individuals (usually females) are choosy in their selection of mates, the gene frequencies may become altered. Darwin called this sexual selection.

Nonrandom mating seems to be quite common. Breeding territories, courtship displays, "pecking orders" can all lead to it. In each case certain individuals do not get to make their proportionate contribution to the next generation.

Assortative mating

Humans seldom mate at random preferring phenotypes like themselves (e.g., size, age, ethnicity). This is called assortative mating. Marriage between close relatives is a special case of assortative mating. The closer the kinship, the more alleles shared and the greater the degree of inbreeding. Inbreeding can alter the gene pool. This is because it predisposes to homozygosity. Potentially harmful recessive alleles &mdash invisible in the parents &mdash become exposed to the forces of natural selection in the children.

Figure 18.6.1: Assortative mating. (Drawing by Koren © 1977 The New Yorker Magazine, Inc.)

It turns out that many species - plants as well as animals - have mechanisms be which they avoid inbreeding. Examples:

  • Link to discussion of self-incompatibility in plants.
  • Male mice use olfactory cues to discriminate against close relatives when selecting mates. The preference is learned in infancy - an example of imprinting. The distinguishing odors are controlled by the MHC alleles of the mice and are detected by the vomeronasal organ (VNO).

Natural Selection

If individuals having certain genes are better able to produce mature offspring than those without them, the frequency of those genes will increase. This is simply expressing Darwin's natural selection in terms of alterations in the gene pool. (Darwin knew nothing of genes.) Natural selection results from differential mortality and/or differential fecundity.

Mortality Selection

Certain genotypes are less successful than others in surviving through to the end of their reproductive period. The evolutionary impact of mortality selection can be felt anytime from the formation of a new zygote to the end (if there is one) of the organism's period of fertility. Mortality selection is simply another way of describing Darwin's criteria of fitness: survival.

Fecundity Selection

Certain phenotypes (thus genotypes) may make a disproportionate contribution to the gene pool of the next generation by producing a disproportionate number of young. Such fecundity selection is another way of describing another criterion of fitness described by Darwin: family size. In each of these examples of natural selection, certain phenotypes are better able than others to contribute their genes to the next generation. Thus, by Darwin's standards, they are more fit. The outcome is a gradual change in the gene frequencies in that population.


X Linked Hardy Weinberg Equilibrium Problem - Biology

Allele frequencies (or percentages, if you prefer) in a population will remain in Hardy-Weinberg Equilibrium (HWE) from generation to generation if the following assumptions are met:

  1. Natural selection is not occurring
  2. Migration (Gene Flow) is not occurring
  3. Mutation is not occurring
  4. Genetic Drift is not occurring (drift is less likely in populations of large size)
  5. Mating occurs at random

Although these assumptions are rarely true in the natural world, they allow us to calculate an expected allele frequency. Significant differences between the observed and expected frequencies indicate that "something" (i.e. one or more of the above) is going on, and therefore tell us that "microevolution" is occurring.

Calculating Expected Allele and Genotype frequencies:

In the simplest possible situation we have a single gene with only two alleles. These alleles might be A and a, or A1 and A2. Let's say that A or A1= tall, and a or A2= short. Don't worry for now whether the alleles are dominant and recessive or co-dominant. They will have frequencies p and q in a population. (Because there are only two possibilities and they have to add up to 100%, p + q = 1.)

If we know the allele frequencies, we can predict the genotype frequencies. The expected genotype frequencies of the two alleles are calculated as shown. This ought to look familiar: it's our old friend the Punnet's Square. Allele A or A1 has a frequency of p, and allele a or A2 has a frequency of q. Multiply the allele frequencies to the get the probability of each genotype.

p q A1 p p 2 pq A2 q pq q 2

In other words, p 2 + pq + pq + q 2 = 1, or 100%. The expected frequencies of the genotypes are therefore:

Genotype Expected Frequency
AA or A1A1 p * p = p 2
Aa or A1A2 pq + pq (or 2pq)
aa or A2A2 q * q = q 2

Let's take a look at some graphs of this to make it a little easier to see. For values of p from 0 to 1, in intervals of 0.1, here's what we get:

Red represents the frequency of the AA or A1A1 genotype, green is the Aa or A1A2 genotype, and blue is the aa or A2A2 genotype.

All of the above has to do with the allele and genotype frequencies we would expect to see. Next, let's look at the real world situation so we can compare.

Calculating Observed Allele and Genotype Frequencies:

In a real world population, we can only see phenotypes, not genotypes or alleles. However, in a population of genotypes AA, Aa and aa, the observed frequency of allele A equals the sum of all of the AA genotype plus half of Aa genotype (the A half). The observed frequency of allele a is therefore half of the Aa individuals (the a half) plus all of aa individuals . If you know one value, you can of course just subtract it from 1 (100%) to get the value of the other. In other words, the observed frequency of A = 100%(AA) + 50%(Aa) and a = 50%(Aa) and 100%(aa)

Phenotype Genotype Makeup Frequency
Tall AA 100% A p 2
Tall A a 50% A and 50% a 2pq
Short aa 100% a q 2
or
Phenotype Genotype Makeup Frequency
Tall A1A1 100% A1 p 2
Medium A1 A2 50% A1 and 50% A2 2pq
Short A2 A2 100% A2 q 2

Tip: If the alleles are codominant, each phenotype is distinct (you can distinguish between tall, medium and short) and your job is easier. If the alleles are dominant and recessive, we can't visually tell the homozygous AA from the heterozygous Aa genotypes (both are tall), so it's best to start with the homozygous recessive (short) aa individuals. Count up the aa types and you have the observed q 2 . Then, take the square root of q 2 to get q, and then subtract q from 1 to get p. Square p to get p 2 and multiply 2*p*q to get the observed heterozygous Aa genotype frequency.

If observed and expected genotype frequencies are significantly different , the population is out of HWE.

Genotype Frequencies
AA Aa aa
Observed
Expected
Difference
or
Genotype Frequencies
A1A1 A1A2 A2A2
Observed
Expected
Difference


Question: Why might observed and expected phenotype frequencies differ? Imagine the following scenarios where natural selection is at work. Situation one favors only one tail of the distribution. Perhaps the tallest, perhaps the shortest, but not both. This is directional selection. Now imagine that both tails of the distribution are selected against, and only the middle is favored. This is called stabilizing selection. Next imagine that the extremes on both ends are favored. This is called disruptive selection. In each of these scenarios, what would happen over time?

Before (dotted line) and after (yellow shaded area) directional selection, stabilizing selection, and disruptive selection.

One common misconception is that dominant alleles will rise in frequency and recessive alleles will decline in frequency over time. In reality, allele frequencies will not change from one generation to the next if the assumptions listed above are not violated. A good example of this is human ABO blood type. Type O blood is recessive but it remains the most common.

In the hwe.xlsx Excel Spreadsheet, there are three examples to help make this more concrete.

Example 1 : Allele A is dominant and allele a is recessive. Set the original frequencies of p (allele A) and q (allele a) at 0.6 and 0.4 in Generation 1. These are highlighted in blue. All other numbers are calculated from these two original data points. The frequency of genotype AA is determined by squaring the allele frequency A. The frequency of genotype Aa is determined by multiplying 2 times the frequency of A times the frequency of a. The frequency of aa is determined by squaring a. Try changing p and q to other values, ensuring only that p and q always equal 1. Does it make any difference in the results?

Example 2 : Alleles A 1 and A 2 are co-dominant. In this case, A 1 is at a frequency of 0.25 and A 2 is at a frequency of 0.75.

Example 3 : Alleles A and a are dominant and recessive. Note that allele A is at very low frequency despite being dominant. Does it increase in frequency?

The second sometimes confusing thing about HWE is that after all of the examples above, you may wonder if it is possible for the observed and expected frequencies to differ. Here's an example where they do:

In a population of snails, shell color is coded for by a single gene. The alleles A 1 and A 2 are co-dominant. The genotype A 1 A 1 makes an orange shell. The genotype A 1 A 2 makes a yellow shell. The genotype A 2 A 2 makes a black shell. 1% of the snails are orange, 98% are yellow, and 1% of the snails are black.

Observed frequency of A 1 allele = 0.01 + 0.5(.98) = 0.50 = 50%

p 2 = Expected frequency of A 1 A 1 = 0.25

2pq = Expected frequency of A 1 A 2 = 0.50

q 2 = Expected frequency of A 2 A 2 = 0.25

Phenotype Orange Yellow Black
Genotype A 1 A 1 A 1 A 2 A 2 A 2
Observed 1% 98% 1%
Expected 25% 50% 25%
Difference -24% +48% -24%

There are significantly fewer orange and black snails than expected, and significantly more yellow snails than expected. It appears that this is a case of stabilizing selection, since both tails appear to be strongly selected against.