FACULTY OF APPLIED SCIENCES AND COMPUTING

BABS2213 PRINCIPLES OF GENETICS

PRACTICAL 3

TITLE: HARDY WEINBERG EQUILIBRIUM

NAME: CHOONG MEL JUNE

GROUP: RBS 2 A1

STUDENT ID: 14WAR10521

DATE: 4 JUNE 20140

DEMONSTRATER: DR LOH KHYE ER

Objectives:1. To derive the Hardy-Weinberg Equation.2. To calculate the genotypic and allelic frequencies.3. To discuss the conditions under which a population is in Hardy-Weinberg Equation.4. To determine if a population is in Hardy-Weinberg Equation.5. To design and perform a test of a disruption to Hardy-Weinberg Equation.6. To identify why the Hardy-Weinberg Equation is difficult or impossible to demonstrate

with living organisms.

Hypothesis:Hardy and Weinberg Equilibrium showed that if there was no migration, no mutation and no selection in a large population, the frequencies of any pair of gene allele will tend to remain constant from generation to generation.

Introduction:A population is a group of organisms of the same species that live and breed in the same

area. Alleles are alternate forms of genes. In standard Mendelian genetics, two alleles - one from each parent - control an inherited trait or characteristic (e.g. seed pod color). A gene pool is the collective term representing all the genes, including all the different alleles, found in a population.

The written expression of the DNA code, called the genotype, is normally represented using two letters, each letter representing one allele (e.g. AA). Dominant alleles are represented by capital (uppercase) letters. Recessive alleles are represented by lowercase letters. Genotypes with two dominant alleles (e.g. AA) are referred to as homozygous dominant. Genotypes with two recessive alleles (e.g. aa) are referred to as homozygous recessive. Genotypes with one dominant and one recessive allele (e.g. Aa) are referred to as heterozygous.

The physical expression of the genotype (e.g. brown hair, blue eyes) is called the phenotype. In standard Mendelian genetics, the heterozygous condition (e.g. Aa) retains the homozygous dominant phenotype because the dominant allele masks the phenotype of the recessive allele. An example of this in humans would be a heterozygote for brown eye color. The person would carry both a dominant brown allele “A” and a recessive blue allele “a” yet have brown eyes. The dominant brown allele masks the recessive blue allele.

The population is considered the basic unit of evolution. The small scale changes in the genetic structure of populations from generation to generation are called microevolution. Microevolution is the study of the change in the frequency of an allele in a population from one generation to the next. Mathematicians Hardy and Weinberg explained how an allele could change in a population by first showing how it would not change, the Hardy-Weinberg principle.

Genetic equilibrium is the state in which allele frequencies remain constant. In 1908, English mathematician G. H. Hardy and German physician W. Weinberg independently developed models of population genetics that showed that the process of heredity by itself did not affect the genetic structure of a population. It can be said the Hardy-Weinberg Principle is a

model used to help clarify evolutionary change by determining what happens if no change occurs. When no change occurs and an environment is stable, genetic equilibrium is maintained. The Hardy-Weinberg theorem states that the frequency of alleles in the population will remain the same from generation to generation. Furthermore, the equilibrium genotypic frequencies will be established after one generation of random mating. This theorem is valid only if certain conditions are met:

- Random mating – all individuals in a population have equal opportunity to pass on their alleles to offspring.

- Large populations – genetic drift, or random changes in allele frequency, has less effect on large populations than small populations.

- No mutations - if genes randomly mutate, new alleles may be introduced into the population and allele frequencies will change.

- No migration – individuals cannot move into or out of the population.

- No natural selection - all genotypes must have equal probabilities of reproduction.

Two equations are used to model the Hardy-Weinberg Principle. p + q = 1; where p represents the dominant allele, “A,” and q represents the recessive

allele, “a.” This equation, used to calculate allele frequency, equals 1 or 100% of the population.

p2 + 2pq + q2 = 1; where p2 represents the homozygous dominant genotype, 2pq represents the heterozygous genotype, and q2 represents the homozygous recessive genotype. This equation, used to calculate genotype frequency, equals 1 or 100% of the population.

I. Testing the Hardy-Weinberg Equilibrium

Materials:1. 400 of white colour beads. (Two different colours of beans that are approximately the

same size)2. 400 of black colour beads. (Two different colours of beans that are approximately the

same size)

Methodology:400 of black colour beads and 400 of white colour beads were mixed together into a beaker.

One pair of bead was picked up without looking.

The colour of the pair of bead was observed and recorded down.

The beads were placed back into the beaker and mixed well.

The steps 2, 3 and 4 were repeated 100 times and all the colour of beads were observed and recorded down.

Results:Assumption:Black bead = Dominant allele, AWhite bead = Recessive allele, a

Genotype Observed number, O Average observed number, O Genotype Frequency Expected

number, EG1 G2 G3 G4

AA 17 30 23 18 22 p2= 0.52= 0.25 25Aa 57 50 49 53 52 2pq= 2(0.5)(0.5)= 0.5 50aa 26 20 28 29 26 q2= 0.52= 0.25 25

Total, n 100 100 100 100 100 1 100

Calculation: - p = 400/800 = 0.5 - q = 400/800 = 0.5

- AA + 2 Aa + aa = 1- p2 + 2pq + q2 = 1

Applying chi square formula

X2=(O1−E1 )2

E1+

(O2−E2 )2

E2+

( O3−E3 )2

E3

X2= (22−25 )2

25+ (52−50 )2

50+ (26−25 )2

25

X2=0.36+0.08+0.04

X2=0.48

degree of freedom=2

P value=0.80 0.70

critical value=0.05

Interpretation: Failed to reject the null hypothesis, as the P value is more than the critical value of 0.05 at degree of freedom to 2, the hypothesis that there was no migration, no mutation and no selection in a large population, the frequencies of any pair of gene allele will tend to remain constant from generation to generation. There is 70% ~ 80% of the time that the deviation of the observed number from the expected is due to the chances.

Discussion:The Hardy-Weinberg theorem provides a mathematical formula for calculating the

frequencies of alleles and genotypes in populations. We begin with a population with two alleles at a single gene locus - a dominant allele, A, and a recessive allele, a - then the frequency of the dominant allele is p, and the frequency of the recessive allele is q. Therefore, p + q = 1. The frequency of one allele, p, is known for a population, the frequency of the other allele, q, can be determined by using the formula q = 1 - p. During sexual reproduction, the f frequency of each type of gamete produced is equal to the frequency of the alleles in the population. If the gametes combine at random, the probability of AA in the next generation is p2, and the probability of aa is q2. The heterozygote can be obtained two ways, with either parent providing a dominant allele,

Male Female

pA qa

pA ppAA pqAa

qa pqAa qqaa

so the probability would be 2pq. These genotypic frequencies can be obtained by multiplying p + q by p + q. The general equation then becomes

(p + q)2 = p2 + 2pq + q2 = 1To summarize:

p2 = frequency of AA2pq = frequency of Aaq2 = frequency of aa

When a population meets all of the of the Hardy-Weinberg conditions, it is said to be in Hardy-Weinberg equilibrium. How far a population deviates from Hardy-Weinberg equilibrium can be measured using the “goodness of fit” or chi-squared test (χ2).

Mathematically the chi-squared test is represented: χ2 = Σ [(observed value – expected value)2 / expected value]

Since we have three genotypes, therefore we have 3 minus 1, or 2 degrees of freedom. Degrees of freedom is a complex issue, but we could look at this in simple terms: if we have frequencies for three genotypes that are truly representative of the population then, no matter what we calculate for two of them, the frequency of the third must not be significantly different for what is required to fit the population.

Looking across the distribution table for 2 degrees of freedom, we find our chi-squared value of 0.48 is more than that required to satisfy the hypothesis that the differences in the O and E data did not arise by chance. Since the chi-squared value falls above the 0.05 (5%) significance cut-off, we can conclude that the population does differ significantly from what we would expect for Hardy-Weinberg equilibrium.

Conclusion:The hypothesis is accepted. At df = 2, there is 70% ~ 80% of the time that the deviation of the observed number from the expected is due to the chances. There was no migration, no mutation

and no selection in a large population, the frequencies of any pair of gene allele will tend to remain constant from generation to generation.

II. Genetic Drift

Materials:1. 400 of white colour beads. (Two different colours of beans that are approximately the

same size)2. 400 of black colour beads. (Two different colours of beans that are approximately the

same size)

Methodology:400 of black colour beads and 400 of white colour beads were mixed together into a beaker.

One pair of bead was picked up without looking.

The colour of the pair of bead was observed and recorded down.

One pair of beads with different colours (white and black) or black in colour were placed back into the beaker and mixed well. One pair with white colour beads represented lethal allele were

excluded and taken out from the beaker.

Steps 2, 3 and 4 were repeated 100 times and all the colour of beads were observed and recorded down.

The whole experiment was repeated again without replace the excluded white beads of the first batch.

Results:

Assumption: The homozygous recessive (aa) gene is lethal.n1 = First batch of 100 picks; n2 = Second batch of 100 picks

GenotypeObserved number, O

Expected number, E Expected genotype

frequency during n1

Expected genotype frequency during n2n1 n2 n1 n2

AA 30 36 25 44 p2= 0.52= 0.25 0.442Aa 49 47 50 45 2pq= 2(0.5)(0.5)= 0.5 0.446aa 21 17 25 11 q2= 0.52= 0.25 0.112

Total, n 100 100 100 100 1 1

Random mating Genotype frequency Progeny frequency

AA Aa aaAA x AA 0.33 x 0.33= 0.109 0.109 - -AA x Aa 2(0.33 x 0.67) = 0.442 0.221 0.221 -Aa x Aa 0.67 x 0.67= 0.449 0.112 0.225 0.112Total 1 0.442 0.446 0.112

Calculation:- p = 400/800 = 0.5 - q = 400/800 = 0.5- AA + 2 Aa + aa = 1- p2 + 2pq + q2 = 1- AA = p2 = (0.5)2 = 0.25- Aa = 2pq = 2 (0.5) (0.5) = 0.5- aa = q2 = (0.5)2 = 0.25

Expected genotype frequency during n2:- aa (Lethal allele) = failed to reproduce- AA = 0.25/0.75 = 0.33- Aa = 0.5/0.75 = 0.6

Discussion:Genetic drift describes random fluctuations in the numbers of gene variants in a

population. Genetic drift takes place when the occurrence of variant forms of a gene, called alleles, increases and decreases by chance over time. These variations in the presence of alleles are measured as changes in allele frequencies.

Normally, genetic drift occurs in small populations. If genetic drift continues, involved allele is either lost by a population or left the only allele present in a population at a particular locus. Both possibilities decrease the genetic diversity of a population. Genetic fixation, the loss of all but one possible allele at a gene locus in a population, is a common result of genetic drift in small natural populations. Genetic drift is a significant evolutionary force in situations known as the bottleneck effect and the founder effect.

Besides, in this experiment, the homozygous recessive (aa) gene is lethal. A lethal allele's phenotype, when expressed, causes the death of an organism. Lethal alleles arise when a mutation to a normal allele disrupts the function of an essential gene. Without this essential gene, the organism dies. Lethal alleles can be embryonic or postnatal. Embryonic lethals cause the death of the fetus, and fertility studies are often required in order to positively determine that an embryonic lethal exists. An example of an embryonic lethal is the AY allele in mice. Meanwhile for postnatal lethal alleles cause abnormalities in the progeny that cause them to die early on in development. An example is parrot jaw. Lethal alleles can be dominant or recessive and can be sex linked or autosomal. If the allele is dominant, then both homozygous dominant and heterozygous individuals will die. If it is a recessive allele, then only the homozygous recessive individuals will die.

In this experiment, 400 black beads represent dominant trait and 400 white beads as recessive trait, and the homozygous recessive 2 white beads is the lethal allele. This allele will cause an organism to die thus failed to reproduce to next generation and failed to contribute to the gene pool.

In a narrower sense, genetic drift refers to the expected population dynamics of neutral alleles (those defined as having no positive or negative impact on reproductive fitness), which are predicted to eventually become fixed at zero or 100% frequency in the absence of other mechanisms affecting allele distributions. So, in this part of experiment, if we continue after 100 times, genetic drift will occur. From the beginning, n1 and n2 are close with the expected number, n1 and n2. This is because there are large population that support the Hardy-Weinberg Equilibrium and the deviation and error are low by first generation, n1 with the ratio of 1:2:1. But once the offspring ends with a lethal allele, it is no longer survived in the population.

If the experiment further continues, the homozygous recessive 2 white beads (lethal allele) will eventually no survived in this population. As the population is getting smaller, genetic drift will having effect. And genetic drift causes populations to lose genetic variation. This is because the number of black beads has increased and the number of white beads has reduced as it is not replacing back. The homozygous trait and heterozygous have an increase of genetic ratio and at the same time the homozygous recessive decrease. Hence, the black phenotype will increase if continue the experiment with the homozygous dominant becomes the major allele in that population with the heterozygous and homozygous recessive will reduce and become extinction.

Conclusion:The hypothesis is accepted. As stated in Hardy-Weinberg Equilibrium, large population can avoid the affect the genetic drift. However, if the experiment continues without replacing back the white beads, the hypothesis is rejected. This is due to the large population is getting smaller and genetic drift occurred. Genetic drift reduces the amount of genetic variation in a population. With less genetic variation, there is less for natural selection to work with. And, alleles face a greater chance of being lost. As Hardy Weinberg Equilibrium states that it is only valid when there is no genetic drift.

III. Detection of Gene Frequencies in Human Population

Materials:1. PTC paper strip2. 33 members of course mates

Methodology:(A) PTC-tasting

The whole course members tasted phenylthiocarbamide (PTC) by using PTC paper.

The results were recorded down the tasters as homozygous dominant or heterozygous and non-tasters were homozygous recessive.

(B) Tongue-rollingCourse members who were able to roll their tongue were recorded as homozygous dominant or

heterozygous.

The course members who were not able to roll their tongue were recorded as homozygous recessive.

(C) Facial dimplesThe course members with facial dimples were recorded down as homozygous dominant or

heterozygous.

The course members without facial dimples were recorded down as homozygous recessive.

Results:

Phenotype Genotype Observed number, O

Estimated genotypic frequencies

Estimated number

PTC- Taster TT 18 0.1 3Tt 0.44 15

Non- PTC Taster tt 15 0.45 15

Tongue roller RR 28 0.37 12Rr 0.5 16

Non-tongue roller rr 5 0.15 5

Facial dimples DD 9 0.02 1Dd 0.25 8

No Facial dimples dd 24 0.73 24

Calculation:(A) PTC – taster - q2 = 15/33

q = √ 15/33q = 0.67

- p = 1 – q p = 1 – 0.67 p = 0.33

Estimated genotypic ratio of- TT = p2 = (0.33)2 = 0.1- Tt = 2pq = 2 (0.33) (0.67) = 0.44- tt = q2 = (0.67)2 = 0.45

Estimated number of- TT = 0.1 × 33 = 3- Tt = 0.44 × 33 = 15- tt = 0.45 x 33 = 15

(B) Tongue Roller- q2 = 5/33

q = √ 5/33q = 0.389

- p = 1 – q p = 1 – 0.389p = 0.611

Estimated genotypic ratio of - RR = p2 = (0.611)2 = 0.373- Rr = 2pq = 2 (0.611) (0.389) = 0.475- rr = q2 = (0.389)2 = 0.151

Estimated number of- RR = 0.373 × 33 = 12- Rr = 0.475 × 33 = 16- rr = 0.151 x 33 = 5

(C) Facial Dimples- q2 = 9/33

q = √ 9/33q = 0.853

- p = 1 – q p = 1 – 0.853p = 0.147

Estimated genotypic ratio of - DD = p2 = (0.147)2 = 0.02- Dd = 2pq = 2 (0.147) (0.853) = 0.25- dd = q2 = (0.853)2 = 0.73

Estimated number of- DD = 0.02 × 33 = 1- Dd = 0.25 × 33 = 8- dd = 0.73 x 33 = 24

Discussion:In this experiment, the class population has higher dominant trait in PTC taster and

tongue roller, and higher recessive trait in facial dimples.

PTC is known as Phenylthiocarbaminde Taste Paper. Besides, it is also known as Phenylthiocarbamide, or phenylthiourea, is an organic compound that either tastes very bitter, or is virtually tasteless, depending on the genetic makeup of the tasterThe genetic taste phenomenon of PTC was discovered in 1931 when a DuPont chemist named Arthur Fox accidentally released a cloud of a fine crystalline PTC. . The ability to taste PTC is a dominant genetic trait. The test to determine PTC sensitivity is one of the most common genetic tests on humans. The genetic correlation was so strong that it was used in paternity tests before the advent of DNA matching. There are three SNP's (single nucleotide polymorphisms) along the gene that may render its

proteins unresponsive. There is conflicting evidence as to whether this trait is a result of either dominance or incomplete dominance. Any person with a single functional copy of this gene can make the protein and is sensitive to PTC. Hence, PTC tasting is largely determined by a single gene, TAS2R8, with two common alleles, and the allele for tasting is mostly dominant over the allele for non-tasting. However, both classical family and twin studies, and modern molecular genotyping, show that there are other genes or environmental factors that influence PTC tasting. As a result, there is a continuous range of PTC tasting, not absolute separation into tasters and non-tasters.

Next, the ability to roll one's tongue is a genetic trait that historically with the concepts of dominant and recessive traits. However, the dominant nature of the "tongue-rolling gene" has been disproved. From some researches, family studies clearly demonstrate that tongue rolling is not a simple genetic character, and twin studies demonstrate that it is influenced by both genetics and the environment. The percentage of the human population that has this skill ranges from 65 to 81 percent. A slightly higher percentage of females can do this than males. 

When some people smile, they have dimples in one or both cheeks. Other people don't have dimples. This is occasionally said to be a simple genetic trait, which dimples are controlled by one gene with two alleles, and the allele for dimples is dominant. Dimples are caused by a fault in the subcutaneous connective tissue that develops in course of the embryonic development. A variation in the structure of the facial muscle may also cause dimples. Transfer of dimples from parents to children occurs due to just one gene. The dimple creating genes   are present in the sex cells prior to the process of reproduction. Each parent provides one of these genes to the child. So, if both the parents have dimples, the children have 50-100% chances of inheriting dimple genes. If, however, only one parent has dimple genes, the chances of the children inheriting the genes are 50%. If neither of the parents has the dimple genes, their children will not have dimples. But this is not always be, because their inheritance isn't completely predictable, dimples are considered an “irregular” dominant trait. Having dimples is probably controlled mainly by one gene but also influenced by other genes. However, the presence of dimples may change during an individual's lifetime, and there is no published evidence for a genetic basis for dimples. Therefore, dimples cannot be said as basic genetics.

Conclusion:The hypothesis is rejected. This is because the Hardy-Weinberg Equation is difficult or impossible to demonstrate with living organisms as living organisms may be mutated and they are natural selection. For example, tongue rolling and dimples. The genes can be mutated by other genes or influenced by environment.

IV. Determining Gene Frequencies Where Three Alleles are Involved

Materials:1. Sterile lancet2. Sterile cotton3. 70% ethyl cotton4. Anti-A and anti-B antiserum5. Microscopic slide

Methodology:A drop of each anti-A serum and anti-B serum were put on a clean slide.

Finger index was cleaned by using a piece of 70% ethyl alcohol paper.

The finger index was pricked by using sterile lancet and drop of blood was added to each test sera.

The blood was mixed with the serum after few minute and the result was recorded down.

Results:Phenotype of Blood group

Genotypes Observed number

Observed genotypic frequencies

Estimatednumber

A IA IA 7 0.017 1IA IO 0.194 6

B IB IB 8 0.022 1IB IO 0.220 7

AB IA IB 0 0 0O IO IO 18 0.546 18Total: 33 1 33

Calculation:

Estimated number

- Blood type A: p2 + 2pq = 7/33- Blood type B: q2 + 2qr = 8/33- Blood type AB: 2pq = 0/33- Blood type O: r2 = 18/33

r2 = frequency of the O phenotype,r = √ r2

r = √ 18/33r = 0.739

p = (√ p2 + 2pr + r2) – rp = (√ 7/33 + 18/33) – 0.739p= 0.131

q = (√ q2 + 2qr + r2) – rq = (√ 8/33 + 18/33) – 0.739q= 0.149

Estimated genotype frequency:- AA: p2 = (0.131)2 = 0.017- AO: 2pr = 2 × 0.131 × 0.739 = 0.194- BB: q2 = (0.149)2 = 0.022- BO: 2qr = 2 × 0.149 ×0.739 = 0.22- OO: r2 = (0.739)2 = 0.546

Estimated genotype number:- AA: 0.017× 33 = 1- BB: 0.022 × 33 = 1- AO: 0.194 × 33 = 6- BO: 0.22 × 33 = 7- OO: 0.546 × 33 = 18

Discussion:The blood groups refer to the presence on human red blood cells of certain antigens, the

blood group factors. One very important group of factors present on the red blood cells is the ABO system. The ABO group of a person depends on whether his/her red blood cells contain one, both, or neither of the 2 blood group antigens A and B. Hence, there are 4 main ABO groups: A, B, AB and O.

Antibodies (agglutinins) for the antigens A and B exist in the plasma and these are termed anti-A and anti-B. The corresponding antigen and antibody are never found in the same individual since, when mixed, they form antigen-antibody complexes, effectively agglutinating the blood.

In this experiment, we used anti-A serum and anti-B serum. Anti-A serum and anti-B serum are used in the forward typing of ABO blood grouping. Anti-A sera contain antibodies to the A blood group, while Anti-B sera contain antibodies to the B blood group. So, when doing a blood type, if anti-B serum reacts with the patient's blood, this means that the person is group B or their blood cells at least contain the B antigen. When the patient's blood containing the B antigen is mixed with anti-B serum this causes clumping. They could be type B. If the blood causes clumping with anti-A serum and anti-B serum, this means the person is group AB blood type. However, there is no agglutination in anti-A serum and anti-B serum in type O blood. As blood type O contains anti-A antibodies and anti-B antibodies, which is no containing antibodies, it won’t causes clumping with anti-A serum and anti-B serum.

By referring diagram below:1. Blood group type B2. Blood group type A3. Blood group type AB4. Blood group type O

ABO blood group is an example of three allele involved. In the class population, blood groups of O have the highest probability compared with other blood types. Meanwhile, there is no population that have blood group of AB. When two copies of gene (type A with type O or type B with type O) are fused to form the next generation, the one with blood group type A is IAIA or IAIO while blood group B forms genotype of IBIB or IBIO. Both of this is dominant. However when genotype IAIB gene form, this is a codominant trait which indicates in AB blood group. If an individual consist of genotype IOIO, the individual are said to be homozygous recessive and show as blood group O.

Conclusion:The hypothesis is rejected. Although p + q + r = 1 as Ab group is the codominance, so 2pq is not included, but, in reality, no population satisfies the Hardy-Weinberg Equilibrium completely because human always choose their mates (no random mating) and always migration. It can be seen that phenotype of Blood AB group is absence in this experiment.

References:1. Crow, J. F. Hardy, Weinberg and language impediments. Genetics 152, 821-825 (1999).2. Edwards, A. W. F. & Hardy, G. H. 1908 and Hardy-Weinberg

Equilibrium. Genetics 179, 1143-1150 (2008).