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LSM 1102 (MOLECULAR GENETICS) POPULATION GENETICS: PRACTICAL 8

THE HARDY -WEINBERG PRINCIPLE

The basic law of population genetics was formulated in 1908 by an English mathematician, G.H. Hardy, and a German physician, Wilhelm Weinberg. This principle or law is concerned with the frequency of alleles of a gene in a randomly interbreeding group of plants or animals. Hardy and Weinberg demonstrated that in such natural populations equilibrium is reached. At this equilibrium the frequencies of the different alleles of a gene remain constant generation after generation, if no disturbing effects occur, such as those caused by mutation, selection, random genetic drift, migration, or meiotic drive. Consider the case in which a gene exists as just two alleles--A and a. Let p equal the frequency of the A allele and q equal the frequency of the a allele. Then, p + q = 1. (Why?) It follows that if random mating occurs, the population should consist of p2 AA individuals + 2 pq Aa + q2 aa individuals (see Table 1). If a population consists of p2 AA + 2pq Aa + q2 aa, the next generation, according to the Hardy-Weinberg law, may be expected to consist of exactly the same frequencies of each zygotic combination. The generation-to-generation constancy that does exist has been tabulated; the frequencies of different mating combinations are shown in Table 2, and the frequencies of the different types of offspring resulting from such matings are summarized in Table 3.

Table 1. The Random Combination of Gametes Results in a Population Consisting of p2 AA + 2pq Aa + q2 aa.

Sperms

Ova p (A) q (a) p (A) p2 (AA) pq (Aa) q (a) pq (Aa) q2 (aa)

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Table 2. The Determination of the Frequency of Different Matings in a Population

Consisting of p2 AA + 2 pq Aa + q2 aa. Males Females p2 (AA) 2 pq (Aa) q2 (aa) p2 (AA) p4 (AA × AA) 2 p3 q (AA × Aa) p2 q2 (AA × aa) 2 pq (Aa) 2 p3 q (AA ×Aa) 4 p2q2 (Aa ×Aa) 2 pq3 (Aa ×aa) q2 (aa) p2q2 (AA ×aa) 2pq3 (Aa ×aa) q4 (aa ×aa) Table 3. The Frequency of Different Types of Progeny Resulting From Random Matings

in a Population in Hardy-Weinberg Equilibrium

Resulting Progeny From Matings Summary of frequency of mating AA Aa Aa

p4 (AA ×AA) p4 4 p3q (AA ×Aa) 2 p3q 2 p3q 2 p2 q2 (AA×aa) 2 p2q2 4 p2q2 (Aa×Aa) p2q2 2 p2q2 p2q2 4 pq3 (Aa×aa) 2 pq3 2 pq3

q4 (aa ×aa) q4 Totals p4+ 2p3q + p2q2 2p3 q+ 4p2q2 + 2pq3 p2q2 + 2pq3 + q4

Totals factored p2 (p2 + 2pq + q2) 2pq(p2 + 2pq + q2) q2(p2 + 2pq + q2) Totals* p2 2pq q2

*p2 + 2pq +q2 =1. Therefore, totals in this row are possible. A study of these tables indicates that if a population is in Hardy-Weinberg equilibrium, it will stay in that equilibrium generation after generation. Neither the gene frequency nor the zygotic frequencies will change as long as random mating continues and there are no disturbing effects caused by mutation, migration, selection, or random genetic drift. Objectives of the Investigation Upon completion of this investigation, the student should be able to 1. calculate the gene (allele) frequencies for a population sample in which each of the

genotypes AA, Aa and aa has a unique phenotype 2. determine whether a sample in which each of the genotypes AA, Aa and aa has a

unique phenotype represents a population in Hardy-Weinberg equilibrium. 3.

calculate the frequencies of the alleles A and a when dominance is complete (i.e., AA = Aa in phenotype), given the frequencies of A- and aa individuals in a population sample;

4. calculate the frequencies of the A, AB, and a alleles, given a population sample in which the frequencies of individuals having blood types A, B, AB, and O are known;

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I. DETERMINING GENE FREQUENCIES WHEN CODOMINANCE EXISTS

The Hardy-Weinberg law need not be used to calculate gene frequencies in cases in which genes exhibit codominance or incomplete dominance. For such cases, gene frequencies can be determined directly from the phenotypes of the individuals involved. Suppose, for example, that you wish to determine the frequencies of the alleles R (red) and r (white) in a population of shorthorn cattle consisting of 63 RR (red), 294 Rr (roan), and 343 rr (white) cattle. Suppose further that you wish to know whether the population is in Hardy-Weinberg equilibrium. To make these determinations, proceed as follows: A. Gene Frequencies

In this population of 700 cattle, 63 possess two R genes and 294 possess one R gene. Thus, among 1400 genes (2× 700) are (63× 2) + 294 = 420 R genes. The frequency of R = 420 = 6 = 3 = 0.3 1400 20 10 Likewise the frequency of r = 294 + 343 (2) = 294 + 686 = 980

1400 1400 1400 = 14 = 7 = 0.7 20 10

B. Hardy-Weinberg Equilibrium

If the population is a Hardy-Weinberg equilibrium, (0.3) × (0.3) = 0.90 = 9% of the population may be expected to be RR (red). Similarly, 2 × (0.3) × (0.7) = 0.42 = 42% may be expected to be Rr (roan), while (0.7) × (0.7) = 0.49 = 49% may be expected to be rr (white). Applying these percentages to the population of 700 cattle reveals that it is in perfect Hardy-Weinberg equilibrium.

9% of 700 = 0.09(700) = 63; 42% of 700 = 0.42(700) = 294; 49% of 700 = 0.49(700) = 343.

The same technique determining gene frequencies and then substituting these frequencies in the Hardy-Weinberg zygotic distribution might be used to determine Whether a human population is in equilibrium for the M and N blood groups. The three possible blood types are type M (genotype MM), type N (genotype MNMN), and type MN (genotype MMN).

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II. DETERMINING GENE FREQUENCIES USING THE HARDY-WEINBERG LAW

When dominance and recessiveness affect a pair of alleles, it is impossible to detect all three genotypes by their phenotypes and to estimate gene frequencies directly. If, however, you assume the population to be in a Hardy-Weinberg equilibrium, you can estimate gene frequencies knowing that q2 = frequency of homozygous recessive individuals. You can then calculate the frequency (p) of the document allele by subtraction, because you know that p + q = 1. Knowing the gene frequencies, you can then calculate the frequencies of the homozygous dominant individuals (p ) and heterozygous dominant individuals (2pq).

Rh blood group

For example, the class of Genetics students in 1996, showed that out of a total of 191 students 5 of them (2.6%) were Rh-negative (rr), you could equate q2 and 26%.

q2 = 0.026

026.0=q

q = 0.16 = frequency of the allele, r

Then p = 1 – q = 1 – 0.16 = 0.0.84 = frequency of the allele, R

It then follows that the 97.4% of this sample that is Rh-positive can be divided as follows:

2pq = 2(0.16)(0.84) = 0.268 = 26.8% of this population expected to be

heterozygous, Rr.

p2 = (0.84)(0.84) = 0.706 = 70.6% of this population expected to be

homozygous, RR.

PTC tasting

Now you can use the Hardy-Weinberg equilibrium to determine the frequencies of the two alleles that regulate the ability to taste the organic compound phenylthiocarbamide (PTC). Estimating the frequencies of these alleles in your genetics class should give you some idea of their frequencies in the population in general.

Studies of individual human families (pedigree studies) have indicated that the ability to taste PTC is due to a dominant gene and that inability to taste the chemical (taste blindness) occurs in the homozygous recessive individuals. Thus, there are two phenotypes – “tasters” and “nontasters” – but three genotypes (TT, Tt, and tt).

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A. Procedure You will be given a small piece of filter paper that has been impregnated with PTC. You will use this paper to determine whether you are a taster. Before you place the PTC paper on your tongue, take a piece of untreated filter paper and taste it. This will serve as a control so that you will be readily able to determine the difference between the taste of the filter paper and the taste of PTC.

1. Are you a taster of PTC? ___________ If so, how did PTC taste?

Was it sweet, sour, salty, bitter, or what? ______________________

2. You now know your phenotype. Do you know your genotype? __________

_________________________________________________________________

a. If not, why not? _____________________________________________

b. If not, how could you go about finding out what your genotype is? _________________________________________________________

__________________________________________________________

3. Record your phenotype on the table provided on the chalkboard. After all of the

data are recorded, copy the total class data in the following table. Frequency Phenotype Number (Decimal Fraction) Tasters ________ __________________ Non-tasters ________ __________________ Total ________ __________________ 4. Now calculate the frequency of the recessive allele, t, applying the Hardy

Weinberg equilibrium to the class data. The frequency of t for the class data is __________________________________________________________________

5. The frequency for the T allele is thus ___________________________

Using the Hardy-Weinberg equilibrium, you may expect the frequency (p2) of TT individuals to be ___________________________________

The frequency (2pq) of Tt individuals is _________________________

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B. Chi-Square Test

Studies of the frequencies of tasters and non-tasters have shown that among the North American white population about 70% can taste PTC, whereas about 30% are taste-blind. Other populations show different frequencies of two phenotypes. For example, some studies show that about 90% of American blacks are tasters and only 10% are non-tasters. Now substitute your data in Table 4 and calculate chi-square (χ2) on the assumption that your class is a sample from a population consisting of 70% tasters and 30% non-tasters. 1. How many degrees of freedom do you have in interpreting χ2?

__________________________________________________________ Does the class seem to be a representative sample of the U.S white population? If not, indicate some reasons why the class may not fit this population. ______________________________________________________________ ____________________________________________________________ Null hypothesis, Ho: Local sample fits the expected population ratio of 70% tasters: 30% non-tasters (based on North American white population) Alternative hypothesis, H1: Local sample does not fit the expected population ratio of 70% tasters: 30% non-tasters

Table 4. Calculation of chi-square test for the 1996 class data on the ability to taste PTC

Phenotype Observed Number (O)

Expected Number (E) O – E (O – E)2 (O-E)2/E

Taster Non-taster Totals χ2 = What is your conclusion from the results of the χ2 test with regards to the class sample compared with the US populations for the PTC gene? ________________________________________________________________________

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III . DETERMINING GENE FREQUENCIES WHEN THREE ALLELES ARE INVOLVED: ABO human blood group The two preceding examples the genes involved existed as two alleles (R, r; and T, t). In many situations, however, three or more alleles exist in a population. Consider the A, B, O blood groups, with the three alleles A, AB, and a. Let p = frequency of the A allele, q = frequency of AB allele, and r = frequency of the an allele. Then p + q + r = 1. (Why?) The frequencies of the different blood groups are tabulated in Table 5 for a randomly mating population with the gene frequencies p, q, and r.

Table 5 may be summarized as follows: the population should consist of p2 (AA) + 2 pr (Aa) + q2 (ABAB) + 2qr (ABA) + 2pq (AAB) + r2 (AA). The frequencies of the different phenotypes, then, are as follows:

Phenotype Genotypes Frequency

A AA and Aa p2 + 2pr B ABAB and ABa q2 + 2qr AB AAB 2pq O aa r2

______________________________________________________________________ *Note that A and AB are both dominant to a but are codominant with respect to each other. Gene frequencies may now be estimated. 1. Calculation of the frequency of the a allele:

r2 = frequency of the O phenotype. r = square root of the frequency of the O phenotype.

2. Calculation of the frequency of the A allele:

p2 + 2pr + r2 = (p + r) 2 = frequency of the A phenotype plus the frequency of the O phenotype.

(p + r) = square root of the frequency of the A phenotype + the frequency of the O phenotype. p = [√ p2 + 2pr + r2 ] - r

3. Calculation of the frequency of the AB allele: This may be obtained most conveniently, by subtraction, since p and r have already been determined. q = 1 – (p + r)

Apply this information to the calculation of p, q and r for the sample that your class represents. To do this, type each person’s blood and then determine the frequencies of the different blood types. Table 6. Random Recombination of Gametes in a Population in which a Gene Exists as Three Separate Alleles

Sperms Ova

p (A)

q (AB)

r (a)

q (A) p2 (AA) pq (AAB) pr (Aa) q(AB) pq (AAB) p2 (AB AB) pr (ABa) r (a) pr (Aa) qr (AB a) r2(aa)

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A. Background

The study of the physiology and genetics of the A, B, O blood groups were initiated by an Austrian, Karl Landsteiner, early in this century. The results of his studies are found in any beginning Genetics textbook and in most general biology textbooks. Landsteiner (Table 7) found that certain antigens are associated with the human red blood cell (erythrocyte). The antigens are capable of stimulating antibody production when they are injected into animals such as rabbits or guinea pigs. The antibodies so produced are then capable of reacting with the specific antigen that caused their production. Thus, if red blood cells from a type A person were injected into a guinea pig, the guinea pig would produce antibodies to A antigen. Blood serum from the guinea pig would contain these antibodies, which, when mixed with type A human blood, could cause agglutination (clumping) of the human erythrocytes. Table 7 shows that the four basic blood groups are controlled by three alleles of a single gene. The A and AB alleles are both dominant to the a allele but are codominant with respect to each other. Thus Aa has type A blood, ABa has type B blood, but AAB possesses both of the antigens and has type AB blood. Notice also that the blood serum of a type A person has antibodies to type B blood, the blood serum of a type B person has antibodies to type A blood, the blood serum of a type AB person has neither antibody, and the blood serum of a type O person has antibodies to both A and B antigens.

Table 7. Four Blood Types, Each Associated With a Different Antigen Combination and all Genetically Controlled

Blood Type

Antigen on RBCs

Antibodies Naturally Present in

Blood Serum

Possible Genotypes

Approximate Frequencies In Singapore Population

Approximate Frequencies In U.S.White Population*

A A β (anti-B) AA, Aa 20% 41% B B α (anti-A) ABAB, ABa 20% 9%

AB A and B Neither α nor β AAB 5% 3% O Neither A nor B both α and β aa 55% 47% Total 100% 100%

The possible reactions are as follows: a. No agglutination – type O blood b. Agglutination with anti-A only – type A c. Agglutination with anti-B only – type B d. Agglutination with both – type AB

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When all of the class data have been recorded on the chalkboard, complete the following table.

Frequency Blood Type Number (Decimal Fractions)

A _________ ____________________ B _________ ____________________ O _________ ____________________ AB _________ ____________________

B. Gene Frequencies

Now, using this information, show your calculations and determine 1. the frequency of the a allele. ______________________________

2. the frequency of the A allele. ______________________________

3. the frequency of the AB allele. _____________________________

4. What basic assumption are you making about the human population when

you calculate gene frequencies in this fashion?

____________________________________________________

C. Calculating chi-square for ABO blood group Using your class data, determine whether the class is a random sample of the Singapore population. Do this by completing Table 7 and calculating χ2.

Null hypothesis, Ho: Sample fits the expected population ratio of 20% A group: 20% B group: 5% AB group: 55%O group Alternative hypothesis, H1: Sample does not fit the expected population ratio

Table 8. Calculation of chi-square on the ABO blood group data on the Assumption that

Class (1996) is a Representative Sample of the Singapore Population Blood type Observed

No. (O) Expected No.

(E) O-E (O-E)2 (O-E)2/E

A B

AB O

Totals χ2 =

1100

1. How do you go about determining the expected numbers in each blood type?

____________________________________________________________

2. How many degrees of freedom do you have in interpreting χ2?

____________________________________________________________

3. What is your interpretation of the χ2 you have just calculated? ________________________________________________

____________________________________________________________

4. Is the class distribution of blood types comparable to that of the Singapore

population? ____________ If not, why not?

____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________

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Table 9. 1996 – 1997 class data for the ABO Blood Group, RH blood group and PTC Tasting

ABO BLOOD GROUP

RH BLOOD GROUP

PTC

Group O A B AB RH+ Taster Non-Taster

No. of students

1 5 2 3 0 10 7 3 10 2 6 3 4 1 14

RH-

12 2 14 3 4 5 0 1 9 7 3 10 4 3 2 0 2 7 4 3 7 5 5 3 4 1 13 12 1 13 6 6 4 2 1 13 12 1 13 7 3 2 3 4 12 11 1 12 8 2 4 5 0 11 10 1 11 9 7 3 3 0 13

1 RH(-)

10 3 13 10 5 6 3 0 12 2 RH(-) 11 3 14 11 5 0 3 4 11 9 3 12 12 4 3 3 1 11 7 4 11 13 8 4 0 0 12

1 RH(-)

10 2 12 14 9 2 2 2 14 14 1 15 15 8 1 4 1 14 12 2 14 16 6 1 2 1 10

1 RH(-)

8 2 10 Total 86 45 41 19 186 5 RH(-) 156 35 191

191

191 191

Table 10. Table of χ2 values

df

P=.99 .95 .80 .50 .20 .05 .01

1 .000157 .00393 .0642 .455 1.642 3.841 6.639 2 .0201 .103 .446 1.386 3.219 5.991 9.210 3 .115 .352 1.005 2.366 4.642 7.815 11.345 4 .297 .711 1.649 3.357 5.989 9.488 13.277 5 .554 1.145 2.343 4.351 7.289 11.070 15.086 6 .872 1.635 3.070 5.348 8.558 12.592 16.812 7 1.239 2.167 3.822 6.346 9.803 14.067 18.475 8 1.646 2.733 4.594 7.344 11.030 15.507 20.090 9 2.088 3.325 5.380 8.343 12.242 16.919 21.666 10 2.558 3.940 6.179 9.342 13.442 18.307 23.209 15 5.229 7.261 10.307 14.339 19.311 24.996 30.578 20 8.260 10.851 14.578 19.337 25.038 31.410 37.566 25 11.524 14.611 18.940 24.337 30.675 37.652 44.314 30 14.953 18.493 23.364 29.336 35.250 43.773 50.892

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REFERENCES

Ayala, F.J. 1982. Genetic variation and evolution. Carolina Biology Reader Series.

Burlington, N.C.: Carolina Biological Supply Co. Binkley, S.W. 1981. Blood typing for the classroom. Carolina Tips 44(12):49-52. Bodmer, W.F., and L.L. Cavalli-Sforza. 1976. Genetics, evolution and man. San

Francisco: W.H. Freeman, Castle, W.E. 1903. The laws of heredity of Galton and Mendel, and some laws

governing race improvement by selection. Proceedings of the American Academy of Arts and Science 39(8):223-242.

Cavalli-Sforza, L.L. 1974. The genetics of human populations. Scientific American 231(3):80-89.

Mertens, T.R. and Hammerstein, R.L. 1995. Genetics, Laboratory investigations. 8th Ed. McMillan Publishing Co., New York. 10th ed.

Hardy, G. 1908. Mendelian proportions in a mixed population. Science 28:49-50. Hartl, D.L. 1981. A primer of population genetics. Sunderland, Mass.: Sinauer

Associates. Mourant, A.E., A.C. Kopec, and K. Domaniewska-Sobczak. 1976. The distribution of

human blood groups. 2d ed. New York: Oxford University Press. Snustad, D.P., M.J. and Simmons. 1999. Principles of Genetics. John Wiley and Sons. 2nd Ed Zar, J.H. 1996.Biostatistical analysis. 3rd Ed., Engelwood Cliffs, NJ: Prentice-Hall.

POPULATION GENETICS DEFINITIONS 1. Genetic population 2. Hardy-Weinberg equilibrium 3. Genetic equilibrium 4. Multiple alleles 5. Sex-linkage 6. Random mating 7. Non-random mating 8. Two-way recurrent mutation 9. Mutation rate 10. Genetic variation 11. Genetic fitness 12. Selection coefficient 13. Overdominance 14. Genetic polymorphism 15. Transitional polymorphism

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TUTORIAL PROBLEMS 1. In a population of 2,000 gaboon vipers a genetic difference with respect to venom

exists at a single locus. The alleles are incompletely dominant. The population shows 100 individuals homozygous for the t allele (genotype tt non-poisonous), 800 heterozygous (genotype Tt mildly poisonous) and 1,100 homozygous for the T allele (genotype TT deadly poisonous). a) What is the frequency of the t allele in the population? b) Are the genotypes in Hardy-Weinberg equilibrium?

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2. Three alleles are found at a autosomal locus coding for the enzyme, malate dehydrogenase (MDH) in the spotted chorus frog. These three alleles are codominant. Chorus frogs were collected from a breeding pond and the genotype of each frog was determined by electrophoresis. The following numbers were found:

M1 M1 8 M1 M2 35 M2 M2 20 M1 M3 53 M2 M3 76 M3 M3 62 Total 254

a) Calculate the frequencies of the M1, M2 and M3 alleles in this population. b) Using the chi-square test determine whether the MDH genotypes in this population

are in Hardy-Weinberg proportions.

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3. In some populations approximately 8% of males have a particular type of colour blindness (an X-linked recessive trait). What proportion of females would you expect to have this trait? What proportion of females would you expect to be carriers for this allele?

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4. About one normal allele in 300,000 mutates to the X-linked recessive allele for haemophilia in each human generation (µ). Assume for the purposes of this population that one h allele in 30,000 mutates to the normal alternative in each generation (v). The mutation frequencies are indicated in the following equation:

µ h+ h v where v = 10u, What allelic frequencies would prevail at equilibrium under mutation pressures alone in these circumstances? Let pe and qe be the equilibrium gene frequencies of h+ and the h genes, respectively.

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5. A selectively neutral, X-linked recessive character appears in 0.40 of the males and 0.16 of the females, in a randomly interbreeding population. What is the frequency of the X-linked recessive gene that codes for the trait? How many females are heterozygous for it? How many males are heterozygous for it?

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TUTORIAL ESSAY QUESTIONS

1. Discuss the effects oif systematic forces that affect gene frequencies in large randomly mating populations

2. What are the primary effects of the following evolutionary forces on the allelic and

genotypic frequencies of a large randomly mating population? a) Mutation b) Natural selection

DISCUSS DURING Tutorial/Lab Session THE MAIN POINTS. ATTEMPT THESE ESSAY TYPE QUESTIONS AT HOME.