Report 9: Population Genetic HARDY WEINBERG

Population genetics: Hardy Weinberg equilibrium.

Early in the twentieth century mathematician Godfrey Hardy and physician Wilhelm Weinberg independently developed a model describing the relationship between the frequency of the dominant and recessive alleles (hereafter, p and q) in a population. They reasoned that the combined frequencies of p and q must equal 1, since together they represent all the alleles for that trait in the population.

Hardy and Weinberg represented random mating in the population as the product (p + q)(p + q),which can be expanded to p² + 2pq + q². This corresponds to the biological fact that, as a result of mating, some new individuals have two p alleles, some one p and one q, and some two q alleles. P² then represents the fraction of the population that is homozygous dominant while 2pq and q² represent the heterozygous and homozygous recessive fractions, respectively.

A significant question in population genetics, therefore, is determining the frequency of the dominant and recessive alleles in a population (for example, the frequency of blood type O allele in the United States), given the frequency of the phenotypes. Note that phenotypic and allelic frequencies are related but are not equal. Heterozygote show the dominant phenotype, but carry a recessive allele. Therefore, the frequency for the recessive allele is higher than the frequency of the recessive phenotype.

            There are 5 assumptions in the Hardy Weinberg equilibrium:

1.            No mutation.

2.            No migration.

3.            No natural selection.

4.            No genetic drift.

5.            Random mating.

Results:

A. PTC taster.

1. Are you a taster of PTC?

 Yes.

2. If so, how did PTC taste? Was it sweet, sour, salty, bitter, or some other combination tastes?

Bitter.

3. You now know your phenotype. Do you know your genotype?

Yes I noe my phenotype and I don’t know my genotype.

1. If not, why?

 It may can be either TT @ Tt.

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

 find out phenotype of family member. Then, calculate probability or percentage allele frequencies from the family members.

Record your phenotype in the table provided on the chalkboard. After all data are recorded, copy the total class data into the following spaces

Phenotype	Number	Frequencies (decimal fraction)
Tasters	46	46/47 = 0.979
Nontasters	1	1/47 = 0.021
Total	47	1

The frequency of t for the class data is: √0.021 = 0.145.

The frequency for the T allele is thus 0.855.

                  p + q = 1; q = 0.145              p = 1 - 0.145

                                                                     = 0.855 

                                                                  p = T = 0.855

Means the frequency for T (P) = 0.86                  

0.14 + 0.86 = 1 and (0.86)² + 2(0.86)(0.14) + (0.14)² = 1

From Hardy-Weinberg equation, P² of TT = (0.86) ²

                                                                     = 0.7396

The frequency 2pq = 2(0.86) (0.14) = 0.2408                

Chi-square test.

How many degrees of freedom do you have in interpreting X²?

df = n – 1

= 2 – 1

= 1

Does the class seem to be a representative sample of the U.S. white population?

No.

If not, indicate some reasons why the class might not fit this population 

because U.S. population has many nontaster while our sample only has one nontaster.

Phenotype	Observed no.	Expected no.	O – E	(O – E) 2	(O – E) 2/ E
Taster	46	70/100 x 47 = 32.9	13.1	171.61	5.216
Nontaster	1	30/100 x 47 = 14.1	-13.1	171.61	12.171
Totals	47				X2 = 17.387

X² Table = 3.841

X² Calculated = 17.387

X² Calculated > X² Table. So, the U.S. hypothesis is rejected.

B. Blood group.

Phenotype	Genotype	Number	Frequency
A	AA and Aa	9	0.158
B	AA AB and AB a	20	0.351
AB	AAB	6	0.105
O	aa	22	0.386
		Total = 57

1. Frequency of the a allele.

Freq of a allele =     

                         =         2(22) + 9       

                                       2(57)

                         =         0.465

2. Frequency of the A allele.

Freq of A allele = 2(9) + 6

                             2(57)

                      =  0.211

3. Frequency of the A^B allele.

Freq of A^B allele = 20 + 6

                                  114

                             =  0.228

4. What basic assumption are you making about the human population when you calculate allele frequencies in this fashion.
• No mutation.
• No migration.
• No natural selection.
• No genetic drift.
• Random mating.

5. Is this a reasonable assumption for the trait of blood types? Justify the answer.

No. Because genotype frequencies in the next generation are different from the first generation.

DISCUSSION:

            An understanding of evolution depends upon knowledge of population genetics. One of the more difficult concepts to understand when studying population genetics is Hardy-Weinberg Equilibrium. The Hardy-Weinberg equilibrium is the statement that allele frequencies in a population remain constant over time, in the absence of forces to change them. Its name derives from Godfrey Hardy, an English mathematician, and Wilhelm Weinberg, a German physician, who independently formulated it in the early twentieth century. The statement and the set of assumptions and mathematical tools that accompany it are used by population geneticists to analyze the occurrence of, and reasons for, changes in allele frequency. Evolution in a population is often defined as a change in allele frequency over time. The Hardy-Weinberg equilibrium, therefore, can be used to test whether evolution is occurring in populations.

p + q = 1
p= frequency the dominant allele A
q= frequency the recessive allele a

Then the second equation are came out:

p² + 2pq + q² = 1 
 p²= frequency for the genotype homozygous dominant (AA)
 2pq= frequency for the genotype heterozygous (Aa)
q²= for the genotype homozygous recessive (aa)

The chi-square test can determine if observed and expected data are in agreement. Therefore, we can use a chi-square test to determine whether a population exhibits Hardy Weinberg equilibrium for a particular gene. To do so, it necessary to distinguish between homozygotes and heterozygotes, either phenotypically or at the molecular level. This is necessary so that we can determine both allele and genotype frequencies. In this experiment, for the PTC tester for the class the result for X² is 17.387. the result was too big then the X² chi-squarewich is 3.8441. from this show that the result was rejcted. It is because the result had very big different then the US tester.The second test the blood type, the result also rejected it is because the result for the class was too big from the expected result. The result  of blood type for our class is  X²=63.5587 and the real result that we need to get is X² chi-square = 7.815

CONCLUSION

As a conclusion, we are able to calculate the gene (allele) frequencies for a population sample in which each of the genotypes AA, Aa, aa has a unit phenotype that regulate the ability to taste PTC. Beside that, we are able to determine whether a population sample in which each of the genotypes AA, Aa and aa has a unique phenotype represents a population in a Hardy-Weinberg equilibrium. Lastly, we can calculate the frequencies of the I^A I^B and i alleles, given a population sample in which the frequencies of individuals having blood types A, B, AB and O are known.

REFERENCES

Brooker. R. J. 2009. Genetics: Analysis & Principles. Third Edition. McGraw-Hill Higher Education.

Eldra P. S. Linda R. B. Diana W. M. 2005. Biology. Seven Edition.

Brooks/cole-Thomson Learning.

Kenneth R.M. and Joseph L. 2004. Prentice Hall Biology. Pearson Prentice Hall.

http://www.talkorigins.org/faqs/genetic-drift.html

http://www.conservapedia.com/Genetic_drift

http://users.rcn.com/jkimball.ma.ultranet/BiologyPages/H/HardyWeinberg.html

http://www.tiem.utk.edu/bioed/bealsmodules/hardy-weinberg.html