Question

Rules for Assigning h5762a Answer ALL of the following: What are the assumptions of the Hardy-Weinberg Law? a) b) People with sickle cell anaemia are homozygous (HbS/HbS) for an abnormal form of haemoglobin which distorts the shape of the red blood cells. People with the sickle cell trait have a less severe form of this disease. These people are heterozygous at this locus, having one HbS allele and one normal haemoglobin allele (HbA). The frequency of sickle cell anaemia and sickle cell trait was determined in a group of 900 adults from Mahuta village in Northerm Nigeria, a region where malaria is endemic and the following genotypes were observed: HbA/HbA HbA/HbS HbS/HbS Total 405 450 45 900 i) Is this population in agreement with the Hardy-Weinberg equilibrium? (A table of chi-squared values is provided. See overleaf.) i) Explain your answer. Describe the various stages and mechanisms involved in speciation. Include in 4. 5 8 9 alt gr ctrl

JUST PART B PLEASE AND THANK YOU

TABLE OF CHI SQUARED VALUES. D.F = DEGREES OF FREEDOM

df				<0.10	<0.05	<0.01	<0.001
1				2.706	3.841	6.635	10.8
2				4.605	5.991	9.210	13.8
3				6.251	7.815	11.345	16.3
4				7.779	9.488	13.277	18.5
5				9.236	11.070	15.086	20.5
6				10.645	12.592	16.812	22.4
8				13.362	15.507	20.090	26.1
10				15.987	18.307	23.209	29.6
12				18.549	21.026	26.217	32.9
20				28.412	31.410	37.566	45.3
25				34.382	37.652	44.314	52.6

Answer 1

a) according to Hardy-Weinberg equilibrium allelic frequencies in a large population will remain constant when there are mutations, no migration, random mating, no natural selection. so the assumptions here are 1) population is large, 2) random mating, 3) No migration, 4) no natural selection, 5) no mutation

b) HbA/HbA= 405

HbA/HbS=450

HbS/HbS=45

Let's calculate the frequency of HbS allele

genotypic frequency of HbS/HbS= number of HbS/HbS/total number

=45/900

=0.05

Genotypic frequency of HbS/HbA= number of HbA/HbS/900

=450/900

=0.5

allelic frequency of HbS= Genotypic frequency of HbS/HbS+ ( genotypic frequency of HbS/HbA)/2

=0.05+0.25

= 0.3

Allelic frequency of HbS+ Allelic frequency of HbA=1

Allelic frequency of HbA= 1-0.3

=0.7

Expected number of HbA/HbA= Expected genotypic frequency of HbA/HbA* total number

= ( allelic frequency of HbA)^2*900

= (0.7)^2*900

=441

Expected number of HbA/HbS= Expected genotypic frequency of HbA/HbS*900

= (2* allelic frequency of HbA*allelic frequency of HbS*900)

=2*0.7*0.3*900

= 378

Expected number of HbS/HbS= Expected genotypic frequency of HbS/HbS*900

= ( allelic frequency of HbS)^2*900

=(0.3)^2*900

=81

For chi-square test null hypothesis= Population is in Hardy-Weinberg equilibrium

Genotype	Expected ( E)	Observed (O)	(E-O)^2/E
HbA/HbA	441	405	2.9387
HbA/HbS	378	450	13.714
HbS/HbS	81	45	16

Chi-square value= 2.9387+13.714+16

=32.6527

this is one locus 2 allele case so df=2-1

=1

for P<0.05 and df=1 value from Chi-square table=3.841 so we reject null hypothesis so this population is not in Hardy-Weinberg equilibrium

i) from the chi-square analysis population is not in Hardy-Weinberg equilibrium

ii) from chi-square analysis for P<0.05 and df=1 value from Chi-square table=3.841 so we reject null hypothesis so this population is not in Hardy-Weinberg equilibrium