The map shows the allele frequency of sickled hemoglobin (HbS) and malaria endemicity in Africa.
1 answer:
When an area is malaria-free, the HbS allele frequency is between 0 and 2.02. Thus, option "A" is correct.
<h3>How, explain your answer briefly?</h3>In the two maps that show the allele frequency of sickle haemoglobin (HbS) and malaria endemicity in Africa, it can be seen that in the regions with no malarial outbreak or malaria-free areas coincide with the grey to lighter shades of red in the heat map of the allele frequency of HbS. The colour coding in the heat map for the allele frequency tells us that the frequencies range from 0 to 2.02 in the malaria-free areas.
Hence the answer is A)When an area is malaria-free, the HbS allele frequency is between 0 and 2.02.
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Answer:
See the answer below
Explanation:
The Chi-square () is calculated by the formula:
=
where O = observed frequency and E = expected frequency
A standard dihybrid cross is expected to give 9:3:3:1 phenotypic ratio.
Phenotype O E
Round yellow 104 9/16 x 160 = 90 = 2.18
Round green 27 3/16 x 160 = 30 = 0.30
wringled yellow 27 3/16 x 160 = 30 = 0.30
wrinkled green 2 1/16 x 160 = 10 = 6.40
Total value to two decimal places = 9.18
Degree of freedom = (n-1) = 4 -1 = 3
Critical value at 95% probabillity level = 7.815
<em>Since the calculated </em><em> si more than the critical </em>
<em> value, </em><em>the null hypothesis is rejected. </em>