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egoroff_w [7]
3 years ago
5

प्र.नं. १ समूह'क' मा दिइएकोश

Biology
2 answers:
ioda3 years ago
6 0

plz attach photo .....

Explanation:

फोटो हाले बल्ला बुझिन्छ । यसरी त के के हो के के।।।।

GREYUIT [131]3 years ago
5 0

मिल्छ भने क्रिपया फोटो राखेर प्रश्न न सोध्नु होला

question sodda photo rakhera sodnu vako vaye bujhna sajilo hunthiyo

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What are seasons? What causes them?
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A man with type O blood and a woman with type AB blood get married. What is the probability that they will have a child with typ
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4 0
2 years ago
A population of 150 individuals has an allele frequency of 0.2 for the dominant allele (H) and a frequency of 0.8 for the recess
Amanda [17]
Hardy-Weinberg Equation (HW) states that following certain biological tenets or requirements, the total frequency of all homozygous dominant alleles (p) and the total frequency of all homozygous recessive alleles (q) for a gene, account for the total # of alleles for that gene in that HW population, which is 100% or 1.00 as a decimel. So in short: p + q = 1, and additionally (p+q)^2 = 1^2, or 1
So (p+q)(p+q) algebraically works out to p^2 + 2pq + q^2 = 1, where p^2 = genotype frequency of homozygous dominant individuals, 2pq = genotype frequency of heterozygous individuals, and q^2 = genotype frequency of homozygous recessive individuals.
The problem states that Ptotal = 150 individuals, H frequency (p) = 0.2, and h frequency (q) = 0.8.
So homozygous dominant individuals (HH) = p^2 = (0.2)^2 = 0.04 or 4% of 150 --> 6 people
Heterozygous individuals (Hh) = 2pq = 2(0.2)(0.8) = 0.32 or 32% of 150
--> 48 people
And homozygous recessive individuals (hh) = q^2 = (0.8)^2 = 0.64 = 64% of 150 --> 96 people
Hope that helps you to understand how to solve these types of population genetics problems!
6 0
3 years ago
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