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telo118 [61]
3 years ago
7

With respect to the three genes mentioned in the problem, what are the genotypes of the homozygous parents used in making the ph

enotypically wild f1 heterozygote?
Biology
1 answer:
Fantom [35]3 years ago
7 0
The genotypes of the homozygous guardians utilized as a part of making the phenotypically wild F1 heterozygote is cl h +/cl h + and + sp/+ sp. I hope the answer will help you. For more questions, feel free to ask in Brainly. 
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Use the following information to answer the following question.
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Answer: If this population were in equilibrium and if the sickle-cell allele is recessive, the proportion of the population susceptible to sickle-cell anemia under typical conditions should be 0.20

Explanation: Hardy-Weinberg law provides an equation to relate genotype frequencies and allele frequencies in a randomly mating population. The equation is;

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For 2 alleles such as A and a, where

p² = homozygous dominant

q² = homozygous recessive and

2pq = heterozygous

From the question, it is said that the sickle-cell allele (SS) constitutes 20% (that is, 20/100) of the hemoglobin alleles in the human gene pool and it is also said to be the homozygous recessive allele.

Therefore, q² = 20/100 = 0.20

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In the year 2500, five male space colonists and five female space colonists (all unrelated to each other) settle on an uninhabit
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Answer:

A) 0.1 a, 0.9 A

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According to the given information, all the individuals in the founder population had free earlobes. Since the allele for the free earlobes (A) is dominant over the one for the attached earlobes (a), the genotype of each of the homozygous dominant founder individual would be "AA". Humans are diploid and one individual has two alleles for each gene in its genome. Total number of alleles for the earlobe in the founder population = 10 x 2 = 20.

Frequency of dominant allele, A in the founder population = total number of homozygous dominant individuals x 2 + the total number of heterozygous genotype / total alleles for the earlobe in the population. There were 8 homozygous dominant and 2 heterozygous dominant individuals in the population.  

So, frequency of dominant allele = 8 x 2 + 2 / 20 = 16 + 2 / 20 = 18/20 = 0.9

Since p+q=1; frequency of recessive allele "a" = 1-p = 1-0.9 = 0.1

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Answer:

Explanation:

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