the following data were obtained from a lebanese population. blood type percent of individuals o 36% a 13% b 45% ab 6% a. calculate the frequencies of the ia, ib and io alleles. assume that the population is in hardy-weinberg equilibrium. b. in this population, a couple with type b blood plan to have a child. what is the probability that they will have a son with type o blood?

Answers

Answer 1

a. To calculate the frequencies of the IA, IB, and I alleles, we can use the following formulas:

p + q + r = 1

p2 + 2pq + q2 = 1

where the frequencies of the IA, IB, and I alleles are represented by p, q, and r, respectively.

Given the level of people with each blood classification, we can switch them over completely to frequencies by partitioning by 100:

p(O) = 0.36

p(A) = 0.13

p(B) = 0.45

p(AB) = 0.06

The complement of the sum of the IA and IB alleles can be used to determine the frequency of the I allele:

r = 1 - p - q

r = 1 - 0.13 - 0.45

r = 0.42

The equation for heterozygotes can be used to determine the frequency of the IA and IB alleles:

2pq = p(A) * p(B) * 2

p + q = 1 - r

p + q = 1 - 0.42

p + q = 0.58

2pq = 0.58 * 0.45 * 2

2pq = 0.522

p = sqrt(2pq - q2)

p = sqrt(2 * 0.522 - 0.452)

p = 0.174

q = sqrt(2pq - p2)

q = sqrt(2 * 0.522 - 0.172)

q = 0.398

Therefore, the frequencies of the IA, IB, and I alleles in this population are:

p(IA) = 0.174

p(IB) = 0.398

p(i) = 0.428

b. The genotype IBi is shared by the couple with type B blood. We must take into account the potential genotypes of their offspring as well as the corresponding blood types in order to determine the probability of having a son with type O blood.

The Punnett square for this cross would look like this:

| IB | i

--|-----|-----

IB| IBB | iB

i | iB | ii

Each crate addresses a potential genotype of their posterity, and the letters address the alleles acquired from each parent. The primary box (IBB) addresses a kid with genotype IBIB and blood classification B, the subsequent box (iB) addresses a kid with genotype IBi and blood classification B, the third box (iB) addresses a kid with genotype IBi and blood classification B, and the fourth box (ii) addresses a youngster with genotype ii and blood classification O.

The likelihood of having a kid with type O blood is equivalent to the recurrence of the ii genotypes:

P(ii) = p(i) * p(i)

P(ii) = 0.428 * 0.428

P(ii) = 0.183

Subsequently, the likelihood that the couple will have a child with type O blood is roughly 0.183, or 18.3%.

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Answer:

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Answers

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Explanation:

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Answer:

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Answer:

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Answers

Answer:

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Answer:

Marble D

Explanation:

According to this question, a group of four students is participating in a marble race. Each marble is said to have different mass but the same force were applied to each of them.

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Answers

Answer:

isotonic; the same rate

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Answers

Answer:

substitution

Explanation:

has a chance to cause silent mutation, which doesnt do anything to a protein.

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A substitution of one base for another within the sequence of a protein will cause an insignificant/silent mutation within the protein sequence and this mutation will not be evident in the protein.

When a base is deleted either near the start of a gene or near the end of the coding sequence, it will have a significant effect because this will change the sequence of the DNA ( due to the deletion of at least one nucleotide ).

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Answers

Answer:

There are pros and cons in genetic modification.

Explanation:

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Answers

The moon is abiotic because it is a non living thing

Answer:

The moon is abiotic because it doesnt move and it's not a living thing.

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B. Carbon dioxide is produced

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D. Lactic acid is produced

Answers

Answer:

the correct answer is A :Glucose is broken down into pyruvate

Answer:

If I'm correct, the glucose is broken down into pyruvate and energy.

Explanation:

I hope this helps! ^^

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Answer:

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Explanation:

sorry

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As the plants is a hybrid fruit tree, it is not possible to produce plumcot plants through fertilization of their seeds.

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