"If the frequency of an allele in a population is 0.999, what is the expected frequency of homozygotes for this allele in a population at Hardy-Weinberg Equilibrium? (please give your answer to three decimal places)"

Answers

Answer 1
Answer:

Answer:

The answer to the question is

The expected frequency of homozygotes for the allele in a population at Hardy-Weinberg Equilibrium is 0.998

Explanation:

From the question, we have the frequency of an allele  =  0.999

By Hardy-Weinberg Equilibrium

p² + 2pq + q² = 1 and p + q = 1

p = dominant allele frequency  in the population

q = recessive allele frequency  in the population

p² = homozygous dominant individuals  percentage

q² = homozygous recessive individuals  percentage

2pq = heterozygous individuals percentage

Therefore the frequency of the homozygotes for the allele = 0.999² =  0.998001 ≅ 0.998 to three decimal places

The Hardy-Weinberg equilibrium states that the genetic disparity in a population will remain unchanged from generation to generation in the absence of evolutionary disturbing influences


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. In a replicating DNA molecule, the region in which parental strands are separating and new strands are being synthesized is called a(n)

Answers

Answer:

replication fork

Explanation:

The deoxyribonucleic acid (DNA) is a double-stranded helix composed of two long chains of nucleotides. The replication fork is a Y-shaped structure by which both DNA strands are separated in order to be replicated during cell division. In a cell, DNA replication starts at specific sites in the genome referred to as 'origins of replication'. A replication fork is generated by helicase enzymes that unwind and separate the DNA double helix strands by interrupting hydrogen bonds that hold the two DNA strands together. These DNA strands act as templates for the leading and lagging DNA strands. During DNA replication, the leading strand is synthesized continuously in the same direction as the replication fork, while the lagging DNA strand is synthesized in a direction away from the replication fork, in small pieces of DNA called Okazaki fragments.

MULTIPLE CHOICEQuestion 15
If a parent cell has 46 chromosomes, how many
chromosomes will be present in each daughter cell after
mitosis?
А
12.
B
23
46
D
9
2

Answers

Answer:

B

Explanation:

Its 23 in mitosis cell division

In mice, the autosomal locus coding for the β-globin chain of hemoglobin is 1 m.u. from the albino locus. Assume for the moment that the same is true in humans. The disease sickle-cell anemia is the result of homozygosity for a particular mutation in the β-globin gene.a. A son is born to an albino man and a woman with sickle-cell anemia. What kinds of gametes will the son form, and in what proportions? (2 points)b. A daughter is born to a normal man and a woman who has both albinism and sickle-cell anemia. What kinds of gametes will the daughter form, and in what proportions? (2 points)c. If the son in part a grows up and marries the daughter in part b, what is the probability that a child of theirs will be an albino with sickle-cell anemia? (1 point)

Answers

Answer:

Explanation:

The disease sickle cell anemia is caused by homozygosity for a mutation:

  • SS or Ss genotypes: normal
  • ss: anemia

Albinism is a recessive trait:

  • AA or Aa: normal
  • aa: albinism

The genes S/s and A/a are linked and separated by 1  map unit.

Remember that 1 mu means that 1% of the gametes produced by an individual will be recombinant.

a) Albino man (Sa/Sa) X woman with sickle-cell anemia (sA/sA)

Son: Sa/sA

The gametes he can produce are:

  • Parentals: Sa and sA
  • Recombinant: SA and sa

Since the frequency of recombination is 1% (or 0.01), each recombinant gamete has a frequency of 0.005 (since there are two possible recombinant gametes).

The parental gametes will appear with a frequency of 0.99, each of them with a frequenct of 0.495.

b) Normal man (SA/SA) X woman with anemia and albinism (sa/sa)

Daughter: SA/sa

The gametes she can produce are:

  • Parentals: SA (0.495) and sa (0.495)
  • Recombinant: Sa (0.005) and sA (0.005)

c) Son (Sa/sA) X daughter (SA/sa)

In order for them to have an albino child with anemia (sa/sa), the gamete sa from each parent had to be produced and fused into the zygote. Since the production of gametes by each parent is an independent event, the probability of having sa/sa offspring can be calculated as:

Prob. sa gamete son × Prob. sa gamete daughter = 0.005 × 0.495 = 0.0025

The probability that a child of theirs will be an albino with sickle-cell anemia is 0.25%

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