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sashaice [31]
3 years ago
15

Dolly was a female domestic sheep, and the first mammal to be cloned from an adult somatic cell, using the process of __________

________.
A) DNA transfer
B) cell transfer
C) oocyte transfer
D) nuclear transfer
Biology
2 answers:
ELEN [110]3 years ago
4 0
The answer is D)Nuclear transfer
mezya [45]3 years ago
4 0

Answer: The correct answer for the blank is-

D) nuclear transfer.

Dolly was a female domestic sheep, and the first mammal that was cloned using an adult somatic cell ( particularly a mammary gland cell from a Finn Dorsett sheep). She was produced using a technique, which is termed as somatic cell nuclear transfer.

In this technique, the nucleus from an adult somatic cell was transferred to an unfertilized oocyte ( which was first enucleated that is nucleus was removed). After this, unfertlized egg was activated using electrical stimuli, which initiated cell division and produced Dolly.

Thus, option D) nuclear transfer is the right answer.

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I say D because ribosomes make proteins and the endoplasmic reticulum does transport things around the cell

Explanation:

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5 0
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3 years ago
If the frequency of homozygous dominant is 60%, the frequency of heterozygous is 20%, and the frequency of homozygous recessive
Simora [160]
When a genetic population follows Hardy-Weinberg Equilibrium (HW), it states that certain biological tenets or requirements must be met. Given so, then HW states that 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 decimal. 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 = frequency of homozygous dominant individuals, 2pq = frequency of heterozygous individuals, and q^2 = frequency of homozygous recessive individuals.
So the problem states that homozygous dominant individuals (p^2) account for 60%, or 0.60. Thus the square root (sr) of p^2 = p or the dominant allele frequency in the population. So sr(p^2) = sr(0.60) -->
p = 0.775 or 77.5%
Homozygous recessive individuals (q^2) account for 20%, or 0.20. Thus sr(q^2) = q or the recessive allele frequency in the population. So sr(q^2) = sr(0.20) --> q = 0.447 or 44.7%
But since 44.7% + 77.5% = 122.2%, which is not equal to 1, we have a situation in which the allele frequencies do not match up, therefore this population cannot be determined using the Hardy-Weinberg Equation.
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