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Mashcka [7]
3 years ago
14

A paleontologist has recovered a bit of tissue from the 400-year-old preserved skin of an extinct dodo bird. The researcher woul

d like to compare a specific region of the DNA from the sample with DNA from living birds. Which of the following would be most useful for increasing the amount of dod DNA available for testing?
a)RFLP analysis

b)electroporation

c)polymerase chain reaction (PCR)

d)gel electrophoresis

e) Southern blotting
Biology
1 answer:
slava [35]3 years ago
8 0

Answer:

c)polymerase chain reaction (PCR).

Explanation:

Because the paleontologist recovered only a bit of tissue  and it is very old, it is very likely that the DNA in the sample is very small and part of it is degraded. Anyway, the paleontologist must first amplify the DNA sample to obtain many identical copies of the specific region of the DNA they want to compare. the above is done through a polymerase chain reaction (PCR).

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A signal transduction pathway is initiated when a _____ binds to a receptor.
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Signal molecule

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Some snail species' shells can have right-handed direction of shell coiling or left-handed direction of shell coiling. The direc
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The cross between Dd egg-snail and a dd s snail mate will result in snails with genotype 1/2 Dd:1/2 dd and phenotype all right-coiling.

<h3>What is a genotype?</h3>

The genotype refers to the allele combinations that can be found for one or more loci in an individual.

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Learn more about genotypes here:

brainly.com/question/22117

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3 0
3 years ago
Read 2 more answers
In a population of Mendel's garden peas, the frequency of the dominant A (yellow flower) allele is 80%. Let p represent the freq
DENIUS [597]

Answer:

If the frequency of the dominant allele in the pea population is 0.8, the genotype frequencies in the population would be 0.64 AA, 0.32 Aa, and 0.04 aa.

Explanation:

Hardy-Weinberg equation tells us that:

p2 + 2pq + q2

p2 is the frequency of AA plants

2pq is the frequency of Aa plants, and

q2 the frequency of aa individuals.

A allele in the population is 80%, the frequency is 0.8.

p represent the frequency of A allele. p is 0.8

Therefore calculating for q ( frequency of a allele).

p + q = 1.0

q = 1.0 − p

q = 1.0 − 0.8

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Now we can calculate the predicted frequencies of the different genotypes, remembering that p2 is the frequency of the AA genotype, 2pq is the frequency of the Aa genotype, and q2 the frequency of aa genotype.

p2 = 0.8 x 0.8 = 0.64

2pq = 2 x 0.8 x 0.2 = 0.32

q2 = 0.2 x 0.2 = 0.04.

Thus, we would expect to see genotype frequencies of 0.64 AA, 0.32 Aa, and 0.04 aa in the pea plants.

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