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Phoenix [80]
3 years ago
8

A paleontologist has recovered a bit of tissue from the 400-year-old preserved skin of an extinct dodo (a bird). To compare a sp

ecific region of the DNA from a sample with DNA from living birds, which of the following would be most useful for increasing the amount of dodo DNA available for testing?
a. SNP analysis
b. polymerase chain reaction (PCR)
c. electroporation
d. gel electrophoresis
Biology
1 answer:
AlladinOne [14]3 years ago
8 0

Answer:

b. polymerase chain reaction (PCR)

Explanation:

  • PCR is the technique that generates millions of copies of DNA starting from a small amount of template.

In this reaction, a molecule of double stranded DNA is denatured and the chains separated. Then, the single strands of DNA are hybridized with a small complementary fragment of DNA called a primer, which is then extended by an enzyme called DNA polymerase by adding complementary nucleotides to the template strand. The result will be two double stranded DNA molecules. If you repeat the process many times, <u>you can amplify the original DNA material exponentially</u>, and the final PCR product will be millions of DNA molecules that can be used for testing.

  • A SNP analysis is a method that allows scientists to detect variations of a single nucleotide in the DNA sequence of different individuals.
  • Electroporation is a technique that creates temporary pores in a cell membrane, allowing the delivery of drugs or DNA to a cell.
  • Gel electrophoresis is a method to analyze the PCR product in which the DNA molecules are separated by size.

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Answer:

The genotypic frequency = 1:1

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Explanation:

Given that:

The allele → R = Red beetles

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Since the gene color shows a codominant allele

The Red Beetle = RR

The blue beetles will be = BB

The heterozygous beetle will be = RB

∴

The punnet square showing the crossing of RB × RR is:

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R                RR               RB

R                RR               RB

The result shows that we have two red beetles and two heterozygous beetles.

Hence;

The genotypic frequency = 1:1

The phenotypic frequency = 1:1

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