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klio [65]
3 years ago
13

Meselson and Stahl cultured E. coli for several generations in a medium with a heavy isotope of nitrogen, 15N. They transferred

the bacteria to a medium with a light isotope of nitrogen, 14 N. After two rounds of DNA replication, half the DNA molecules were light (both strands had 14N) and half were hybrids (15N-14N). What did the researchers conclude from these results?
Biology
1 answer:
Sidana [21]3 years ago
8 0

Answer:

DNA replication follows semi conservative mode of replication

Explanation:

The results of Meselson and Stahl's experiments can be explained by semi conservative mode of DNA replication. According to it, the two DNA strands separate and both are used for synthesis of new DNA strands such that the new DNA molecule has one parental and one new strand.

In the experiment, originally the medium was 15N so all the DNA molecules had 15N. Later the medium was changed to 14N. In first round of replication in new medium, the two 15N strands in a DNA molecule separated and formed new strands using 14N. So all the DNA molecules had one 15N and one 14N strand.

In the second round of replication, the 15N-14N hybrid DNA molecule separated again. Now, 15N gave rise to 14N strand and 14N strand also gave rise to 14N strand such that half of the DNA molecules were 15N-14N hybrids and half were 14N-14N. Hence, researchers concluded that DNA replicated according to semi conservative mode of replication.

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True or False?<br> Dominant alleles will always increase in frequency. Explain your answer.
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A man and a woman are both carriers for two autosomal recessive disorders, PKU (chromosome 12) and cystic fibrosis (chromosome 7
ValentinkaMS [17]

Answer:

Pr= (3/16)

Explanation:

It was stated that both the man and the woman are both carriers for two autosomal recessive disorders, PKU (chromosome 12) and cystic fibrosis (chromosome 7).

∴ let Aa¹Bb° represent the traits in the Man

Where;

Aa¹ = carrier of PKU

Bb° = carrier of cystic fibrosis

Also, let Cc¹Dd° represent the traits in the woman

Where;

Cc¹ = carrier of PKU

Dd° = carrier of cystic fibrosis

Now, if Aa¹Bb° self-crossed, we'll have the F1 progeny as AB, Ab° , a¹B and a¹b°

also, if Cc¹Dd° self-crossed, we have CD, Cd° , c¹D and c¹d° as their F1 progeny

In the F2 generation, the dihybrid cross between the F1 generations will be:

AB, Ab° , a¹B, a¹b°    × CD, Cd° , c¹D, c¹d°

ACBD, ACb°D, a¹CBD, a¹Cb°D

ACBd°, ACb°d°, a¹CBd°, a¹Cb°d°

Ac¹BD, Ac¹b°D, a¹c¹BD, a¹c¹b°D

Ac¹Bd°, Ac¹b°d°, a¹c¹Bd°, a¹c¹b°d°

Only (a¹CBD, Ac¹BD, a¹c¹BD)  shows the probability that she will have PKU but not CF.

6 0
3 years ago
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