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Vladimir79 [104]
2 years ago
13

Assume that you are trying to insert a gene into a plasmid. Someone gives you a preparation of genomic DNA that has been cut wit

h restriction enzymes X. The gene you wish to insert has sites on both ends for cutting by restriction enzyme Y. You have a plasmid with a single site for Y, . but not for X. Your strategy should be to a.insert the fragments cut with restriction enzyme X directly into the plasmid without cutting the plasmid. b.cut the plasmid with restriction enzyme X and insert the fragments cut with restriction enzyme Y into the plasmid. c.cut the DNA again with restriction enzyme Y and insert these fragments into the plasmid cut with the same enzyme. d.cut the plasmid twice with restriction enzyme Y and ligate the two fragments onto the ends of the DNA fragments e.cut with restriction enzyme X. cut the plasmid with restriction enzyme X and then insert the gene into the plasmid.
Biology
1 answer:
DENIUS [597]2 years ago
3 0

Answer:

Cut the DNA again with restriction enzyme Y and insert these fragments into the plasmid cut with the same enzyme.

Explanation:

Plasmid may be defined as the extra chromosomal DNA present in bacteria that can replicate independently of the chromosomal DNA. Plasmid can be inserted from bacteria to another bacteria.

The restriction enzymes are used to cut the plasmid and the insertion of the desired gene into the plasmid to obtain the desired product. The plasmid do not have insertion site for X as shown in the question. This can be overcome if the DNA is cut again  with Y enzyme as this will create the common site for plasmid and the DNA fragments that contains the portion of X gene can be inserted into plasmid.

Thus, the correct answer is option (c).

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m_a_m_a [10]

Answer:

The children that the couple may have a 25% chance of having cystic fibrosis and 75% chance of being carriers for the CF gene

Explanation:

We know that both parents are in the same situation. Let's start for example with the mother. She has one parent and one sibling with cystic fibrosis (CF). So, supposing that the allele of the gene that causes the disease is <em>"f"</em>,  as the disease is autosomal recessive, both, the parent and the sibling have the genotype ff.

Also, knowing that the other parent has a normal phenotype, but the sibling has CF, we deduce that the other parent of the mother must have a <em>Ff</em>.

So, <em>despite the mother's phenotype is normal, her genotype is Ff</em> (see the Punnett square 1) Couple's parents in the attached image). In other words, she is a carrier for the CF gene.

As the parents' situation is analogous, <em>we can deduce that the father also has a genotype Ff  </em>and is also a carrier for the CF gene.

Now, if we look at the Punnett square for them  (see the square 2) Couple in the attached image), we see that all their children will always have at least one copy of the CF allele, <em>f</em>.

More specifically, their children have 25%  chance of having cystic fibrosis and 75% chance of having a normal phenotype but being carriers of the CF gene.

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Why do scientists use the scientific method
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