Difference between expression vector and cloning vector

1 answer:

8 0

Cloning vectors are the DNA molecules that can carry a foreign DNA segment into the the host cell.The vectors used in recombinant DNA technology can be plasmids, cosmids, bacteriophages etc.

Plasmids: Self replicating, circular, extra chromosomal DNA present in bacteria. Plasmids have only one or two copies per cell.

Bacteriophages: Virus infecting bacteria. Bacteriophages have high number per cell, so their copy number is also high in genome.

Cosmids: Hybrid vectors derived from plasmids which contain cos site of lambda phage.

Feature required to facilitate cloning into vector: Origin of replication, Selectable marker, unique restriction sites.
The purpose of cloning vector is often to make numerous copies of the inserted gene.

Expression Vectors: The cloning vector containing suitable expression signals to have maximum gene expression, is called expression vector.
The following expression signals are introduced into gene cloned vectors to get maximum expression:

Insertion of a strong promoter.

Insertion of a strong termination codon.

Adjustment of distance between promoter and cloned gene.

Insertion of transcription termination sequence.

Insertion of a strong translation initiation sequence.

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3 years ago

Find the recursive rule, explicit rule, and f(20)

DiKsa [7]

Answer:

Recursive:

$f(1)=35, f(n)=f(n-1)+10$

Explicit:

$f(n)=35+10(n-1)$

And the 20th term is 225.

Step-by-step explanation:

We have the sequence:

35, 45, 55, 65.

Notice that each subsequent term is 10 more than the previous term.

Therefore, our common difference is (+)10.

Recursive Rule:

The standard format for the recursive rule is:

$f(n)=a, f(n)=f(n-1)+d$

Where a is the initial term and d is the common difference.

From our sequence, we know that a the initial term is 35.

And as determined, our common difference d is 10.

Substitute. Hence, our recursive rule is:

$f(1)=35, f(n)=f(n-1)+10$

Explicit Rule:

The standard format for the explicit rule is:

$f(n)=a+d(n-1)$

Where a is the initial term and d is the common difference. So, let’s substitute 35 for a and 10 for d. Hence, our explicit formula is:

$f(n)=35+10(n-1)$

Now, let’s find the 20th term. We will utilize the explicit rule since the recursive rule can get tedious. Substitute 20 for n because we would like to 20th term. Thus:

$f(20)=35+10(20-1)$