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erastova [34]
2 years ago
7

You have a 15kb plasmid. EcoRI cuts the plasmid into 7kb and 8kb pieces, and BamHI cuts the plasmid into 1kb and 14kb pieces. If

you do a double digest with both EcoRI and BamHI on your plasmid, which of the following band patterns are possible on your gel?a. Bands at 1kb, 6kb and 8kb.b. Bands at 1kb, 7kb and 8kb.c. Bands at 1kb, 6kb and 14kbd. Bands at 7kb, 8kb and 14kb.
Biology
1 answer:
Harman [31]2 years ago
5 0

Answer:

a. Bands at 1kb, 6kb and 8kb

Explanation:

The EcoRI and BamHI are the restriction enzymes which cut the DNA sequence especially a plasmid at specific sites called the restriction sites.

The restriction enzymes produces bands of specific length therefore these restriction enzymes are used to estimate the approximate length of the DNA.

In the given question, the

1. EcoRI- produces two strands of 7 kb and 8 kb

2. BamHI- produces two strands of 1kb and 14kb

This shows that the length of DNA sequence is 15kb

But when the DNA strand are digested with both the enzymes simultaneously then it will produce three bands as:

i) 14 kb can be broken down in 2 bands of 6 kb and 8 kb

ii) 1 kb band is already produced by the Bam HI.

This shows that 1+6+8= 15 kb

Thus, Option-A is correct.

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Yellow smooth  - 9

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Green smooth- 3

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true breeding smooth green plant ( RRGG) and true breeding wrinkled yellow (rrgg)

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F1 offspring will be RrGg , Thus all F1 offspring will be heterozygous smooth and yellow.  

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F2 Generation –  

RrGg x RrGg

Gametes will be RG, Rg, rG, rg

RG         Rg          rG           rg

RG RRGG RRGg RrGG RrGg

Rg RRGg RRgg RrGg Rrgg

rG RrGG RrGg rrGG rrGg

rg RrGg Rrgg rrGg rrgg

R is dominant over r and g is dominant over G

Genotypes are –  

RRGG - 1  (Smooth Green)

RRGg-2  (Smooth yellow)

RrGG-2  (Smooth Green)

RrGg-4  (Smooth yellow)

RRgg- 1  (smooth yellow)

Rrgg – 2  (Smooth yellow)

rrGG – 1  (wrinkled Green)

rrGg – 2  (Wrinkled yellow)

rrgg – 1  (wrinkled yellow)

Yellow smooth  - 9

Yellow wrinkle - 3

Green smooth- 3

Green Wrinkle - 1

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