Question

1. During the process of electrophoresis, the ________ functions like a molecular sieve, separating the samples according to their size.

A) agarose gel

B) sample mixture

C) positively charged electrode

D) negatively charged electrode

2. The restriction enzyme SacI has a recognition sequence of GAGCT^C, where the caret (^) indicates the cut site. Examine the DNA molecule below.

AGAGCTCAGTCGAGAGCTCAGATCGATAGGAGCTCAGATCTCGATCACCTC
TCTCGAGTCAGCTCTCGAGTCTAGCTATCCTCGAGTCTAGAGCTAGTGGAG

How many separate molecules of DNA would you end up with if you treated the above DNA molecule with SacI?

A) four

B) three

C) five

D) two

3. The restriction enzyme BamHI recognizes the DNA sequence GGATCC and always cuts between the two G nucleotides. How many bases long is the sticky end of a DNA molecule that has been cut with BamHI?

A) four

B) three

C) five

D) two

Answer 1

1.Option A

Agarose is a component of agar which is produced after removal of agaropection from the agar. It is extracted from an algal species called as red seaweed. It consists of monomers of agarobiose. Agarose is commercially available for electrophoresis experiments and for microbiology experiments for plating the bacteria. It is porous in nature and thus can act as a sieve causing the molecules like proteins and DNA to move according to their sizes (molecules of lower size move faster than the heavier ones).

2. Option C

AGAGCT ^

CAGTCGAGAGCT^

CAGATCGATAGGAGCT^

CAGATCTCGATCACCTC
TCTCGAGTCAGCTCTCGAGTCTAGCTATCCTCGAGTCTAGAGC^

TAGTGGAG

These are the cutting sites of the enzyme Sac I giving rise to 5 fragments.

3. Option A

5' GGATTC 3' is the recognising sequence for Bam HI enzyme and it cuts at the position between GG. Now let's see the cutting mechanism. As DNA is double stranded molecule, it will have a complementary sequence on the other strand.

5' G G A T C C 3'

3' C C T A G G 5'

Now after it is cut using the restriction enzyme Bam HI, it looks like:

5' G ^ G A T C C 3'

3' C C T A G ^ G 5'

Forming ends like these:

5' G G A T C C 3'

3' C C T A G         G 5'

The ends that are free without base pairing but they have hydrogen bonds tha are free after they are cut by the restriction enzymes. In this case GATC and CTAG are the sticky ends hence they are of 4 base pairs long.