Question

1.Consider a plasmid (circular DNA) which has 1000 bp and a G+C content of 50%. How many phosphorous atoms are there?

2. Consider a plasmid (circular DNA) which has 1000 bp and a G+C content of 50%. How many nitrogen atoms are there?

3. Suppose the G-C content in the plasmid (1000 bp) was GREATER than 50%. Which (if any) of your above answers would change, and how? (increase/decrease/no change?)
a) # of phosphodiester bonds
b) # of H-bonds
c) # of purines
d) # of phosphorous atoms
g) # of nitrogen atoms

4. EcoRI is a 6-cutter restriction enzyme that creates 4-base, 5’ overhangs (“sticky ends”) and recognizes the sequence: 5' GAATTC 3'
Based on this information, EcoRI must break the phosphodiester bond between:

5. A particular double-stranded plasmid (circular DNA molecule) has exactly 3,300 bp (base pairs). If 1,715 of the bases are cytosine (C), how many of each of the other three bases are present?

6. Consider a circular dsDNA with 25% A and a random DNA sequence. If this molecule includes 100 copies of the sequence 5’ TGG 3’, approximately how many times should the sequence 5’ CTAT 3’ be represented?

7. A different dsDNA has 30% A and 100 copies of the sequence 5’ TGG 3’ Assuming a random DNA sequence, ~ how many times should the sequence 5’ CTAT 3’ be represented in this case?

8. The restriction enzyme SauIIIA recognizes the 4 bp sequence GATC, and the "6-cutter" enzyme BamHI recognizes the sequence GGATCC. Digestion of a particular bacteriophage genome produced a total of 160 SauIIIA fragments. Approximately how many BamHI fragments would you expect if:
A) the phage genome has a “G+C” content of 50%?
B) the phage genome is 80% G+C?

Answer 1

1. A circular plasmid with 1000 bp has 50% G-C content so the number of phospho diester bonds will be - 1000 for one strand(as its a circular DNA molecule, otherwise it would have been 999) , as the plasmid is double stranded so total number of phosphodiester bonds in this molecule is 2000, so total number of phospsorus molecules will be 2000.

2. As we know

A has - 5 nitrogen atoms.

T has - 2 nitrogen atoms.

G has - 5 nitrogen atoms.

C has - 3 nitrogen atoms.

This is a 1000 bp plasmid having 50% G-C content.

so number of G-C pairs - 500

number of A-T pairs -500

number of

A-250 , So Nitrogen number- (250*5)=1250

T-250, So Nitrogen number- (250*2)=500

G-250,So Nitrogen number- (250*5)=1250

C-250, So Nitrogen number- (250*3)=750

So, total number of Nitrogen atoms , (1250+500+1250+750) = 3750.

3. If the amount of G-C content was higher than 50%

Number of

a. Phosphodiester bonds - Will remain unchanged because the number of bp remains unchanged.

b. H-bonds - Will change ( increase, because A-T pairs have 2 H bonds and G-C pairs has 3 bonds)

c. Purines - Will not change because (A and G are purines and T and C are pyrimidines so whatever the base pair may be there is always one purine ad one pyrimidine, that is the basic rule of formin the double helix)

d. Phosphorus atoms - Will remain unchanged because the number of bp remains unchanged.

e. Nitrogen atoms - Will change (because although A and G both has 5 nitrogen molecules, C has 3 nitrogens while T has 2).

4. ECo RI must break the Phosphodiester bond between - 5' G-AATTC 3'

3' CTTAA-G 5'

So the bond is broken between G and A in the 5' to 3' strand and between A and G on the reverse strand.

5. Out of 3300 bp 1715 bases are cytosine, so the G-C content will be 1715 or 51%

So A-T will be 49% or 1671 So A - 1671

T-1671, G-1715.

6.