Question

6. Consider the enzyme HindIII cutting a DNA molecule of 40% GC

a)  What sequence does this enzyme recognise?

b)  What proportion of this DNA is made up of adenine bases?

c)  How often would HindIII cut this molecule (in bp)?

d)  If this DNA molecule was 21,000 bp in length, how many fragments would be produced on average?

Could you please explain your reasoning for the answers so that i may actually understand it. so far I know HindIII enzyme recognises AAGCTT, but thats the extent of my understanding of what is actually being asked of me.

Answer 1

Q76: (a) Hind Ill recognises the sequence " AA G CTT". And it cuts at SAAG CTT? 3 T T C G AAS (b) Gaven, "GC" is 40% of the DNA That means "AT" is 60% of the DNA as, %Get AT 40 = 100%. w Art's So in the DNA I No. of A = No. of t . 800, Now, if we replace I with "A" then, 2A = 60% >> A = 30%. Thus, 30% of DNA is made up of bases, A

(C). For efficiently digestion efficiently digestion by Hindu Least 3 bp are requireg at end. Means at both end there should be 3 bp. extra nucleotides for efficient cut by Hindu 51 x Y Z AA GETI и стг/ хүz ? ZYX TT G C AA ZYX 3 - A 3bpi. lle 3 bpm > so, in the Diva Hindu require, restriction digestion will involve. at least 12 bp => This 12 bp means 12% of the nucleotide bp are involved and Total is 100%. y so, in every (126p) the Hind IT will cut the DNA. A

BM we (d) The DNA molecule is 21000 bp. so, Number of cuts that we will - get by Hind Ih is? - 21000 bp -> 1750 wts. 12 bp C minimumored required Wehen, 1 cut is made we get 2 fragments - Similarly, After 1750 wts we will get 1751 fragments . (*) This calculation is gone assuming that Hind II restriction site is present through out the DVAG cases