Question

How many Guanines are found in a dsDNA that has 700 hydrogen bonds and 10% Adenine?
a)195
b)206
c)200
d)210
please help me out guys?​

Answer 1

Answer:

hope you got it amigo

Adenines use two H-bonds and guanines use three H-bonds and there are total 700 H-bonds. Thus,

2x + 3y = 700

Also, 10℅ of total H-bonds are made of Adenine's. So,

2x = 10℅ of 700 = 70

Which means x = 35 & y = 210.

Where x = no. Of adenines and y = no. Of guanines.

And according to Chargaff's law, no. Of purines = no. Of pyrimidines or A=T & C=G. So this question does follow the rule and is correct.

Answer 2

Answer:

Here is your answer please make it as brainlyest

Explanation:

How many guanines are present if there are 700 hydrogen bonds and 10% of them are adenine? How does the Chargaff law apply here? Is it correct?

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From your question it seems unclear whether 10% of the bases are Adenine or if 10%of the total H-bonds are from Adenine.Anyway,I will answer it in both the cases.

Taking the first case,from Chargaff's law we know,no of A equal to no of T,and no of G equal to no of C.So,if A=10%of bases, then T=10% as well.A+T=20% of the bases.The remaining 80%=G+C

As from Chargaff's law,no of G equal to no of C,then in this DNA sample,G=C=40%of the bases. There are in total 700 H-bonds.So,taking the total no of bases in the DNA sample as n,we get,A=(10/100)n=n/10 and G=40%of n,I.e.2n/5

Now,as each A forms 2 H-bonds and each G forms 3 H-bonds,we can write,

(2xn/10)+(2nx3/5)=700

now solving we get,n=500,as G=40%of n,so G=200.

Coming to the next case,where 10%of the H-bonds are from A,the no of H-bonds from A are,10%of 700,I.e.70.The remaining 630are from each G forms 3H-bonds,you can now easily calculate the no of G needed to form 630and that's 210no of G.

Chargaffs laws are correct assuming the model of DNA being only the Watson Crick type I.e.B form of DNA, also Hugsteen pairing in DNA makes the structure deviate from the Watson Crick proposed form,so in these cases Chargaffs law might not be completely accurate. Pls please mark it