Respuesta :
Answer:
In sample A,
As A comprises 21.1% thymine, so A = T = 21.1%
Also, A+T = 42.2%
Therefore, G+C = 57.8%
So, G = 57.8/2 = 28.9%
Also, G = C = 28.9%
In sample B,
Sample B comprises 27.1 % thymine, therefore A = T = 27.1%
Also, A+T = 54.2%
Thus, G+C = 45.8%
So, G = 45.8/2 = 22.9%
Also, G = C = 22.9%
Since, the content of GC is more in the sample A, that is, 57.8 %. Thus, the sample A exhibits higher temperature to denature in comparison to B, which is 45.8%.
Answer:
Sample A contains 21.1% thymine which means it contains the same percentage of adenine.
So, the sample contains 21.1% adenine and 21.1 % thymine. The sum of both of them is 21.1+ 21.1= 42.2 %.
The rest of the sample contains guanine and cytosine. The amount of guanine and cytosine is 100- 42.2=57.8 %. The guanine percentage is 28.9% and same is the cytosine.
The sample B contains 27.1% of thymine which means adenine percentage is also the same. The total of adenine and thymine is 54.2%
The guanine and cytosine percentage is 45.8 % which means 22.9% each.
The GC content in the sample B is 45.8% and AT content is 54.2%
The GC content in the Sample A is 57.8% and AT content is 42.2 %
The sample B will require more energy to denature the sample than sample A because it has more GC content in it. This is because the GC has 3 hydrogen bonds in between them and AT has two hydrogen bonds between them.