Hey everyone, this is a super simple question.
THIS IS FACTUAL. I UNDERSTAND THIS PART:
So C forms a triple hydrogen bond with G, hence when the DNA ds is rich with CG pairing, it'll require higher temperatures to denature (i.e. the double stranded DNA will then denature at a higher temperature than would an AT rich double strand of DNA).
AND THIS IS MY QUESTION:
So why would a SINGLE STRAND formed of lots of Cs and Gs require higher temperatures to denature? (i.e. the base composition on this single strand is mostly be made of C and G).
Saying this implies that something inherently within the C or G dNTPs makes them require higher heat. BUT I thought the whole point behind Cs and Gs requiring higher temperatures to denature was the HYDROGEN BONDING occurring between them i.e. necessitating that they be double stranded, and not single stranded like what the question/answer was about!
If you're confused by my question, this is from an old EK passage - EK 1001's, chapter 2, passage 204 (page 27), question #147, with answers on p. 198. Thanks a ton.
THIS IS FACTUAL. I UNDERSTAND THIS PART:
So C forms a triple hydrogen bond with G, hence when the DNA ds is rich with CG pairing, it'll require higher temperatures to denature (i.e. the double stranded DNA will then denature at a higher temperature than would an AT rich double strand of DNA).
AND THIS IS MY QUESTION:
So why would a SINGLE STRAND formed of lots of Cs and Gs require higher temperatures to denature? (i.e. the base composition on this single strand is mostly be made of C and G).
Saying this implies that something inherently within the C or G dNTPs makes them require higher heat. BUT I thought the whole point behind Cs and Gs requiring higher temperatures to denature was the HYDROGEN BONDING occurring between them i.e. necessitating that they be double stranded, and not single stranded like what the question/answer was about!
If you're confused by my question, this is from an old EK passage - EK 1001's, chapter 2, passage 204 (page 27), question #147, with answers on p. 198. Thanks a ton.