nucleic acids Flashcards
Explain how the data of four bases support the concept of complementary base pairings
- % A is approximatley equal to % T and % G is approximatley equal to % C
- due to complementary base pair on antisense strand
Chicken anaemia virus (CAV) is a single stranded DNA virus. Expalin how the resulsts from CAV would differ from a double stranded DNA helix
- % A won’t equal %T and % G won’t equal %C
- as there are no complementary base pairing due to being single stranded
The double helix takes 3.4 nm make one complete turn and the base pairs are 0.34 nm apart how many base pairs would you expect in three complete turns of the helix
- 3.4/0.34 = 10
- 10 x 3 = 30
Describe how a polymer of RNA would be different from a poymer of DNA
- RNA is single stranded whereas DNA is double stranded
- Rna would have the base uracil whereas DNA has the base thymine
name the three components of DNA
- phosphate group
- nitrogenous base
- deoxyribose sugar
What bands would show for the second generation frown on N14 medium for Meselshon Stahl experiment
- a light band of DNA
- intermediate band of DNA
Draw a diagram of tRNA
- clover leaf shape
- anticodon on bottom
- show some hydrogen bonds
Meselsohn and Stahl concluded that bacteria replicated by semi conservative replication. How do the results from the first generation support this conclusion
- intermediate weight band seen
- indicating the DNa molecule contained one strand from the heavy parent DNA and one newly syntheised light DNA strand
Explain how your results for the second generation rule out dispersive replication
- The DNA extracted formed an intermediate weight band halfway up the tube and a lighter band towards the top of the tube
- because half of the DNA was intermediate weight and half light, this rules out disperive replication
What would you expect to see if the experiment was continued for a further two generations using the N14 medium
- the proportion of light DNA was intermediate weight and half light , this rules out dispersive replication
Explain why it would not be possible to estimate tge proportion of missing nucleotides if the organisms contained single stranded DNA
- Complementary base pairing rule would not apply
Scientissts have sugested that bacteria living in hot springs have a much higher proportions of guanine and cytosine than bacteria living in more temperate environments. using your knowledge of DNA structure , suggest a reason for this observation
- three hydrogen bonds between G to C only two between T to A
- more hydrogen bond in double helix would require more energy to break and denature molecule
Where in eukaryotic cells does transcription take place
nucleus
Where in eukaryotic cells does translation take place
ribosomes
Describe the steps involved in synthesising haemoglobin in eukaryotic cells
- DNA acts as a template for the production of mRNA
- DNA helicase unwinds the double helix and seperates the two strands by breaking the hydrogen bonds between the complementary base pairings
- Free RNA nucleotides pair by RNA polymerase
- The cycle repeats until it reaches the stop codon
- mRNA strands leave the nucleus via the nuclear pores and moves to the ribosomes
- In eukaryotes, pre - mRNA is spliced to remove the itron non coding region before passing to the ribosomes
- translation involves the transfer RNA
- exposed bases called the anticodon which are compllementary to mRNA codon
- attatchment to the relevant amino acid to the attatchment site i called amino acid activation known as polypeptide
- Translation produces a polypeptide but further modification is needed in order to produce a protein with a secondary,tertiary, quanternary structure
- modificaition occur in the golgi body
- to form haemoglobin two alpha chains and two beta chain need to be asembled together with iron as prosthetic group
