Molecular Genetics

Question 1

point mutations (silent, missense, nonsense)

Answer 1

• silent mutation - mutation does not change which amino acid is encoded, as the mutation happens to code for the same amino acid, due to redundancy
• missense mutation - codes for the wrong amino acid;
• nonsense mutation - changes codon to a stop codon

Question 2

Griffith’s experiments 1928

Answer 2

Genetic traits are molecules

Griffith was looking for a vaccine for pneumonia by studying mice infected with 2 different strains…
he proposed that cells contained a non-living ‘TRANSFORMING PRINCIPLE’ that could transform one cell into another.
Dead matter without a living spirit can change the heritable nature of a living organism.
Genetic traits are encoded in molecules (but it was unclear if this was nucleic acids or proteins)

Question 3

Avery, MacLeod and McCarty’s experiment 1944

more experiments with mice and pneumonia

Answer 3

DNA is the molecule that carries info ab gen traits (but no one cares)

Aimed to identify transforming principle identified by Griffith;
Grew large batches of S-strain bacteria and tested which of different cellular components was the transforming principle.
Treated cell extracts with:
Proteinase - transformation still occurs
Nuclease - no transformation
Implies that DNA is the transforming principle
BUT - No-one believed them!

Question 4

Hershey-Chase experiment

the last thing you want in your Hershey-Chase bar is bacteriophage viruses…

Answer 4

DNA is the molecule that stores info ab gen traits (everyone cares now)

8 yrs after Avery et al
Bacteriophages (viruses) we created to have radioactive Sulfer (found in proteins, not nucleic acids) and the other stream to have radioactive Phosphorus (found in nucleic acids and not so much in proteins)
Only the radioactive Phosphorus was passed on, meaning nucleic acids (DNA) was the molecule of inheritance

Question 5

Chargaff’s rules (1950)

Answer 5

composition of DNA varied between species, BUT

As and Ts and Gs and Cs found in same molar amounts for all species; ie there is complementary base pairing

Question 6

Nucleotide structure

Answer 6

Nucleotide = nucleoside + monophosphate
nucleoside = nitrogenous BASE + pentose sugar (ribose OR deoxyribose)
pentose sugars have 5 carbons, referred to by prime designation, written as an apostrophe: e.g. 1’ (one prime) - 5’ prime
The nitrogenous base is bonded at 1’
3’ and 5’ are significant in the structure of the DNA and RNA polymers

Question 7

Nucleotide bases

Answer 7

Purines - these are larger;
- A and G
Pyrimidines - these are smaller;
- T and C

Question 8

Franklin and Wilkins (early 1950s)

Answer 8

x-ray crystallogram experiment suggested the DNA structure was helical

Question 9

Watson and Cricks revelations

Answer 9

Used X-ray data from Rosalind Franklin (Maurice Wilkins and Raymond Gosling)
Knowledge of chemical composition of DNA
Built models
Had no real evidence for their structure!
their epiphany was putting the strands antiparallel

Question 10

central dogma

Answer 10

hierarchy of DNA, RNA, and proteins
DNA stores the genetic information (genotype), RNA transcribed the genetic information, proteins are translated from the RNA resulting in a phenotype

Question 11

polymers, monomers and bonds of biological polymers

Answer 11

DNA, deoxyribosnucleotide, phosphodiester bond
RNA, ribonucleotide, phosphodiester bond
Protein, amino acid, peptide bond

Question 12

Beedle and Tatum’s experiments (1940s)

Answer 12

found that proteins are the link between genotype and phenotype; one gene = one protein

Question 13

phenotypic expressions of proteins

Answer 13

Proteins can be the molecules visible in the phenotype.
e.g.: beta globin
Proteins can be enzymes that make molecules visible in the phenotype.
e.g.: tyrosinase
Proteins can be the signals to make visible changes in phenotype.
e.g.: SRY gene
Different alleles produce different proteins.
Different proteins produce different phenotypes.

Question 14

Nitrogenous bases in DNA versus RNA

Answer 14

G - Guanine; a purine; forms hydrogen bonds with C
C - cytosine; a pyrimidine; bonds with G
A - adenine; a purine; bonds with T
T - thymine; a pyrimidine; bonds with A

In RNA, thymine is replaced by Uracil (U)

Question 15

differences between DNA and RNA

Answer 15

Thymine base in DNA; Uracil base in RNA
DNA has a Hydrogen group at 2’ locus on pentose sugar; RNA has a Hydroxyl group at this same locus
- the lack of this extra oxygen in DNA allows it to be wound very tightly
Pentose sugar of deoxyribose in DNA; ribose in RNA
RNA is single stranded

Question 16

dNTP and rNTP

Answer 16

nucleotide monomers of DNA and RNA respectively
deoxyribonucleoside triphosphate
ribonucleoside triphosphate
three phosphates at the phosphate group - carries energy for bonding

Question 17

polymerisation (of RNA)

Answer 17

directional, from 5’ end to 3’ end
energy for bonding stored in the triphosphate group
monomers join onto the 3’ hydroxyl group
catalysed by RNA polymerase II

Question 18

transcription

Answer 18

INITIATION, ELONGATION, TERMINATION
INITIATION
promoter - a DNA sequence at the start of a gene (including “TATA box”)
transcription factors - proteins that bind to the DNA at the promoter site, telling the RNA polymerase II to transcribe now
RNA polymerase II - an enzyme that catalyses the polymerisation of RNA
ELONGATION - addition of complementary rNTPs
transcription sequence - the sequence of DNA to be transcribed
terminator - the sequence on the DNA that signals the end of the gene
TERMINATION - Release of RNA polymerase and completed RNA from the DNA template at the terminator

Question 19

codon

Answer 19

a sequence of 3 bases on a strand of RNA that corresponds with the translation of specific amino acids, start codons and stop codons

Question 20

Enzymes of DNA synthesis (replication)

Answer 20

Jobs to be completed:

1. s.t. to separate strands;
2. s.t. to initiate synthesis;
3. s.t. to make it bidirectional…

(first step) - Primase (an RNA polymerase) lays down RNA primers (“tails”), complemetary to short sequences of the template strand;

• DNA Polymerase III catalyses DNA polymerisation in the 5’ to 3’ direction; ALSO, it “erases” in the 3’ to 5’ direction, ie it has a proofreading function and can go backwards, can’t initiate AND can’t tell the diff btwn DNA and RNA;
• Helicase “unzips” the DNA, ie separates the H-bonds;
• DNA polymerase I removes RNA primers and fills in with DNA;
• Ligase joins up any gaps in sugar phosphate backbone between DNA laid down by Pol III and Pol I, using ATP.

Question 21

DNA replication - leading and lagging strands

Answer 21

both strands are both;
helicase unzips in both directions;
Pol III can follow helicase on one strand (the leading strand), where replication is continuous;
Pol III cannot follow helicase on the other strand because of directionality, hence synthesis is discontinuous on the lagging strand.

Question 22

summary of DNA replication (including Okazaki fragments)

Answer 22

• Primase synthesises an RNA primer
• DNA polymerase III extends the primer
• Completion of an Okazaki fragment
  Release of DNA polymerase III
• Synthesis of the next Okazaki fragment by DNA polymerase III
• Replacement of the RNA primer by DNA polymerase I
• Joining the backbone by DNA ligase

Question 23

summary of enzymes in DNA replication (initiation, extension, and fixing the lagging strand)

Answer 23

INITIATION:
- Helicase
- Primase
EXTENSION:
- DNA Pol III
FIXING LAGGING STRAND:
- DNA Pol I
- Ligase

Question 24

accuracy of DNA replication AND mutation rate

Answer 24

High accuracy due to:
- proof reading by DNA polymerases and DNA repair mechanisms that happen after DNA replication

Error rate in humans is equivalent to 50-200 new mutations in humans each generation , most in non-coding DNA (1-2 mutations in coding DNA)

mutation rate is a specific constant that appears to have been selected for – if there were too many, most offspring would be at a selective disadvantage; too few and the species would not be able to adapt to change

Question 25

3 types of RNA involved in protein synthesis (translation)

Answer 25

mRNA: encodes a protein

tRNA: adaptor molecules that convert a sequence of codons to a sequence of amino acids

rRNA: a component of the ribosome

Question 26

protein synthesis (3 stages)

Answer 26

INITIATION
Assembly of the ribosome on the mRNA at the start codon

ELONGATION
A three step cycle:
- Entry of tRNA into the A site
- Formation of peptide bond 
- Translocation (movement of the ribosome by one codon)

TERMINATION
Release of the completed protein at a stop codon

Question 27

Signal selection and amplification

Answer 27

Transcription:
- Signal Selection:
Promoter determines where, when and how many transcripts
- Signal Amplification:
multiple mRNA molecules synthesised from one gene

Translation:
Signal Amplification
Multiple protein molecules synthesised from one open reading frame