Unit 15- Proteins

Question 1

1 gene, 1 enzyme hypothesis

Answer 1

• gene function by encoding enzymes, and each gene encodes a sep enzyme
• aka 1 gene, 1 polypep hypothesis

Question 2

Beadle and Tatum exper

Answer 2

• isolated auxotrophic mutants in Neurospora
• iradiated spores and placed on complete medium
• moved to min medium where only auxotrophic survive
• moved nonauxo to other mins w/ 1 amino acid. arginine was mutant as only place they survived

Question 3

structure and function of proteins

Answer 3

• proteins are polymers consisting of amino acids linked by peptide bonds
• AA seq is primary structure
• structure folds to create secondary and tertiary structures.
• 2+ polypeptide chains associate to form quaternary structure.

Question 4

dehydration reaction

Answer 4

• removes H2O to form peptide bond between AA

Question 5

codons

Answer 5

a triplet RNA code
- 64 possible codons
(3 stop and 61 sense codons)

Question 6

degenerate code

Answer 6

AA may be specified by more than 1 codon

Question 7

synonymous codons

Answer 7

codons that specify the same AA

Question 8

isoaccepting tRNAs

Answer 8

diff tRNAs that accept the same AA but have diff anticodons

Question 9

wobble hypothesis

Answer 9

base at 5’ end of tRNA anticodon can pair with several diff bases in a codon (only at 3rd base)

Question 10

reading frame

Answer 10

way in which the seq can be read in groups of three. Each diff 3 part seq can encode a diff AA seq

Question 11

nonoverlapping

Answer 11

a single nucleotide may not be included in more than 1 codon

Question 12

translation of mRNA

Answer 12

• Location: ribosome
• N terminus: amino acid end at end of chain
• C terminus: carboxyl end in the robosome

Question 13

binding of AA to tRNA

Answer 13

Aminoacyl-tRNA synthetase: loads AA onto tRNA (tRNA charging)
- determines specificity between an AA and its tRNA
- 20 diff types (like 20 AA)

CCA: AA acceptor stem at 3’ end

Question 14

initiation

Answer 14

• initiation factor: IF-3 (and IF-2, and IF-1)
• TRNA loaded with N-formylmethionine attached = fmet-tRNA
• Energy molecule: GTP

Question 15

initiation recognition seqs

Answer 15

The Shine-Dalgarno consensus seq in bacterial cells is recognized by the small unit of ribosome

Question 16

steps of initiation

Answer 16

1. If-3 binds to the small subunit
2. small subunit binds to mRNA
3. tRNA charged with N-forylmethionine forms a complex with F-2 and GTP
4. bings to intiation codon and IF-1 joins small subunit
   5, initiation factors dissociate and GTP hydrolized to GDP
5. large subunit binds

Question 17

elongation

Answer 17

-Elongation factors: Tu, Ts, G (EF-Tu; EF-Ts; EF-G)
- exit site = E
- Peptidyl site = P
- Aminoacyl site = A

Question 18

steps of elongation

Answer 18

1. met tRNA in P site of ribosome
2. EF-Tu, GTP, and charged tRNA form a complex and enter the A site
3. GTP –> GDP and EF-Tu-GDP released (regenerates)
4. peptide bond forms between AA and P site tRNA releases AA
5. ribosome moves down mRNA and repeats
   (tRNA in P site now in E site and removed. tRNA in A site now in P site)

Question 19

Termination

Answer 19

release factors bind to termination codon, causing release of:
- peptide from last tRNA
- tRNA from ribosome
- mRNa from ribosome

Question 20

steps of termination

Answer 20

1. reach stop codon
2. RF-1 attatches to A site and RF-3 with GTP binds to ribosome
3. everything released and GTP hydrolized

Question 21

polribosome

Answer 21

an mRNA with several ribosomes attached

Question 22

quality control

Answer 22

• messenger RNA surveillance (detect and deal with errors in mRNA)
• posstrnaslational modifications (folding: molec chaperones)
• nonstandard AA synthesis