Molecular Biology (C3) Flashcards

Question

Nonsense Mutation | type of point mutation

Answer 1

stop codon replaces a regular codon and prematurely shortens the protein

Answer 2

codon changed to new codon for same AA, no change in protein's AA sequence

Answer 3

addition of 1 or more nucleotides to DNA sequence

Answer 4

removal of nucleotides from DNA sequence

Answer 5

mutations that cause a change in the reading frame very serious can lead to premature termination of translation not all insertions/deletions are frameshift mutations, if insert/delete 1 whole codon or several whole codons, you add/remove AAs from polypeptide w/o changing reading frame

Answer 6

when a segment of a chromosome is reversed end to end | chromosome undergoes breakage and rearrangement within itself

Answer 7

when a segment of a chromosome is duplicated

Answer 8

result when recombination occurs b/w nonhomologous chromosomes can create gene fusion, where new gene product is made from parts of 2 genes that were not previously connected - common occurrence in many types of cancer can be balanced (no genetic info lost), or unbalanced (genetic info lost/gained)

Answer 9

mobile genetic elements within their genomes jumping genes all contain a gene that codes for a protein called a transposase (cut and paste activity), where it catalyzes that mobilization of the transposon (excision from donor site) and intergration into new genetic location (acceptor site) 3 types: 1) IS element 2) complex transposon 3) composite transposon can cause mutations if jump to important part of genome can affect gene expression or cause mutations can cause structural changes to chromosomes when they work in pairs (if chromosome has 2 transpoonson with same direction, transposons can line up beside each other so they are parallel, this causes chromosomal segment b/w them to loop around, recombination occurs b/w the transposon and this causes deletion of the DNA b/w the transposons, original chromosome completely loses the DNA segment b/w the transposons. segment of DNA that's lost takes 1 transposon w/ it and can jump back into genome somewhere else, causing chr rearrangement)

Answer 10

if a deletion removes the normal copy of the gene, the only remaining copy is the mutated version makes the locus hemizygous: only 1 gene copy in the diploid organism *diploid organisms have 2 copies of each gene, and generally a mutation in 1 copy is tolerated as long as the other copy of the gene is normal

Answer 11

a diploid organism has only a single functional copy of a gene, and this single copy is not enough to support a normal state

Answer 12

huge group of genetic diseases that involve disorders of metabolism most due to single mutation in single gene that codes for some sort of metabolic enzyme

Answer 13

driven by mutation accumulation | inherited or carcinogen exposure

Answer 14

Bad Bases (mismatched, oxidized, crosslinked, etc.): 1) mismatch repair pathway 2) nucleotide excision repair Brokem Chromosomes (physical damage, X rays): 3) homology-directed repair 4) non-homologous end-joining ``` DNA Rearrangement (transposons): 5) generally don't lead to repair mechanisms ```

Answer 15

during, or shortly after replication | parent strand is methylated, while daughter strand is not, can identify daughter/parent

Answer 16

can happen at any time in cell cycle (ideally before replication) remove bad base, replace with good base

Answer 17

must happen after replication (when the sister chromatid is present) use (identical) sister chromatid as template to fix broken chromosome

Answer 18

can happen at any time in cell cycle ligate broken ends together mutagenic b/c usually lose some bases can also result in translocations even poorly repaired chromosomes are better than a broken chromosome not as good as H-D R

Answer 19

rRNA: ribosomal mRNA: messenger tRNA: transfer hnRNA: heterogenous nuclear (preprocessed mRNA - initial transcript) miRNA & siRNA: micro, small interfering (complementary to RNA, binds to mRNA and prevents translation)

Answer 20

``` Similarities: - START site - 5'-3' direction - DNA template Differences: - STOP site - no primer - no editing a. transient, b. but not passed to offspring, c. more tolerant of mutations ```

Answer 21

transcription is the primary point of regulation for translation regulation: 1) promoter: - strong: high affinity for RNA pol - lot of RNA transcribed - weak: low affinity for RNA pol - less RNA transcribed 2) DNA binding proteins: - repressors - enhancers

Answer 22

transcription and translation at same time in same place no mRNA processing polycistronic = many diff proteins from single mRNA 1 RNA pol

Answer 23

``` transcription and translation at diff times, in diff places mRNA processing: - a. 5' G-cap - b. 3' poly-A tail - c. splicing monocistronic = one mRNA, one protein 3 RNA pol: 1) RNA pol I - rRNA 2) RNA pol II - mRNA 3) RNA pol III - tRNA ^R,M,T ```

Answer 24

at least 20 tRNA amd at most 61 aminoacyl tRNA synthetase 2 ATP to load AAs example of tRNA actually carrying met (AA): met-tRNA^met

Answer 25

first 2 codon-anticodon pairs bind normally (normal Watson-Crick pairing) 3rd anticodon more flexible also, an A on tRNA can get converted to inosine allowing for even more flexibility wobble base pairing happens where there is G, U, I at the 5' end of the anticodon can use same tRNA to translate different codons - fewer tRNAs are necessary

Answer 26

Prok: large = 50S small = 30S total = 70S Eukaryote: large = 60S small = 40S total = 80S prok are odd, euk are even ribozymic activity of the ribosome found in large subunit of both prok and euk ribosomes

Answer 27

P site = growing Protein held here A site = new AA added here E site = Exit site - shift mRNA down 1 codon - translocation 2 ATP to attach AA to free-floating tRNA 2 ATP for peptide bond to form

Answer 28

a. no tRNA that recognizes a STOP codon b. instead, bind a "release factor" c. release factor breaks the bond b/w the final tRNA and the final AA (i.e., "releases" the completed protein)

Answer 29

AA x4 = # ATP needed ``` tRNA loading - 2 ATP per tRNA Initiation - 1 ATP A site binding - 1 ATP per tRNA Translocation - 1 ATP each time Termination - 1 ATP ```

Answer 30

1) AA attached to AMP to form aminoacyl AMP - nucleophile = acidic oxygen of the AA - leaving group (LG) = PPi 2) pyrophosphate LG is hydrolyzed to 2 orthophosphates (provides energy to attach AA to its tRNA molecule) - rxn highly favourable 3) tRNA loading (unfavourable rxn) driven forward by destruction of high-energy aminoacyl-AMP bond created in step 1 requires 2 ATP equivalents b/c uses 2 high-energy bonds ATP equivalent = single high-energy P bond 2 ATP equivalents = hydrolyzing 2 ATP to 2 ADP+2Pi OR ATP to AMP+2Pi 2 functions: 1) specific + accurate AA delivery 2) thermodynamic activation of the AA

Answer 31

specific to each AA | at least 1 aminoacyl-tRNA synthetase for each AA

Answer 32

1) Protein folding: - aided by chaperonins 2) Covalent modification: - disulphide bonds, glycosylation, phosphorylation, etc. 3) Processing: - cleavage to form active protein (e.g., zygomens)

Answer 33

no promoter, has ribosome-binding site --> Shine-Dalgarno sequence (-10) 3 distinct stages: 1) initiation: initiator tRNA (tMet-tRNAfMet) - initiator tRNA sites in P site of 70S ribosome, H-bonded to start codon (AUG) - only initiates translation when preceded by Shine-Dal. seq. 2) elongation: - 3 step cycle: 1. aminoacyl-tRNA enters A site and H-bonds with 2nd codon 2. peptidyl transferase activity of the large subunit catalyzes formation of peptide bond b/w fMet and 2nd AA (amino group of AA#2 acts as nucleophile and tRNAfMet is LG; it dissociates from the ribosome) 3. translocation: tRNA#1 moves to E site, tRNA#2 (holding growing polypeptide) moves into P site, and next codon to be translated moves to A site direction of synthesis: N-->C (since N of AA#2 binds to C of AA#1) as polypeptide elongates, its N terminus will come snaking out of the ribosome 3) termination: - stop codon appears at A site - release factor enters A site (causes peptidyl transferase to hydrolyze the bond b/w the last tRNA and the completed polypeptide)

Answer 34

ribosome larger than prokaryote (80S) mRNA must be processed before translation - spliced, with cap and tail Met not fMet mRNA transported from nucleus to cytoplasm

Answer 35

Initiation: 5'UTR seqs. (Kozak seq) - starts at 5' end of mRNA initiation complex: 40S small subunit, Met-tRNAMet, euk. initiation factors (eIFs) - scans mRNA from 5' end, looking for start codon, once found, 60S subunit recruited and translation begins - eIFs: number of proteins closely controlled, affects amount of translation occurring, levels = rate-limiting step for translation - newly formed polypeptide chains emerging from polyribosome in euk are all same (b/c monocistronic)

Answer 36

transcription & splicing - both occur in nucleus

Answer 37

thought that all euk translation started at 5' end of mRNA (assumed euk transcripts to be monocistronic and codedn for 1 polypeptide chain, b/c of important role of 5'mRNA recognition, its called cap-independent translation) - euk sometimes capable of starting translation in middle of mRNA molecule (b/c beginnign of translation doesn't require 5'cap of mRNA) ^transcript must have an interrnal ribosome entry site (IRES)

Answer 38

DNA level | RNA level

Molecular Biology (C3) Flashcards

(62 cards)