DNA Replication And Repair

Question 1

by convention the base sequence of nucleic acid strand is written as

Answer 1

5’ to 3’(if not mentioned, it means its from 5’)

if from 3’ to 5’ it should be mentioned in the sequence

Question 2

m

Answer 2

m

Question 3

chargaff rule

Answer 3

no. of purines= no.of pyrimidines

Question 4

B DNA

Answer 4

watson crick model

right handed double helical structure

Question 5

which bonds are broken during denaturation

Answer 5

H bonds only leading to ssDNA

NOT covalent(including phosphodiesterase bonds)

Question 6

what is positive and negative supercoiling

Answer 6

POSITIVE
if the DNA wound more loosely than the WC dna

NRGATIVE
if the DNA wound more tightly than the watson crick dna.
req for most biologic reactions

Question 7

fxn of topoisomerases

Answer 7

change the amount of supercoiling in dna

Question 8

basic packing unit of chromatin

Answer 8

nucleosome

Question 9

heterochromatin

Answer 9

tightly condensed

areas of chromosomes not expressed

Question 10

nucleosome(10 nm chromatin) is made up of

Answer 10

1. DNA
2. two copies of each of the following histones

H2A
H2B
H3
H4
forming histone octamer

Question 11

where is H1 located and its fxn

Answer 11

ass with linker DNA bw nucleosomes

helps in further packing to solenoid like structure(30nm fibre)

Question 12

how many origin of replication are there in prokaryotic dna

Answer 12

only 1

Question 13

replication is bidirectional

Answer 13

true

at a Origin of replication 2 replication forks are created that move in opposite direction

Question 14

dna pol has proof reading activity whereas rna pol does not

Answer 14

yes

3’ to 5 ‘ exonuclease proof reading activity

Question 15

dna pol requires a primer but rna pol does not?

Answer 15

true

dna pol can not initiate strand synthesis.

they can only extend the strand from 3’ end of primer.

Question 16

role of dna pol I in prokaryotes

Answer 16

removes primer (as in okazaki fragments) due to 5’ exonuclease activity and synthseise new dna from 3’ end of neighbouring okazaki fragments.

the fragments are sealed by dna ligase

Question 17

drugs acting on bacterial and human topoisomerases

Answer 17

quinolones

etoposide, teniposide

Question 18

unwinding of the dna done by

Answer 18

helicase

break the double stranded dna i.e. H bonds

Question 19

quinòlones target

Answer 19

topoisomerase II/ DNA gyrase

resp for negative supercoiling

Question 20

whatvare the different eukaryotic dna pol

Answer 20

alpha, delta- dna synthisis

gamma-mitichondrial dna syn

beta, epsilon-dna repair

Question 21

dna syn in prokaryotes by

Answer 21

dna pol III

Question 22

removal of the rna primers by

Answer 22

eukar- RNAHase

prokar- DNA pol I

both have 5’to 3’ exonuclease activity

Question 23

replacement of rna with dna in primer during replication by

Answer 23

eukar-DNA pol delta

prokar- DNA pol I

Question 24

okazaki fragments are joined via

Answer 24

dna ligase in both eu/prokaryotes

Question 25

moa of AZT

Answer 25

thymidine analog enters cell and converted to triphosphate DNA pol(of RTase of virus) adds it to the dna chain since no OH' group in the 3' position. therefore chain termination.

Question 26

proteins inbolved in the regulation of G1 checkpoint

Answer 26

1. cyclinD , Cdk4(cyclin dep kinase) 2. RB 3. p53

Question 27

role of cycin D and Cdk4

Answer 27

cyclin D binds to Cdk4 and makes the cell to enter from G1 to S phase Cdk4 phosphorylates RB which releases the captured EF2 required for replication

Question 28

role of p53

Answer 28

1. provides time for cell to repair the damage by arresting it in G1 phase p53 inhibits Cdk4 which ll not be able to inactivate RB 2.if excess damge then apoptosis of cell decrease in growth promoting gene and BCL2 (antiapotosis)

Question 29

gene inactivated in Telengia ectasia

Answer 29

ATM encodes a protein kinase essential for p53 activity

Question 30

defect in xeroderma pigmentosa

Answer 30

thymine dimers formed due to UV radiation are not corrected due to def in exi/endonuclease enz dimers interfere with DNA replication and gene expression.

Question 31

defect in HNPCC

Answer 31

mutation in one of the 2 genes-- hMSH2/ hMLH1 leads to defective repair of DNA mismatches formed after DNA replication

Question 32

features of xeroderma pigmentosum

Answer 32

extreme uv sensitivity excessive frekling multiple skin cancers corneal ulveration

Question 33

cytosine deamination defect

Answer 33

cytosine converted to uracil in DNA recognised and removed by uracil glycosylase enzyme. dna pol fills gap dna ligase seals the repaired dna to original strand

Question 34

microsatellites

Answer 34

these are di, tri and tetranucleotide repeats dispersed in dna usually in the noncoding region. if the cell lack mismatch repair then the replicated dna ll vary in the no. of repeats in that locus.This variation is called microsatellite instability. eg: found in tumor cells of HNPCC

Question 35

most abundant rna in the cell

Answer 35

ribosomal RNA 2nd most abundant :tRNA

Question 36

hnRNA represents(heterogenous nuclear)

Answer 36

precursor mRNA/pre mRNA after posttranscriptional processing of it mRNA is formed

Question 37

which rna participates in mRNA splicing/removing of introns

Answer 37

snRNA | small nuclear