“Molecular Basis of Inheritance” is central to understanding how genetic information is stored, replicated, and expressed at the DNA/RNA level. Board and competitive exams repeatedly test concepts like DNA replication mechanics, mutation rates, proofreading, recombination/interference, gene regulation (operons), and experimental interpretation of restriction/replication problems—so mastering this chapter directly improves performance in both conceptual and numericals.
15
Minutes
10
Questions
1 / -0
Marking
Q1. A linear eukaryotic chromosome of length bp is to be replicated during S‑phase lasting 30 minutes. Replication forks move at and each origin produces two replication forks. What is the minimum number of origins required to complete replication within the S‑phase?
8 origins
9 origins
10 origins
7 origins
Q2. Consider an E. coli merodiploid constructed by introducing a plasmid into a chromosomal strain. The chromosomal genotype is and the plasmid carries . In the absence of inducer, what is the expected level of ‑galactosidase synthesis from the chromosomal gene?
Chromosomal will be expressed constitutively at high level because the chromosome lacks functional LacI.
Chromosomal expression will be high because the mutant operator on the plasmid prevents LacI from binding DNA at all, so the chromosomal is not repressed.
Chromosomal expression will be unaffected, but plasmid will show constitutive expression of its lac operon.
Chromosomal will be repressed (very low expression) in the absence of inducer because the plasmid‑encoded LacI acts in trans to repress the chromosomal operator.
Q3. A bacterial DNA polymerase with 3'→5' proofreading has an error rate of errors per base per replication. A mutant lacking proofreading shows an error rate of errors per base. For a genome size of bp, by approximately what fold does the expected number of base‑substitution errors per genome replication increase when proofreading is lost?
100-fold
10-fold
1000-fold
50-fold
Q4. Statement (S): DNA methylation of cytosine residues in CpG islands of gene promoters is commonly associated with transcriptional silencing in eukaryotes.
Reason (R): Methylation causes silencing primarily by directly blocking binding of general transcription factors (such as TFIID/TBP) to promoter DNA.
Both S and R are true and R explains S.
Both S and R are true but R does not explain S.
S is true but R is false.
S is false but R is true.
Q5. In a three‑point test cross for genes arranged A–B–C, the observed recombination frequencies are and . Among progeny, double crossovers were observed. Interference is defined as . What is the interference (nearest value)?
0.50
0.33
0.67
0.20
Q6. A circular bacterial chromosome of length base pairs begins replication from a single origin and proceeds bidirectionally. If each replication fork progresses at , the time required to complete replication (in minutes) is closest to:
Q7. In a three-point cross the recombination frequencies (from two-point tests) between gene pairs A–B and B–C are and respectively. In a three-point testcross with progeny the observed number of double crossover (DCO) progeny is . The interference value () is:
0.20
0.40
0.80
0.60
Q8. DNA polymerase has an error rate of errors per base per replication. For a bacterial population of cells, consider a single gene of length . After one round of replication, the expected number of cells carrying at least one point mutation in that gene can be estimated approximately by . The expected number is:
1500
150
15000
15
Q9. Assertion (A): Proofreading by DNA polymerases reduces the mutation rate during replication by removing incorrectly incorporated nucleotides.
Reason (R): Proofreading requires exonuclease activity present in many DNA polymerases which removes mismatched nucleotides.
Both A and R are true and R is the correct explanation of A.
Both A and R are true but R is not the correct explanation of A.
A is true but R is false.
A is false but R is true.
Q10. A linear DNA molecule of length was analyzed. A single digest with enzyme A yields fragments of and . A single digest with enzyme B yields fragments of and . A double digest with A and B yields fragments of , and . Which of the following maps of restriction sites on the linear molecule is consistent with these results?
B cuts within the fragment generated by A, dividing it into and fragments.
B cuts within the fragment generated by A, dividing it into two fragments while the other A fragment remains .
A and B cut at the same site on the molecule so double digest gives only the same two fragments as single digests.
B cuts outside both A-generated fragments producing an and fragment which is then split by A into and .