ch 6

Question 1

Griffith’s Transformation Experiment (1928):

Answer 1

Studied Streptococcus pneumoniae bacteria.

Question 2

Found that heat-killed virulent (smooth, S) strain could

Answer 2

ransfer genetic material to the non-virulent (rough, R) strain

Question 3

Nucleotides are the building blocks of DNA, consisting of:

Answer 3

phosphate group.
A deoxyribose sugar.
A nitrogenous base (A, T, C, G).

Question 4

Chargaff’s Rule:

Answer 4

A = T, C = G (Base pairing follows a 1:1 ratio).

Question 5

Nucleotides Link via

Answer 5

Phosphodiester Bonds:

Covalent bonds form between the phosphate of one nucleotide and the sugar of the next.

Question 6

Avery, MacLeod, and McCarty (1944):

Answer 6

Showed that DNA, not protein or RNA, was the “transforming principle” by selectively destroying DNA, RNA, and proteins and testing for transformation.

Question 7

Hershey-Chase Experiment (1952):

Answer 7

Found that only DNA entered the bacterial cells, proving it was the genetic material.

Question 8

A-T pairs form

Answer 8

2 hydrogen bonds.

Question 9

C-G pairs form

Answer 9

3 hydrogen bonds (stronger bond).

Question 10

Double Helix Model
Antiparallel Strands:

Answer 10

One strand runs 5’ → 3’, the other 3’ → 5’.

Question 11

Major Steps in Recombination

Answer 11

Double-Strand Break Formation –
End Processing –
Strand Invasion –
DNA Synthesis –
Holliday Junction Formation –
Resolution –

Question 12

Double-Strand Break Formation –

Answer 12

Enzymes cut the DNA.

Question 13

End Processing

Answer 13

The ends of the broken strands are trimmed.

Question 14

Strand Invasion –

Answer 14

One strand invades the homologous chromosome, forming a D-loop

Question 15

DNA Synthesis –

Answer 15

NA polymerase extends the invaded strand.

Question 16

Holliday Junction Formation –

Answer 16

he structure allowing crossover.

Question 17

Resolution –

Answer 17

holliday junction is cut in different ways, leading to crossover or non-crossover products.

Question 18

Only non-sister chromatids of homologous chromosomes participate in

Answer 18

dna steps

Question 19

Each new DNA molecule consists of

Answer 19

one original (parent) strand and one newly synthesized strand.

Question 20

DNA polymerase adds nucleotides only to the 3’ end,

Answer 20

making replication occur 5’ → 3’.

Question 21

Origin of replication is

Answer 21

y specific DNA sequences recognized by a protein complex called the Origin Recognition Complex (ORC)

Question 22

Steps of DNA Replication

Answer 22

Initiation:

Origin of replication is identified.
Helicase unwinds DNA.
Primase synthesizes RNA primers.

Question 23

Elongation:

Answer 23

DNA polymerase adds nucleotides 5’ → 3’.
Leading strand: Continuous synthesis.
Lagging strand: Synthesized in short Okazaki fragments.

Question 24

Polishing (Termination & Proofreading):

Answer 24

Exonucleases remove primers.
DNA Ligase joins Okazaki fragments.

Question 25

Resolving Supercoiling

Answer 25

Topoisomerase prevents excessive twisting ahead of the replication fork.

Question 26

Resolving the Lagging Strand Problem

Answer 26

Okazaki fragments on the lagging strand are later joined by DNA ligase.