Study Unit 1

Question 1

DNA Structure

Answer 1

DNA is a common structure across all organisms

Question 2

DNA Functions

Answer 2

1. Storing genetic information.
2. Copying itself

Question 3

F. Griffith’s Experiment (1928) Objective:

Answer 3

Show that DNA is the genetic material.

Question 4

F. Griffith’s Experiment (1928) Procedure:

Answer 4

Injected mice with virulent and nonvirulent strains of S. pneumoniae.

Question 5

F. Griffith’s Experiment (1928) Results:

Answer 5

-Virulent strain caused pneumonia and death in mice.
-Nonvirulent strain did not cause disease.
-Heat-killed virulent bacteria mixed with nonvirulent bacteria turned nonvirulent cells virulent, causing disease.

Question 6

F. Griffith’s Experiment (1928) Conclusion:

Answer 6

Genetic information remained in heat-killed virulent strain.

Question 7

3. Avery, MacLeod, and McCarty’s Experiment
   Objective:

Answer 7

Confirm that DNA carries genetic information.

Question 8

Avery, MacLeod, and McCarty’s Experiment Findings:

Answer 8

DNA is responsible for transformation in bacteria.

Question 9

Molecular Components
1.1. Nucleotides, Definition

Answer 9

Building blocks of nucleic acids.

Question 10

1.1 Nucleotides Components (monomer/building blocks)

Answer 10

A sugar.
A base.
One or more phosphate groups.

Question 11

1.2 DNA Bases

Answer 11

Adenine (A).
Thymine (T).
Cytosine (C).
Guanine (G)

Question 12

Nucleosides:

Answer 12

A sugar and a base.

Question 13

Nucleotides

Answer 13

A nucleoside with one or more phosphate groups.

Question 14

Phosphodiester Bonds

Function

Answer 14

Link nucleotides together, forming DNA backbone.

Question 15

Phosphodiester Bonds Properties

Answer 15

-Covalent bonds resistant to pH and temperature changes.
-DNA sequence polarity: 5′-3′ direction.

Question 16

DNA Structure and Stability
Watson and Crick’s Model. Discovery:

Answer 16

Structure of DNA (double helix).

Question 17

Watson and Crick’s Model Basis:

Answer 17

-X-ray crystallography (Rosalind Franklin and Maurice Wilkins).
-Biochemistry results from Erwin Chargaff (C=G and A=T).

Question 18

Hydrogen Bonding. Between Complementary Base Pairs:

Answer 18

Contributes to stability of DNA double helix.

Question 19

Hydrogen Bonding Base Stacking

Answer 19

Interactions between bases on the same strand.

Question 20

DNA Replication
Replication Mechanism Complementary Base Pairing

Answer 20

Ensures fidelity of replication.

Question 21

DNA Replication
Replication Mechanism Result

Answer 21

Two new double helices, copies of the original strand.

Question 22

Mutation Cause

Answer 22

Errors in DNA replication.

Question 23

Mutation Types:

Answer 23

Harmful.
Beneficial.
Neutral.

Question 24

Central Dogma
Information Flow

Steps:

Answer 24

DNA is transcribed into RNA.
RNA is translated into proteins.

Question 25

Transcription Locations
Prokaryotes:

Answer 25

Cytoplasm.

Question 26

Transcription Locations
Eukaryotes

Answer 26

Nucleus.

Question 27

DNA vs. RNA

Differences:

Answer 27

DNA: Deoxyribose sugar, Thymine (T), double-stranded.
RNA: Ribose sugar, Uracil (U), single-stranded.
5′ End: DNA (monophosphate), RNA (triphosphate).
Size: DNA (very large), RNA (smaller)

Question 28

Transcription Process

Template

Answer 28

DNA serves as the template for RNA production.

Question 29

Transcription
Initiation:

Answer 29

Transcription factors and RNA polymerase bind at promoter sequences.
Common eukaryotic promoter sequence: 5′ - TATAAA - 3′.

Question 30

Template vs. Nontemplate Strands:

Answer 30

Genes can be coded on either strand of DNA.
Example Question: For Gene A, which strand will be the template?
Answer: The bottom strand.

Question 31

16. Transcription Regulation

Answer 31

Gene Expression: Turned on or off depending on signals.

Question 32

Transcription
Promoter Recognition

Eukaryotes

Answer 32

General transcription factors recognize promoter-specific sequences.

Question 33

Transcription
Promoter Recognition: Prokaryotes:

Answer 33

Sigma factor associates with RNA polymerase to recognize promoter sequences.

Question 34

Transcription Initiation and Elongation: 4 main components and enzymes

Answer 34

RNA Polymerase II
Transcription Bubble
RNA-DNA Duplex
Elongation

Question 35

Transcription Initiation and Elongation: RNA Polymerase II

Answer 35

Unwinds DNA and adds nucleotides.

Question 36

Transcription Initiation and Elongation
Transcription Bubble

Answer 36

Approximately 14 base pairs in length.

Question 37

Transcription Initiation and Elongation
RNA-DNA Duplex:

Answer 37

Approximately 8 base pairs in length.

Question 38

Transcription Initiation and Elongation
Elongation:

Answer 38

RNA Pol II unwinds DNA, allowing nucleotides to be added to the mRNA transcript.

Question 39

RNA Processing (Eukaryotes)
5’ Cap

Function:

Answer 39

Stability and protection from exonucleases.
Ribosome recognition for translation

Question 40

RNA Processing (Eukaryotes)
Poly(A) Tail

Function

Answer 40

Added at the 3’ end of mRNA.
Role in transcription termination and protection from exonucleases.
Targets mRNA for transport to the cytoplasm.

Question 41

RNA Splicing
Process

Answer 41

Introns are removed, and exons are spliced together.

Question 42

RNA Splicing
Alternative Splicing

Answer 42

A single gene may produce different protein products in different cells.

Question 43

RNA World Hypothesis

Idea

Answer 43

The first nucleic acids were RNA molecules.

Question 44

RNA World Hypothesis
RNA Functions

Answer 44

Involved in cellular processes and central dogma steps.
Has enzymatic properties.
DNA is more stable, hence used by cells.