Molecular Genetics Test

Question 1

nucleotide structure

Answer 1

Each nucleotide consists of:

A phosphate group
A deoxyribose or ribose sugar
A nitrogenous base (adenine, thymine, cytosine, guanine)

Question 2

bases, pyrimidines vs purines

Answer 2

Pyrimidines: Cytosine (C) and Thymine (T) (one-ring structure)

Purines: Adenine (A) and Guanine (G) (two-ring structure)

Question 3

Complementary base pairing and Chargaff’s rule

Answer 3

Complementary Base Pairing

A pairs with T (two hydrogen bonds)
C pairs with G (three hydrogen bonds)

Chargaff’s Rule
States that in any given DNA molecule, the amount of adenine equals thymine, and the amount of cytosine equals guanine (A=T, C=G).

Question 4

all enzymes and their functions (terms: replication fork, SSBPs, helicase, topoisomerase (gyrase); primase, DNA polymerase III, DNA Polymerase I, ligase

Answer 4

SSBPs (Single-Strand Binding Proteins): SSBPs are proteins that bind to single-stranded DNA during replication, to help stabilize the newly unwound single DNA strands.

Replication Fork: The replication fork is the Y-shaped structure formed during DNA replication where the DNA is unwound and the two strands are separated to allow for the synthesis of new DNA strands.

DNA helicase - unwinds and separates double stranded DNA as it moves along the DNA. It forms the replication fork by breaking hydrogen bonds between nucleotide pairs in DNA.

DNA primase - a type of RNA polymerase that generates RNA primers. Primers are short RNA molecules that act as templates for the starting point of DNA replication.

Topoisomerase or DNA Gyrase - relieves tension in DNA strands to prevent the DNA from becoming tangled or supercoiled.

DNA polymerases - synthesize new DNA molecules by adding nucleotides to leading and lagging DNA strands.

DPI → Replaces RNA primers with DNA nucleotides and proofreads and corrects typos in the DNA

DPIII → adds DNA nucleotides in the 5’ to 3’ direction

DNA ligase - joins DNA fragments together by forming phosphodiester bonds between nucleotides.

Question 5

Describe the Process of DNA Replication (Initiation)

Answer 5

Unwinding the DNA: The process begins at specific locations called origins of replication. The enzyme helicase unwinds the double helix, creating a replication fork. Single-stranded binding proteins (SSBPs) stabilize the unwound DNA strands.

Relieving Tension: As DNA unwinds, topoisomerase alleviates the tension ahead of the fork, preventing supercoiling.

Primer Synthesis: The enzyme primase synthesizes short RNA primers complementary to the template strands, providing starting points for DNA synthesis.

Question 6

Describe the Process of DNA Replication (Elongation)

Answer 6

Nucleotide Addition: DNA polymerase III attaches to the RNA primer and begins adding nucleotides in the 5’ to 3’ direction. On the leading strand, this occurs continuously.

Lagging Strand Synthesis: On the lagging strand, DNA is synthesized in short segments called Okazaki fragments. Each fragment is initiated by a new RNA primer.

Primer Replacement: Once DNA synthesis is complete, DNA polymerase I removes the RNA primers and replaces them with DNA nucleotides.

Sealing Gaps: The enzyme ligase then seals the gaps between Okazaki fragments (using phosphodiester bonds), ensuring a continuous DNA strand.

Question 7

Describe the Process of DNA Replication (Termination)

Answer 7

-Occurs upon completion of the 2 new strands of DNA are synthesized, 2 double strand DNA molecules are produced that will recoil
into a helix.

• replication machine is dismantled

Question 8

Replication Models

Answer 8

1) Conservative: The original DNA molecule remains intact, and a completely new molecule is synthesized.

2) Semi-conservative: the original molecule is split in half, and the other side is filled-in

3) Dispersive: Each new molecule is comprised of bits and pieces of both new DNA and the original strand

Question 9

Meselson-Stahl Experiment:

Answer 9

The Meselson-Stahl experiment demonstrated that DNA replication is semi-conservative.

Setup: E. coli bacteria were grown in a medium containing the heavy nitrogen isotope N-15, labeling their DNA.

Transfer: The bacteria were then moved to a medium with the lighter isotope N-14.

Sampling: After one and two rounds of replication, samples of DNA were taken.

Analysis: The DNA was separated by density using centrifugation.

Results:
- After one replication, the DNA had an intermediate density (one old strand and one new).
- After two replications, there were both light DNA (two N-14 strands) and intermediate DNA (one N-15 and one N-14 strand).

This proved that the Semi-conservative model was correct

Question 10

Role of telomeres and telomerase

Answer 10

Telomeres are repeating sequences of nucleotides

• If chromosomes were not capped by telomeres, a small portion of a gene near the end of the chromosome could be lost every time DNA replication occurred
• Instead, only portions of telomeres are lost

Telomerase: An enzyme that extends telomeres, particularly active in germ cells and some stem cells, helping maintain chromosome integrity.

Germ line cells are unique because they must be able to continue replicating-

• Must maintain genetic integrity from parent to offspring

-Enzyme called telomerase adds more DNA to shortening telomeres, restoring their length

-Stem cells and some white blood cells also show presence of telomerase

Question 11

RNA vs DNA

Answer 11

DNA:
Sugar: DNA contains deoxyribose, a sugar with one less oxygen atom than ribose.

Strands: DNA is typically double-stranded, forming a double helix.

Bases: The four nitrogenous bases in DNA are adenine (A), thymine (T), cytosine (C), and guanine (G). Adenine pairs with thymine, and cytosine pairs with guanine.

Shape: The double-stranded structure is stable and coiled into a double helix.

RNA:
Sugar: RNA contains ribose, a sugar with one more oxygen atom than deoxyribose.

Strands: RNA is usually single-stranded.

Bases: RNA uses adenine (A), uracil (U), cytosine (C), and guanine (G). In RNA, uracil replaces thymine as the complementary base to adenine.

Shape: RNA is typically single-stranded and can fold into various shapes, depending on its function.

2. Function
   DNA:
   Genetic Material: DNA stores the genetic instructions used in the growth, development, functioning, and reproduction of all living organisms. It is the blueprint for all biological information.

Replication: DNA is replicated when cells divide, ensuring that each new cell receives an exact copy of the genetic information.

RNA:
Protein Synthesis: RNA plays a central role in protein synthesis.

Question 12

Transcription (Initiation)

Answer 12

Promoter region: a sequence of nucleotides in DNA that indicates where the RNA polymerase complex should bind to initiate transcription

-Key element of the promoter in eukaryotes is the TATA box (a portion of DNA with high percentage of Adenine and Thymine bases)

-Prokaryotes have a TATATT sequence instead for this

-RNA polymerase binds to a promoter region on the DNA same purpose

Question 13

Transcription (Elongation)

Answer 13

RNA polymerase complex works its way along DNA molecule

-Without needing a primer to be already in place

-Synthesizes mRNA strand that is complementary to template strand of DNA (T is replaced with U)

-RNA polymerases work in the 5ʼ → 3ʼ direction, using the 3ʼ→ 5ʼ DNA strand as a template strand.

As RNA polymerase moves along the DNA, it unwinds the DNA at the forward end of the enzyme

-RNA strand grows as nucleotides are added, one by one forming a temporary RNA-DNA double helix with the template strand

-As the RNA polymerase passes, the DNA double helix reforms

-Once an RNA polymerase molecule has started transcription and progressed past the beginning of a gene, another molecule of RNA polymerase may start producing another RNA molecule if there is room at the promote

Question 14

Transcription (Termination)

Answer 14

The transcription is terminated when RNA polymerase recognizes a termination sequence.

Them RNA has been completely transcribed

●In eukaryotes, this is pre-mRNA and must be further processed before exiting the nucleus

Question 15

Coding vs Template strands (which goes in which direction)

Answer 15

Coding (sense) is the same sequence as the created mRNA strand but with thymine instead of uracil 5’ to 3’ direction

Template (anti-sense) is the strand used as the template for the mRNA strand 3’ to 5’ direction

Question 16

mRNA processing in eukaryotes (poly A tail, G cap, splicing of exons

Answer 16

Poly(A) tail: a chain of adenine nucleotides added to the 3’ end of pre-mRNA molecule to protect it from enzymes in the cytosol

-Enables mRNA to be translated efficiently and protects from attack by RNA-digesting enzymes in the cytosol

Splicing
The intron sequences are removed from pre-mRNA and exons are joined together to form mature mRNA

G Cap

-Involves covalent linkage of modified G nucleotides onto to 5’ end of pre-mRNA

-The cap is recognized by the protein synthesis machinery

-All eukaryotic mRNA undergo modifications on their ends

-These modifications convert precursors mRNA (pre-mRNA) to mature mRNA

Question 17

Exons vs introns, alternate splicing

Answer 17

Exons are coding regions

Alternate Splicing
Exons may be joined in different combinations to produce different mRNAs from a single DNA gene sequence
-A mechanism called alternative splicing increases number and variety of proteins encoded by a single gene

-Allows more than one possible polypeptide to be made from a single gene

-Alternative splicing helps understand why humans with only 20 000 genes can produce approx. 100 000 proteins

Question 18

Structure of tRNA (anticodon

Answer 18

-3’ end of tRNA binds to specific amino acids

-Anticodon on tRNA complements mRNA codon

Question 19

Structure of ribosome - a, p and e sites

Answer 19

-2 subunits: small and large

-E site → exit site

-P site → polypeptide binding site

-A site → amino acid site

-Composed of proteins and rRNA

-3 tRNA binding sites

Question 20

Translation - Initiation

Answer 20

-mRNA, tRNA and small ribosomal subunit bind with P site at start codon

-Only the first tRNA binds to P at start codon

-Large subunit binds using energy from GTP

Question 21

Translation - Elongation

Answer 21

mRNA read 3 nucleotides at a time in codons

-tRNA brings corresponding amino acid into the A site of the ribosome

-Ribosome catalyzes dehydration synthesis reaction between amino acids in P site and A site

-Peptide bond is formed

-Growing polypeptide now attached to tRNA in A site

-Ribosome moves forward one codon

-Free tRNA in P site exits out the back of ribosome on the E site

-tRNA (with polypeptide) moves into P site

Question 22

Translation - Termination

Answer 22

-Elongation continues until reaching a stop codon

-There is no amino acid for stop codon, just STOP

-Release factor binds and hydrolyzes the bond between the last tRNA and its amino acid

-New protein is free

Question 23

frameshift vs. point mutations, missense, nonsense, silent

Answer 23

Frameshift Mutations
– shifts the reading frame of the codons so that a completely different amino acid sequence is produced from the point of the shift. Will likely not produce a functional protein

point mutations
– change in one nucleotide

○Substitution
○Substitutions will only affect a single codon
■THE FAT CAT ATE THE RAT
■THE FAT HAT ATE THE RAT
○Insertion (Frameshift)

■THE FAT CAT ATE THE RAT
■THE FAT CAT HATE THE RAT
○Deletion (Frameshift)

■THE FAT CAT ATE THE RAT
■THE FAT CAT AT THE RAT

Missense
A substitution that results in a different amino acid is
called a missense mutation

Nonsense
A nonsense mutation produces a premature STOP codon

Silent
If the mutation still produces the same amino acid it is
called a silent mutation. -Has no impact on the protein formed-This happens because the genetic code is degenerate

Question 24

Structure of an operon and role of components (promotor, operator, repressor, genes)

Answer 24

Structure of an Operon:
Promoter: A DNA sequence where RNA polymerase binds to initiate transcription of the operon.

Operator: A DNA region that acts as a switch. It is located between the promoter and the structural genes, and it is where a repressor protein can bind to block transcription.

Repressor: A protein that can bind to the operator and prevent transcription. The repressor is often regulated by an inducer or co-repressor molecule.

Structural genes: These are the genes that code for proteins, often enzymes, that carry out a particular function (e.g., breaking down a sugar or synthesizing an amino acid).

Question 25

Inducible vs. Repressible operons

Answer 25

Inducible operon - Usually turned “off” . Turned on when the metabolic pathway is needed (high lactose) (eg - lac operon)

Repressible operon - Usually turned “on”. Turned off when the metabolic pathway is no longer needed (high tryptophan) (eg - trp operon)

Question 26

lac Operon - be able to describe what it is and how it regulates the expression of genes that code for enzymes that help break down lactose in bacteria.

Answer 26

The lac operon is an inducible operon in E. coli that is involved in the breakdown of lactose. It consists of the following components:

Repressor (LacI): The gene that codes for the lac repressor, which binds to the operator and prevents transcription.

Regulation Mechanism:
When lactose is absent, the repressor (LacI) binds to the operator, blocking transcription of the lac genes.

When lactose is present, lactose (or more specifically, allolactose, an isomer of lactose) binds to the repressor, causing it to change shape and release from the operator. This allows RNA polymerase to transcribe the genes needed to break down lactose.

Question 27

trp Operon
- be able to describe what it is and how it regulates the expression of genes that code for enzymes that help make tryptophan.

Answer 27

The trp operon is a repressible operon in E. coli that regulates the synthesis of the amino acid tryptophan.

Repressor (TrpR): This gene encodes a repressor protein that, in its inactive form, cannot bind to the operator. However, when tryptophan levels are high, tryptophan itself binds to the repressor, activating it.

When tryptophan levels are low, the repressor remains inactive and the operon is transcribed, allowing the cell to produce tryptophan.

When tryptophan levels are high, tryptophan binds to the repressor, activating it. The repressor then binds to the operator, blocking transcription and stopping the production of more tryptophan.

Question 28

How do you create designer plasmids? Use of restriction enzymes. What are sticky ends?

Answer 28

Creating Designer Plasmids:

Restriction Enzymes: These are molecular scissors that cut DNA at specific sequences. They are used to cut both the plasmid and the DNA of interest at the same site, producing sticky ends (overhanging sequences of DNA) that can pair with complementary sticky ends from another DNA molecule.

(Example: If you cut a plasmid and a gene of interest with the same restriction enzyme, the sticky ends of both will be complementary, allowing them to be joined together. )

Sticky Ends: These are single-stranded overhangs created by restriction enzymes that make it easier to join two pieces of DNA. For example, a sticky end from the plasmid can bind to a complementary sticky end from a gene of interest.

Ligation: The gene of interest is inserted into the plasmid, and the DNA ligase enzyme is used to seal the sugar-phosphate backbone, creating a recombinant plasmid.

Question 29

Be able to explain the process of how you create a recombinant plasmid and their use.

Answer 29

Gene Cloning and Study
Recombinant plasmids are used to clone specific genes. This allows scientists to make copies of genes for research, study their functions, and better understand genetic material.

Protein Production
They are used to produce proteins, like insulin or enzymes, by inserting a gene into a plasmid and introducing it into bacteria or other cells, which then produce the protein in large amounts.

Genetic Engineering
Recombinant plasmids enable genetic modifications in organisms (like plants, animals, or bacteria). This allows for the creation of genetically modified organisms (GMOs) with new traits, such as disease resistance or improved growth.

Question 30

PCR (7.1)
What is the purpose?

Answer 30

A process used to artificially multiply a chosen
piece of genetic material.

May also be known as DNA amplification.

One strand of DNA may yield 2^30 strands or
more.

Question 31

PCR How does it work? Briefly describe the process.

Answer 31

One cycle:

1. Double-stranded DNA is
   denatured using heat (95°C) to
   separate strands by breaking
   hydrogen bonds
2. Temperature is lowered (55-60°C)
   which allows the primers to anneal
3. Temperature raised (72°C) and
   Taq polymerase adds nucleotides in
   the 5’–> 3’ to synthesize the new
   DNA strand

Repeat cycle (steps 1-3)

Question 32

PCR Reason for using taq polymerase

Answer 32

Taq polymerase is used because it can withstand high temperatures (won’t denature), making it ideal for the DNA amplification process in PCR (Polymerase Chain Reaction). Unlike human polymerase which would denature under high temperatures.

Question 33

Gel Electrophoresis - Be able to explain the entire process: restriction enzyme use, how DNA is separated on a gel.

Answer 33

Purpose:
Gel electrophoresis is used to separate DNA fragments by size, enabling analysis of DNA samples.

Process:
Restriction Enzymes: DNA is digested with restriction enzymes to cut it into smaller fragments.

Gel Loading: The DNA fragments are loaded into wells in an agarose gel.

Electrophoresis: An electric current is applied, causing negatively charged DNA fragments to move through the gel towards the positive electrode. Smaller fragments move faster than larger ones, so the DNA separates by size.

Question 34

Gel Electrophoresis - The uses….Paternity, forensics, evolutionary relationships etc.

Answer 34

Applications:
Paternity testing: DNA from a child and possible father are compared. Matching bands indicate a genetic relationship.

Forensics: DNA samples from a crime scene can be compared to suspects’ DNA to identify the perpetrator.

Evolutionary relationships: DNA from different species can be compared to determine evolutionary relationships based on similarity in band patterns.

Question 35

Gel Electrophoresis - Be able to analyze a gel to determine paternity OR crime scene data

Answer 35

Analyzing a Gel: To determine paternity or forensic information, you would look at the banding patterns of DNA samples. In paternity tests, bands from the child must match those of the biological father. In forensics, matching bands from crime scene samples and suspect samples can confirm identity.