bio unit unit 3 Molecular genetics Flashcards

Question 1

Q

Discovery of the existence of DNA

Answer

A

DNA was discovered in 1869, by Swiss chemist, Friedrich Miescher–Although it was not known as genetic material
He isolated the nuclei of white blood cells from pus-soiled bandages
It contained nitrogen and phosphorous
He named it Nuclein
Renamed to Nucleic acid

Question 2

Q

Phoebus Leven

Answer

A

In the early 1900’s Phoebus Levene isolated two types of nucleic acid (now called DNA and RNA)
In 1919 he proposed that DNA and RNA are polymers made up of single units (monomers) called nucleotides
Containing four N-containing bases, a sugar molecule, and a phosphate group
DNA is a polymer of nucleotides (not sure of the shape)
DNA has nucleotides consisting of 4 different nitrogenous bases: adenine (A), thymine (T), cytosine (C), guanine (G)
RNA has the same nitrogenous bases, except, thymine (T) is replaced by uracil (U)
each of DNA’s four types of nucleotides consists of Deoxyribose sugar (5-C) attached to a phosphate group and a nitrogenous base
Recall - the carbon atoms are numbered clockwise, starting with the carbon atom to the immediate right of the oxygen atom. The first carbon is called 1′ (1prime), followed by 2′ (2 prime) and so on

Question 3

Q

By the 1900’s, we assumed/knew

Answer

A

Knew about inheritance of traits
Assumed it was linked to chromosomes
Knew chromosomes were composed of nucleic acids and proteins
They assumed proteins within chromosomes carried the hereditary genetic material that caused the inheritance of traits

Question 4

Q

Griffith’s Discovery of Transformation

Answer

A

“The Transforming Principle” –In 1928, Frederick Griffith studied the pathology (disease causing characteristics) of the bacteria Streptococcus pneuomoniae
He studied 2 strains of bacterium in mice:
1) Disease causing S form
2) Harmless R form
Inject mice with R (nonvirulent) cells and the mice lived.
Inject mice with S (virulent) cells and the mice died. Blood samples from the dead mice contained many S cells.
S cells were killed with heat, then injected into mice and the mice lived.
R cells plus heat-killed S cells were injected into mice and the mice died. Living S cells were found in the blood
The conclusion was that some unknown substance from the dead S cells had transformed the harmless R cells into cells capable of causing death.
Descendants of the transformed cells were also pathogenic (to cause disease).
Discovered the process of transformation
TRANSFORMATION: The introduction of foreign DNA, usually by a plasmid or virus, into a cell.
Griffith transformed nonvirulent pneumococcus into virulent pneumococcus with S-strain DNA

Question 5

Q

TRANSFORMATION

Answer

A

The introduction of foreign DNA, usually by a plasmid or virus, into a cell.

Question 6

Q

Chargaff’s Rule

Answer

A

Late 1940’s: Austrian-American biochemist Erwin Chargaff studied and compared DNA from different species.
He observed that nucleotides and their nitrogenous bases are always present in characteristic proportions
Example: –the amount of adenine in any sample of DNA is always approximately equal to thymine
the amount of cytosine is always approximately equal to the amount of guanine
This constant relationship is known as Chargaff’s Rule

Question 7

Q

Chase and Hershey

Answer

A

Experiment involved infecting a bacteria using a virus (bacteriophage) which consisted of two components: DNA and a protein coat
They knew that viruses replicated themselves by inserting their own hereditary material into host cells, so they were attempting to isolate what the hereditary material was that the viruses were inserting
In the Hershey–Chase experiment, radioactive phosphorus in viral DNA and radioactive sulfur in viral proteins were used to trace the transfer of each type of biological molecule into a bacterial host cell
One of the most famous experiments in history of genetics
Ruled out protein in favour of DNA as the hereditary material
Experiments showed that only the DNA and not the protein coat entered the cell (labelled DNA with P isotope and protein with S isotope).
Hershey and Chase’s experiment clearly showed that DNA is the hereditary material

Question 8

Q

Avery, Macleod and McCarty

Answer

A

By 1944 researchers could grow bacteria in liquid cultures
They prepared cultures of heat-killed S-strain bacteria
They added 1 of 3 enzymes to each bacteria (Enzyme destroying proteins, Enzyme destroying RNA, and Enzyme destroying DNA)
When treating the non-virulent R-strain with these modified heat-killed S-strains, the only enzyme-treated bacteria that did not create a virulent R-strain was the DNA-destroyed bacteria

Question 9

Q

Linus Pauling

Answer

A

Developed methods of assembling three-dimensional models based on known distances and bond angles between atoms in molecules
This helped him discover that many proteins have helix-shaped structure

Question 10

Q

By the late 1940’s scientists knew

Answer

A

From Hershey & Chase: DNA is the hereditary material
From Levene: DNA is a polymer of nucleotides and nucleotides have different nitrogenous bases (A,T,G,C)
From Chargaff: DNA is composed of these nucleotides that exist in fixed proportions (A=T, G=C)

Question 11

Q

Rosalind Franklin

Answer

A

Used x-ray diffraction to analyze structure of biological molecules –Was able to obtain highest resolution photographs at that time
Based on her images, was able to conclude that DNA has a defined helical structure and had two regularly repeating patterns
When DNA reacted with water, she concluded that nitrogenous bases were located inside of helical structure, and sugar-phosphate backbone was located on outside of helical structure

Question 12

Q

Watson & Crick (1953)

Answer

A

Proceeded to construct the current accepted molecular structure for DNA
Watson & Crick concluded that DNA has a twisted, ladder-like structure called a double helix
Sugar-phosphate molecules make up sides or “handrails” of ladder
Bases make up the rings

Question 13

Q

Modern DNA Model: the Double Helix

Answer

A

Two polynucleotide strands that twist around each other forming a double helix
Complementary Base Pairing: A-T, C-G
Hydrogen Bonds link complementary base pairs
A and T share 2 hydrogen bonds
C and G share 3 hydrogen bonds
Two strands of DNA are antiparallel
One strand runs in the 5′ to 3′ direction and the other strand runs in the 3′ to 5′ direction
Sugar-phosphate backbone
Contains Major and Minor grooves (not symmetrical)
5’ end finishes with a phosphate group sticking up
3’ end finishes with a hydroxyl group pointing down
In RNA, the 2’ Carbon has a hydroxyl (OH) group attached instead of a H

Question 14

Q

Antiparallel

Answer

A

the two strands of DNA run in the opposite direction

Question 15

Q

Building DNA polymers

Answer

A

Each strand is made from the bonding of a phosphate group of one nucleotide with the #3 carbon of the next nucleotide via a phosphodiester linkage

Question 16

Q

Genes

Answer

A

the basic unit of heredity that determines, in whole or part, a genetic trait;
a specific sequence of DNA that encodes for proteins and RNA molecules, and can contain sequences that influence production of these molecules.
A gene is a specific chain of base pairs that form specific proteins. They are always found together in that order
The space between genes is variable, and may change as it is passed down with little effect on the protein formed

Question 17

Q

Genome

Answer

A

the complete genetic makeup of an organism; an organism’s total DNA sequence

Question 18

Q

DNA Regulation

Answer

A

Regulatory sequence (turning genes on and off)– a sequence of DNA where proteins bind and regulate the activity of a gene (inhibiting or activating it)

Question 19

Q

Nucleoid

Answer

A

the structure that contains the chromosomal DNA.

Question 20

Q

DNA supercoiling

Answer

A

The formation of additional coils in the structure of DNA due to twisting forces on the molecule.

DNA Supercoiling in Eukaryotic Cells
- Histones – a member of a family of proteins that associate with DNA in eukaryotic cells, which acts to help compact the DNA
- Nucleosome – the condensed structure formed when double stranded DNA wraps around an octamer of histone proteins (8).

first, DNA wraps around histones, which are protein balls. When 8 histones come together, they become a nucleosome (is an octomer) that also has DNA wrapped around it. The DNA wrapped nucleosomes form euchromatin, and then heterochromatin, and then chromosomes (Theres 23 pairs in somatic). This makes the DNA very compacted. This means that this process happens 46 times in somatic cells and 23 times in sex cells
Also, little DNA does NOT mean the organism isn’t complex

Question 21

Q

DNA of Prokaryotic Cells

Answer

A

In bacteria, the amount of supercoiling is controlled by two enzymes: topoisomerase I and topoisomerase II.
Additional proteins help stabilize the fold
topoisomerase II – enzyme is essential for bacterial survival
Antibacterial drugs have been developed that specifically target and block activities of this enzymes.
Ex. Quinolones and coumarins

Question 22

Q

Plasmids

Answer

A

one or more small circular or linear DNA molecules. These tend to carry non-essential genes and can be transferred and copied from cell to cell.

Question 23

Q

DNA of Eukaryotic Cells

Answer

A

The total amount of DNA is much greater
Genetics material is located in the membrane bound nucleus
The approximate length of DNA in the nucleus of a single human cell is 2 meters long
a nucleus is 4 micrometers wide.
Most eukaryotes are diploid – they contain two copies of each chromosome/gene.
Some eukaryotes are haploids, such as ferns and algae
The organization of genes on each chromosome can differ
- Chromosome 19 has 72 million base pairs and 1450 genes
- Chromosome 4 has almost 1.3 billion base pairs and about 200 genes.
There is no correlation between an organisms complexity and genome size or number of protein-coding genes

Question 24

Q

Chromatin

Answer

A

Additional compacting of DNA occurs and becomes chromatin
non-condensed form of genetic material that consists of a complex of DNA and proteins
euchromatin (compacted) FIRST
heterochromatin (highly compacted) SECOND
after chromatin is chromosome

Question 25

Q

Life Cycle of a Cell

Answer

A

The process of copying one DNA molecule into two identical molecules is called DNA replication
Occurs during the S phase of interphase in the cell cycle
All cells must reproduce before they die to pass on genetic information. Since they will live for a little bit longer passed cell division, they will need a copy of their own DNA.

In order for cells to divide they must:
- Grow
- Carry out metabolic activity
- Replicate DNA

During division, each daughter cell contains exact same genetic material as parent cell
DNA Replication: process of producing two identical DNA molecules from an original, parent DNA

Question 26

Q

Three Proposed models of DNA Replication

Answer

A

The conservative model results in one new molecule and conserves the old.
The semi-conservative model results in two hybrid molecules of old and new strands.
The dispersive model results in hybrid molecules with each strand being a mixture of old and new strands.

Question 27

Q

Meselson and Stahl Experiments on DNA Replication

Answer

A

Through experimentation showed that semi-conservative model of DNA replication was correct
Used two different isotopes of nitrogen to distinguish between parental and daughter DNA strands
Used different isotopes of Nitrogen to label DNA in a cell
14N (common, lighter) and 15N (rare, aka “heavy”, when decaying, will become 14N)
Used nitrogen because it’s found in all of our nitrogenous bases
after spinning all of the nitrogen 15, they put it in the nitrogen 14 medium
they saw that 15N was still there, but new strands were made
They replicated again and noticed 14N was appearing more than N15.
Since N14 was more than N15, this helped us to know DNA was semi-conservative
If it was conservative, N15 would be the same the whole time
If dispersive, both N15 and N14 would be equal
At first, their experiment was only planned for N15, but then they switched it to N14

Starting the experiment:
- They grew E. coli bacteria in a medium with 15N for 17 generations
- By doing this, the E. coli could only replicate DNA using the heavy 15Nitrogen Isotope within their nitrogenous bases

Question 28

Q

Initiation

Answer

A

Portion of the DNA double helix is unwound to expose the bases for new base pairing

DNA Replication - Initiation
- During S-phase during interphase
- Replication starts at a specific nucleotide sequence called The Origin of Replication
- DNA helicase unwinds the double helix by breaking the H bonds between the complementary base pairs holding the two DNA strands together. It has a hole which has things like teeth that unwind the DNA, unzips DNA. For example, the triple bond between G and C will be broken by helicase
- Behind helicase is the replication fork,
- Single stranded binding proteins (SSBs) keep the individual strands apart by blocking the hydrogen bonding between the bases. Since the strands want to be helicase structure, SSBs help each strand to make the hydrogen bonds not connect
- Topoisomerase II (also called Gyrase)– relieves stress of the unwinding on the parent DNA molecule by cutting and un-twisting the molecule. When unwinding, the ends of the DNA strands will supercoil, which will cause it to break. Can be solved by having gyrase to relieve it by cutting the DNA, usually binds at the ends
- Replication starts at a specific nucleotide sequence - the origin of replication
- As the two strands of DNA are disrupted, the junction where they are still joined is called the replication fork
- In eukaryotes, DNA replication occurs at more than one site at a time, resulting in hundreds of replication forks across a DNA strand. This is important because it speeds up the process.
- When 2 replication forks form, a replication bubble is also formed
- If the helicase is seen on the left, DNA is unziping on the left and is headed towards the left direction, INTO THE FORK

Transcription- mRNA initiation
- DNA transcription only occurs on one strand: the template strand
- There is no need to transcribe from the coding strand because it is identical to the mRNA being formed (except it has Thymine not Uracil)
- RNA polymerase: The main enzyme that catalyzes the formation of RNA from DNA
- DNA is unwound by RNA Polymerase to expose the template strand
- Also remember that the template strand must be 3’-5’. This allows the mRNA to be built 5’-3

Question 29

Q

Semi-Conservative Replication

Answer

A

Mechanism of DNA replication that produces two copies
both are made up of one new strand and one conserved from original DNA

There are 3 basics phases in replication:
- Initiation
- Elongation
- Termination

Question 30

Q

Elongation

Answer

A

Two new strands of DNA are assembled using parent DNA as template
New DNA molecules (each composed of one strand of parent DNA and one strand of daughter DNA) reform into double helices

Process of DNA Replication - Elongation
- DNA polymerase III - main player that adds nucleotides to the new strand of DNA. can only add nucleotides in the 5′ to 3′ direction, and require RNA primase as starting points, and requires condensation reactions to go through the 5’ to 3’ direction. RNA primers are the starting site because DNA polymerase III will immediately know that it should combine to it
- DNA is always synthesized in the 5′ to 3′ direction
- The leading strand is built continuously by Polymerase III toward the replication fork, starting with 1 RNA primer
- The lagging strand is synthesized by Polymerase III discontinuously in short fragments in the opposite direction to the replication fork
- These short fragments are called Okazaki fragments which each require RNA primers
- The enzyme Primase lays down RNA primers that will be used by DNA polymerase III as a starting point to build the new complementary strands

Why do we need RNA Primers?
- Allows DNA Polymerase III to bind to the strand
- DNA polymerase can only add new nucleotides to a free 3′ end of a growing chain of DNA
- DNA polymerase I removes the RNA primers from the leading strand and from the lagging strand’s fragments. It will then fill in the space with DNA nucleotides by extending the neighbouring DNA fragment
- DNA ligase enzyme joins the Lagging strand’s Okazaki fragments into one strand (if making RNA, then it’s RNA ligase)
- Synthesis of one strand of DNA (leading strand) proceeds continuously in the 5′ to 3′ direction
- Synthesis of the complementary strand (lagging strand) is more complex because it is running opposite to the leading strand, and DNA polymerase can ONLY add new nucleotides to a free 3′ end
- To solve this dilemma, the polymerase builds the lagging strand using many small pieces called Okazaki fragments
- DNA polymerase I will come and remove the RNA primers at the end of elogation
- On the leading strand, there’s only 1 RNA primer. If a bubble, there’s 2 on it.
- Primers are always on the 5’ end of the newly synsesized DNA

Transcription- mRNA
- RNA polymerase reads the template strand and adds complimentary RNA nucleotides in the 5’-3’ direction
- Thymine (T) is replaced by Uracil (U)
- No Okazaki fragments form
- As soon as this begins, another RNA polymerase can bind to the promoter region and start building another strand of RNA (rapid production of RNA)
- RNA Polymerase can synthesize new strands much faster than DNA Polymerase could during DNA Replication
- RNA Polymerase does not proofread the RNA! This is because when there’s a mistake, the RNA turns to protein unlike DNA, which would cause the cell to die if there’s a mistake
- During elongation, an RNA polymerase complex moves along the DNA strand, the DNA helix unwinds, and complementary RNA nucleotides are joined together. After the RNA polymerase has passed, the DNA double helix reforms.

Question 31

Q

The Lagging Strand

Answer

A

RNA primase attaches to DNA and synthesizes a short RNA primer (makes an RNA primer, a sequence of about 10 nucleotides, complementary to the parent DNA)
DNA polymerase III then adds nucleotides to the 3′end of the RNA primer
DNA polymerase I comes in and removes the RNA primers and replaces it with DNA
DNA ligase forms a phosphodiester bond between the 3′ OH of the growing strand and the 5′ phosphate in front of it
DNA is further unwound, new primers are made and DNA polymerase III jumps ahead to synthesize another Okazaki fragment
Synthesized discontinuously

Question 32

Q

Termination

Answer

A

DNA Replication
- Replication process is completed
- Two new DNA molecules separate from each other
- Replication machine is dismantled
- Occurs upon completion of the new DNA strands
- New DNA molecules separate from each other
- Replication machine is dismantled
- Everything falls off (helicase, DNA polymerase III, etc.)
- The three DNA polymerase proof read the nucleotides
- While elongating DNA, polymerase III proof reads DNA (initial proofread)
- When polymerase I removes RNA primers to add DNA, the second proofreading happens
- The third time happens when DNA polymerase II is at the end to proofread

Transcription- mRNA
- Specific sequence signals the end – STOP sequence
- When RNA polymerases reach this sequence, they detach
- The newly synthesized RNA is released and ready to be processed into mRNA
DNA double helix reforms

Question 33

Q

Forming mRNA

Answer

A

Recall: ALL nucleic acids are synthesized 5’to 3’
Every new nucleotide is added to a free 3’ –OH group
mRNA is synthesized off of the 3’ to 5’ DNA Strand
The template (antisense) strand is the 3’to 5’ strand of DNA
This is also the leading strand in DNA replication
mRNA is the same sequence as the 5’ to 3’ DNA strand, with U instead of T
the coding (sense) strand is the 5’ to 3’ strand of DNA
Also known as the lagging strand in DNA replication

Question 34

Q

Topoisomerase 2

Answer

A

Releases strain on the parent DNA molecule due to the unwinding process ahead of any replication forks.

Question 35

Q

Correcting Errors

Answer

A

DNA Polymerase I and II proofread newly synthesized DNA
DNA polymerases remove incorrect bases
Mismatch repair involves proteins recognizing mispaired nucleotides and replacing them

Question 36

Q

DNA

Answer

A

DNA stands for deoxyribonucleic acid
DNA controls all the chemical changes which take place in cells
The kind of cell which is formed, (muscle, blood, nerve etc) is controlled by DNA
Before a cell divides, the DNA strands unwind and separate
Each template strand allows DNA Polymerase III to add a new strand by adding the appropriate nucleotides
Result is that there are now two double-stranded DNA molecules in the nucleus
When cell divides, each nucleus contains identical DNA
This process is called replication

Question 37

Q

The link between DNA and Proteins

Answer

A

In 1953, Frederick Sanger showed that proteins consist of amino acids and that each protein consisted of specific amino acid sequences.
By the 1960’s a clear link was formed between genes and proteins
BUT how information was going from DNA to Proteins of amino acids was a still a mystery
RNA is the link between the two
RNA is slightly more stable than DNA because it has a OH group on the second carbon of its sugar

Question 38

Q

RNA

Answer

A

A genetic Messenger between DNA & Protein
Scientists began to look at RNA as an intermediary between DNA and proteins (Jacob and Monod, 1961)
Ribonucleic acid (RNA) is a single stranded polynucleotide.
2’ OH group on ribose sugar (DNA has H)
Nitrogenous base pair of uracil (U) instead of thymine (T)
RNA is synthesized in nucleus and transported to cytoplasm
This keeps the chromosome in the nucleus, safe from harm
The mRNA will also get degraded like DNA, but not as fast, and can make many proteins before degrading

Question 39

Q

mRNA

Answer

A

messenger
RNA that contains the genetic information of a gene and carries it to the protein synthesis machinery in the cytoplasm
mRNA is complimentary to the gene DNA sequence, opposites that complete each other
template strand: 3’ to 5’
coding strand is the same as mRNA strand, except T is coding turn to U in mRNA, both 5’ to 3’
IMPORTANT: body uses template strand to make the complimentary mRNA strand
mRNA holds the instructions for building chains of amino acids that will form proteins
In 1964, Francois Jacob and Sydney Brenner made these conclusions through experimentation:
-When bacteria were infected by a virus, a virus specific RNA molecule was synthesized and became associated with pre-existing bacterial ribosomes.
-The new RNA molecule has a base sequence complimentary to the DNA and carried the genetic information to produce the viral protein
-The viral RNA molecule was newly synthesized and was not a permanent part of the bacterial ribosomes

Question 40

Q

other types of RNA

Answer

A

tRNA (transfer RNA)
rRNA (ribosomal RNA)
snRNA* (small nuclear RNA)
siRNA* (small interfering RNA)
crRNA (CRISPR RNA)

CRISPR gathered lots of attention in the past few years. crRNA can be created by humans to match to a specific region of DNA, and with the protein Cas9, remove it and replace it with DNA of our choosing – Gene Editing (cutting edge technology) Possible key to replacing antibiotics.

Question 41

Q

Genetic Code

Answer

A

The sequence of bases in DNA forms the Genetic Code
A set of rules for determining how genetic information in the form of a nucleotide sequence is converted to an amino acid sequence of a protein
There are 4 nucleotides and 20 amino acids
Nucleotide bases of mRNA are read 3 at a time, this is known as the triplet hypothesis
Each triplet is called a codon (e.g. GCU or UUU)
There are 64 different codons (43 = 64)
Combinations = possible choices to the power of the number of places in a codon
mRNA is read as Codons of 3. Each codon will code for an Amino Acid

Question 42

Q

Reading mRNA Codons as Amino Acids

Answer

A

Start by finding the first letter of the mRNA codon in the “first base” column.
Read across the rows in the “second base” column to find the second letter of the codon. This will take you to four possible amino acids
Finally, read down the “third base” column to find the last letter of the codon.
The last letter of the codon combined with the previous two letters, identifies the amino acid that corresponds to the codon

Question 43

Q

Characteristics of Genetic Code

Answer

A

It is redundant
- More than one codon codes for same amino acid
- E.g. AUU and AUC code for isoleucine

There are 3 codons that do not code for an amino acid
- They are called STOP Codons (end of translation)
- UAA, UAG, UGA

It is continuous (no spaces, no overlap)
It is universal (except for some protists)
AUG is the START codon (methionine)

Question 44

Q

DNA and Enzymes

Answer

A

DNA allows for proteins to be constructed, proteins build cell structures
Proteins can also be enzymes
The DNA controls which enzymes are made, and the enzymes determine what reactions take place
The structures and reactions in the cell determine what kind of a cell it is and what its function is (muscle cell, blood cell, bone cell, sperm cell, nerve cell etc.) 200 Different cell types in the human body!
DNA exerts its control of a cell through the enzymes it codes for
A sequence of triplets in the DNA molecule may code for 1 complete protein
Such a sequence forms a gene
There may be a thousand or more bases in one gene

Question 45

Q

The Central Dogma of Genetics

Answer

A

A rule or set of rules known to be irrevocably true
-so true and known to be, that it will not be changed in the future
DNA is too valuable to leave the nucleus
It could be damaged, leading to cell death
mRNA carries genetic info out of nucleus
Genes on DNA are regulated, allowing for mRNA to be synthesized quickly, slowly, or not at all. This controls the number of different proteins being made in a cell
Defines cell and its functions, makes cells alive because it governs cells
Doesn’t contain replication, only transcription and translation

Question 46

Q

Gene Expression

Answer

A

The two main steps in gene expression are transcription & translation
Transcription: mRNA is synthesized based on a DNA template
Translation: Ribosome assembles amino acids based on the mRNA codons to synthesize a protein
Gene expression refers to the synthesis of protein based on the DNA sequence of a gene
Gene expression in eukaryotes involves transcription to form mRNA, which undergoes processing before being exported to the cytoplasm. Once in the cytoplasm, the mRNA is used as the template for protein synthesis by the ribosomes.

Question 47

Q

What structural characteristic of A-T bonds make it a good starting point?

Answer

A

Only 2 hydrogen bonds hold adenine and thymine together, making it an easier bond to break.
This allows the DNA to unwind easier than at a G-C bond to initiate transcription

Question 48

Q

Binding to the promoter region

Answer

A

Every exon (coding information, exon) starts with a promoter and ends with a terminator
When RNA polymerase comes into the DNA, it looks for the promoter site b/c of the TATA box
A promoter region is where the RNA polymerase complex binds to initiate transcription
It is rich in A-T bonds
example: the TATA box in E. coli (shown above) contains 2 promoter regions
The promoter is upstream (5’) of the start site. This allows it to bind to the correct strand
: the TATA box is an EXAMPLE from E. coli, it is not present in every organism, although it shows fairly high conservation across species, with small variations.

Question 49

Q

POSTTRANSCRIPTIONAL MODIFICATIONS

Answer

A

We need to modify our pre-mRNA molecule to allow it to get out of the nucleus and be translated into a polypeptide of amino acids!
Prokaryotes: mRNA is used immediately
In eukaryotic cells, the primary transcript needs to undergo capping, tailing and base excision before it can leave the nucleus
Modifications convert precursor mRNA (pre-mRNA) to mature mRNA
Increase stability and resistance in the open cell environment, once outside the nucleus

Question 50

Q

Transcription

Answer

A

Transcription of a DNA template produces an RNA molecule that is a copy of the genetic information. The nucleotide sequence of this newly formed mRNA molecule is then translated using the genetic code to produce the protein coded for by the gene.

The entire mRNA transcript consists of two regions:
1. coding region called exons
2. non-coding region called introns

A coding region refers to a sequence of DNA bases that will produce a protein

Question 51

Q

CAPPING

Answer

A

Cap is added to the 5’ end of mRNA strand (7-methyl guanosine, a modified guanine nucleotide triphosphate)
Capping has two functions: IMPORTANT
1. protect RNA from digestive nucleases as it exits the nucleus and enters cytoplasm
2. recognized by ribosome to translation

Question 52

Q

TAILING

Answer

A

~200 adenine ribonucleotides, known as a poly-A tail, is added to the 3′ end of the transcript by an enzyme called poly-A polymerase
Increases stability, prevents degradation of mRNA near the ends

Question 53

Q

BASE EXCISION (Splicing)

Answer

A

SnRNA’s attach on both sides of intron and then gravitate towards each other, pulling the exons towards each other. This forms a splicosome, which has 4 subunits (2 SnRNA and 2SnRNP). This cuts the introns out. RNA ligase glues exons together
Introns are removed, or else they would prevent translation
Spliceosomes remove introns (cut them out) and join the exons together so the transcript is one continuous coding gene
Introns stay inside the nucleus and are degraded into recycled nucleotides
The “primed” mRNA transcript then moves out of the nucleus and into the cytoplasm where it will be translated (to come…)

Question 54

Q

Silent mutation

Answer

A

Mutation that has no effect on amino acid sequence of a protein
Can be a change of 1 or more base pairs

Question 55

Q

Mutagens

Answer

A

Cause mutations
Physical Mutagens
- They physically change the structure of DNA and range from point mutations to loss of large portions of chromosomes
- get into nucleus and change structure of DNA, more prominant than chemical mutagens
- e.g., X rays, UV Rays, Gamma Rays

Chemical Mutagens
- Can enter the nucleus of a cell and induce mutations by reacting chemically with the DNA. They can cause nucleotide substitutions or a frameshift mutation
- e.g., nitrites, gasoline fumes, cigarette smoke, and any other carcinogens (cancer causing substance).

Question 56

Q

TRANSLATION

Answer

A

Synthesis of a protein from an mRNA template
Process involves several key molecules:
mRNA
the small and large subunits of the ribosome
tRNA and the release factor
Process is broken into three stages: initiation, elongation, and termination

Initiation of Translation
- Small ribosomal subunits bind to the 5′ cap of the mRNA transcript and translation commences
- mRNA codon AUG-codes for methionine amino acid (START)
- tRNA with anticodon UAC, carrying methionine will bind AUG site on mRNA
- Large ribosomal subunit binds the small subunit now, and activates the complex

Elongation of translation
- Ribosome moves along mRNA from 5’ to 3’, reading the code in triplet codons
When the start codon is in the P site, tRNA has delivered methionine
- Second codon is now in the A site
- Appropriate tRNA delivers the next amino acid in the protein sequence
- Peptide bond is formed between methionine and the second amino acid in the P site
- Then, the Ribosome subunits shift down one codon (moving in 5′ to 3′ direction of mRNA strand)
- mRNA is read 5’-3’
- The polypeptide chain grows while it is in the P site
- The start codon establishes the reading frame – all codons read by the ribosome to form the protein
- Methionine is transferred to the A-site amino acid, the first tRNA exits, the ribosome moves one codon (3 base pairs) along mRNA
- First tRNA goes to pick up another amino acid (met)
- This puts the amino acid chain in the P-site, and frees up the A site for the next tRNA
- As elongation continues, the growing peptide is continually transferred to the A-site tRNA, the ribosome moves along the mRNA, new tRNAs enter, and peptide bonds are formed
- Process of elongation continues until a stop codon is read in the A site
- Stop codons are UAG, UGA and UAA

Termination of translation
- Elongation ends when a stop codon is encountered in the A-site
- A release factor enters the A-site and causes the ribosome subunits to disassemble (small and large subunits fall apart) and translation is terminated, releasing the mRNA and newly formed protein
- Protein is folded and modified and then targeted to areas of the cell where it is required
- mRNA can be retranslated, or can be degraded immediately
- The anticodon is also in RNA format
- PROTEIN SEQUENCE IS STILL READ FROM MRNA MOLECULE

Review
Initiation
- Ribosome binds to a specific site on the mRNA
Elongation
- Ribosome moves along the mRNA 3 nucleotides at a time (5’ – 3’ direction)
- Each set of three nucleotides (called a codon) codes for one amino acid
- tRNA delivers the appropriate amino acid to the mRNA during translation
Termination
- The ribosome falls off the mRNA when it comes across a stop codon

Question 57

Q

Transfer RNA (tRNA)

Answer

A

specific to mature mRNA. To find it, get rid of introns so that you only have mature mRNA, and then find the complimentary to it. Start with UAC (anticodon of start codon, AUG)
Single stranded RNA folded into a clover shape
Carry Amino Acids to the ribosome
The tRNA molecule is a cloverleaf in 2-dimensional shapes, and a ‘boot’ shape in 3 dimensions
Hydrogen bonds and base pairings twist the clover shape into the boot shape
There are only 20 amino acids but 61 tRNAs b/c of redundancy, which carries the same amino acids with different codons

Anticodon and Acceptor stem on tRNA
- On the bottom stem loop
- The 3 base portion of tRNA that binds to mRNA
- It is complimentary to the mRNA strand
**Written 3’-5’, since mRNA is 5’-3’
- At the opposite 3’ end of the tRNA is the acceptor stem
- This is where amino acids are attached and carried to the ribosome
- The scanner tries to find its complimentary base pairs, when it does, it attaches to amino acid that is on it. This is why it has anticodon If codon, it wouldn’t be able to find its complimentary base pairs

Question 58

Q

Polypeptide chain

Answer

A

A protein is built one amino acid at a time
Amino acids are added to the chain one by one by the ribosome
Forms a peptide bond between each amino acid
Amino acids are carried to the ribosome by tRNA

Question 59

Q

Chromosomal Mutations

Answer

A

A change to the number of chromosomes is always negative, and often lethal
Eg. trisomy 21 = Down’s Syndrome
There can also be mutations of entire chromosome sections, which would affect several/many genes (Deletion, Duplication, Inversion, Translocation)

Question 60

Q

The Structure of Ribosomes

Answer

A

Ribosome exists as 2 subunits
large and small subunits
When both attach together, they create a ribosome
The small subunit binds mRNA
Large subunit binds 3 tRNA molecules at a time
Begins to grow amino acid chain (polypeptide)
Each subunit is made of different proteins and rRNA

P-A-E sites of Large ribosomal subunit
- The P (peptide) site holds the tRNA of the amino acid being added to the polypeptide chain
- Forms the peptide bond to link each amino acid
- The A (amino) site accepts the next tRNA holding the next amino acid molecule
- The E (exit) site removes tRNA molecules that have given their amino acids to the chain

Question 61

Q

DNA MUTATIONS

Answer

A

CHANGES IN THE GENETIC MATERIAL OF AN ORGANISM
TWO CATEGORIES:
Single-Gene Mutations involve changes in the nucleotide sequence of one gene
Chromosome Mutations involve changes in chromosomes, and may involve many genes

Question 62

Q

MISSENSE MUTATION

Answer

A

Mutation that changes the amino acid sequence of a protein
This can be good or bad
It may introduce a new protein that will help an organism survive it’s environment

Question 63

Q

polypeptide bonds

Answer

A

The bond forms between the amino terminal of the first amino acid and the carboxyl terminal of the next amino acid

Question 64

Q

POINT MUTATION

Answer

A

Single-gene mutation resulting from a change in a single base pair
happen at a specific area in the genetic code, happen within the nucleotides
Can involve substitution/insertion/deletion of a single base pair
Substitutions have a fairly minor effect on cell due to redundancy of genetic code
Eg. A mutation of GGA (glycine) to GGG (glycine), will have no effect on the organism

FRAMESHIFT MUTATION
- Insertion or deletion of nucleotides
- Causes the entire reading frame of gene to be altered (yikes!)
- Causes changes to neighbouring triplets as well

Answer 65

A

Once produced, the amino acids sequence (primary protein structure) begins to fold to form a functional 3D protein
The proteins formed by transcription and translation are sent to the Golgi body for further modification

Answer 66

A

Somatic cells vs Gametes. No. Negative mutations that do not provide a selective advantage are typically not passed on. There are DNA repair mechanisms that will repair this mutation before it gets passed along. Even positive mutations (provide an advantage) may be repaired before being passed on.

Answer 67

A

Most genes are not constantly producing proteins – this would require too much energy
Genes are “turned on” (transcribed and translated) and “turned off” throughout the life of an organism at appropriate times
Genes that are constantly expressed are called constitutive genes aka “housekeeping genes”
These genes express proteins important to the survival of the organism
Most genes are regulated, so that the protein is expressed only in certain amounts at certain times when its needed
Gene regulation involves turning on or off specific genes depending on the needs of the organism

Answer 68

A

Mutation that shortens a protein by introducing a stop codon

Answer 69

A

Transcriptional - regulates which genes in DNA are transcribed
Posttranscriptional – mRNA undergoes changes in the nucleus
Translational – controls how often and how rapidly mRNA transcripts will be translated into proteins
Posttranslational– control the rate at which a protein becomes active

Answer 70

A

positive feedback loop because it’s not inhibited and is always off but on whenever needed
An example of an inducible system to regulate gene expression
lac Operon consists of a cluster of three genes that code for proteins involved in the metabolism of lactose: The three genes are lacZ, lacY and lacA
The lac operon contains three genes (Z, Y, and A) needed for the breakdown of lactose in E. coli. All three genes are under the control of one promoter. Therefore, they undergo the same level of regulation. When lactose is not present, the lac repressor inhibits transcription.
CAP = catabolite activator protein (protein that binds to help produce proteins to catabolize, or breakdown, lactose)
used a lot in research b/c simple, not very big, not complex, grows quickly, this helps to manipulate DNA easier

Lac Operon “OFF”
- When lactose is absent this inhibits lac protein expression. A repressor protein (Lac1 protein) binds to the operator site.
- this blocks RNA polymerase from binding to the promoter
- no β-galactosidase is produced
- Negative feedback loop. In conditions of low lactose, there is increased activity of the repressor protein, which is decreasing the output of this operon.

Lac Operon “ON”
- The presence of lactose acts as an inducer. A derivative of lactose, allolactose binds to the Lac1 protein (repressor) so it can no longer bind to the operator of the lac operon DNA
- RNA polymerase can now bind to the promoter region and transcribe the lacZ, lacY, lacA enzymes

CAP is an activator
- When the cell detects low glucose, it will attempt to increase lactose metabolism and will therefore need these lactose metabolizing enzymes in high amounts.
- In conditions of low glucose, there tends to be high cAMP which binds to the CAP protein and allows it to attach to the CAP site and greatly increase the binding of RNA Polymerase to the promotor

Answer 71

A

Photorepair – A specific repair mechanism that repairs damage to DNA caused by exposure to UV radiation. A photolyase enzyme recognizes the damage, binds to the site and corrects it
Excision repair – Non-specific repair mechanism whereby parts of damaged DNA are removed and DNA polymerase puts down new/correct nucleotides

Answer 72

A

A cluster of genes in a DNA strand under the control of one promoter (only in prokaryotes)
acts as simple regulatory feedback loop
consists of a promoter, operator, and two or more protein-coding genes
Promoters and operators do not code for proteins, but serve as binding sites for regulatory proteins
When a repressor protein binds to the operator, transcription of the structural genes is inhibited (control)
Two examples of operons are the lac Operon and the trp Operon
Inhibition and activation are the ways in which genes are regulated and controlled

Answer 73

A

lactose is a disaccharide found in milk/milk products, consists of two sugars (glucose and galactose)
β-galactosidase is the enzyme responsible for the degradation of lactose
It is produced by the lac operon in E. coli

Answer 74

A

negative feedback loop because it’s inhibited and is always on but off if needed
Trp Operon in E. coli contains five genes that are involved in the synthesis of essential amino acid tryptophan.
This operon is normally transcribed, until the cell has sufficient tryptophan. Once enough tryptophan is present for normal cell functioning, the extra tryptophan binds to the repressor protein allowing it to attach to the operator and inhibit transcription.

The trp operon is “OFF”
- Regulates genes for tryptophan (an amino acid) production in prokaryotes
- The trp operon is inhibited when high levels of tryptophan are present
- Tryptophan is a co-repressor because it binds with the trp repressor protein and activates this repressor protein (transcription proceeds)
- This complex will the deactivate (turn off) gene expression of the trp operon

The trp Operon is “ON”
- Lack of tryptophan deactivates the repressor and activates transcription via RNA Polymerase
- RNA polymerase transcribes trp operon genes

Answer 75

A

Both operons are very well studied and understood
Researchers will insert this gene into bacteria along with a gene being researched
To confirm that the genes have been inserted, scientists use a modified form of lactose.
If the gene has been inserted properly, lactose will be digested and turn the bacteria BLUE
This strain can now be used for research of the other gene attached
easy to manipulate and is mainly found in prokaryotic cells

Answer 76

A

Regulation of eukaryotic gene expression occurs at multiple stages of protein production. The regulation points occur in both the nucleus and the cytoplasm.

Answer 77

A

Pre-transcriptional (DNA)
Transcriptional (DNA to mRNA)
Post-transcriptional (mRNA)
Translational (mRNA to protein)
Post-translational (protein)
*Note: genes are not organized into operons in eukaryotes.
Each gene has its own promoter
Distinct transcription factors control each gene

Answer 78

A

every gene has its own polymerase which has its own promoter
DNA level control
Pre-transcription: Highly condensed chromatin acts as a physical barrier to prevent transcription and the formation of pre-mRNA
Transcriptional Control: Certain types of activator proteins also enhance transcription initiation by binding to transcription factors and RNA polymerase
These are known as enhancers
Having multiple activator regions on a gene allows a gene to be highly tuned to specific environmental factors (like fine tuning your speed while on cruise control in a car)

Transcriptional Control – DNA Level Regulation through Activators and Transcription Factors

Transcription factor binding is required for transcription.

Activators enhance rate of transcription

Answer 79

A

When the protein comes off the ribosomes, they get tagged and the proteas get signaled by ubiquitin to eat it. This only happens if protein isn’t needed.
factors can attach to protein to stop it from folding
The two top things are not ideal bc it’s a waste
if needed, other proteins will come and attach to that protein and make it superfold so that it’s functional
Polypeptide level control
Proteins must be activated by modifications to the amino acid chains
E.g. Insulin is initially folded into 3D structure
Removal of amino acids activates it
Cells can tag a protein for degradation
Adding a chain of ubiquitin molecules to a protein can signal for protein degradation
Proteasome complexes notice the ubiquitin tag and will degrade the protein

Answer 80

A

mRNA level control
Alternative splicing can produce different mRNA molecules
5’ Cap and 3’ poly-A tail are purposefully not added (what will happen?)
RNA interference – the regulation of gene expression by small RNA’s (sRNA); it inhibits gene expression by degrading mRNA or inhibiting translation by binding to it, inhibits the mature RNA from making protein
Small RNA’s (micro RNA) and small interfering RNA can inhibit translation and interact with specific mRNA’s
if ribosomal subunits attach, where methionine attaches to p site, translation can be blocked by a site. If a site is blocked, only methionine is in, but won’t be enough to fold so it won’t be functional

Answer 81

A

DNA made in a lab that consists of material from more than one source
Different sources are often different species
Became possible with the discovery of Restriction Enzymes
Used by bacteria to cut up viral infecting DNA and protect their own genomes from the viral DNA
Restriction Endonuclease: a unique restriction enzyme that cuts DNA within the strand.
Does so by recognizing a short sequence (Target sequence) and cleaving it at the restriction site. Different endonucleases recognize different target sequences.

Answer 82

A

Cut DNA at specific sites
leave “sticky ends”  leftover hanging bits
palindrome (reads the same in the 5’ to 3’ direction)
Cuts are made at the restriction sites

EcoRI
- Each restriction enzyme that we discover is known to recognize a specific sequence
- EcoRI is a restriction endonuclease that will recognize this DNA sequence (notice that it is a palindrome
- Recognition sequence always the same 5’ to 3’

Restriction fragments with sticky ends form base pairs w/ other single stranded regions with the correct complimentary sequence
This is very specific, as all the nucleotides must match to bond
Blunt cuts reduce specificity
No sticky ends that can form complimentary base pairs
Two fragments with blunt end can combine
Less efficient
Only 2 nucleotides must match, higher chance that unwanted DNA will bind to this site

Answer 83

A

A restriction endonuclease is selected
Can cut both of the DNA fragments that
- we want combined
Each DNA piece is reacted with the restriction endonuclease to produce cut fragments that will have matching sticky ends
Cut DNA fragments are incubated with DNA ligase (joins strands)

Answer 84

A

Creating recombinant DNA in a lab allows us to express proteins in larger quantities than they normally would be in vivo (occurring naturally in a living thing)
Expression of our new recombinant DNA can be controlled when we pair our recombinant DNA with an inducible gene (lac operon)
Straight forward process, carried out in microlitre quantities

Answer 85

A

Gene Cloning: making multiple identical copies of a gene or segment of DNA using bacteria
Transformation: bacterial DNA takes up the inserted DNA under specific conditions

Answer 86

A

In bacteria, the vector (carrier of gene) is a plasmid (circular DNA)
- Contains an origin of replication (plasmid will be replicated and included in the daughter cells)
- Plasmids are grown to have resistance to the antibiotic Ampicillin.
- Plasmid has a restriction endonuclease site within the lacZ gene… if the new DNA is added here, it will deactivate the lacZ gene
- Selectable markers allow researchers to identify which bacteria are the bacteria that contain the recombinant DNA.
Transformation
- process for which foreign DNA is taken up by bacteria
- Certain chemicals make cell membrane porous, allows it to take in DNA
Bacteria cells grow in a Petri dish
- Growth Gel in the petri dish contains antibiotic Ampicillin and
X-gal to makes bacteria blue when broken down by the enzyme from the lacZ gene
- If the cells turn blue, we know that the lacZ gene is active and our recombinant DNA was not added successfully
- Selectivity is arguably the most important step. We need to be able to identify which colonies have this gene in them
Bacteria with recombinant DNA are ID’d thanks to the selective markers
- Bacteria growing contains recombinant DNA or plasmid
- Blue bacteria contains an active lacZ gene to therefore we know that no gene was inserted in these bacteria
- White colonies contain recombinant DNA
- Selectivity: We will know if the bacteria has the inserted DNA if:
1) it grows (possesses antibiotic resistance gene)
2) it stays white
- if it turns blue, the lacZ gene is active
- if it is active, it was not cleaved by the R.E
- if it was not cleaved, the DNA could not have been inserted
Cells with recombinant DNA are selected and grown in liquid culture
Recombinant DNA molecules are isolated and purified from bacterial cells
Analysis techniques are used to confirm correct recombinant DNA is made

Answer 87

A

DNA Amplification: production of large quantities of DNA from a sample
Polymerase Chain Reaction (PCR): Automated method for amplifying specific regions of DNA from very small quantities
Produces billions of copies of a section of DNA in a test tube within hours
Very important in Forensics = amplify a very very small amount of a suspects DNA into a quantity that is testable for distinguishing markers

The Polymerase Chain Reaction
1. Double strand DNA heated to 94-96⁰C to denature it into single strands (split it up)
2. Cooled to 50-60⁰C to allow RNA primers to anneal(bind) to the strands
3. Heated to 72⁰C, where a special heat-resistant DNA polymerase (Taq Polymerase) can optimally function and add complimentary nucleotides from the primers
4. Cycle repeats 30 cycles to produce 1 billion copies of initial strand

Answer 88

A

Once DNA has been amplified, there are various methods for analyzing and identifying it
One standard method is Gel Electrophoresis: used to separate DNA molecules according to mass/charge
DNA is negative due to its phosphate (negative charges will repel it)
DNA can be cut into different sizes (larger fragments move slower in gels)

Made up of:
- Agarose, a protein gel that allows DNA to separate and move within it
- Electrodes (positive/negative electric charges)
- Separate wells to add samples

Answer 89

A

Gel is treated with ethidium bromide and added to wells
- The dye binds the backbone of DNA and fluoresces(glows) under UV light
- Helps us to see the DNA on the gel
Gel is placed in buffer and electric current is run from cathode(-) to anode(+)
- Negatively charged DNA (due to phosphate groups) will move from the negative end towards the positive end
- Buffer maintains pH of solution (keeps DNA from denaturing)
Smaller DNA fragments move faster/further than larger fragments
- They fit through the agarose gel easier
Gel is removed from buffer and exposed to UV light
- Ethidium bromide in the DNA backbone glows and allows us to see the DNA

A ladder is a lane that contains DNA of known size (base pair length)
We compare the fragments in the gel to this standard

Answer 90

A

a.k.a DNA profiling
Each individual has unique DNA (except identical twins)
In the past Restriction Fragment Length Polymorphism (RFLP) was used
Cut up DNA and run on a gel
Current method is short tandem repeat (STR) profiling

Answer 91

A

the 5’ carbon is on the fifth end, which has a phosphate group
the 3’ end has an exposed OH group on the 3’ end of the carbon
DNA is always built 5’-3’ because the exposed OH on the 3’ can be removed to add more nucleotides there (condensation reaction)
replication makes DNA that has two strands that are anti-parallel, this is because this maximizes hydrogen bonding b/c the two nitrogenous bases are closer together, making it easier to have hydrogen bonds

Answer 92

A

There are 20 amino acids
- 8 are essential (taken in by food)

Answer 93

A

with the start codon (AUG), we don’t produce it, but its the start to every single protein
The reason why some codons are repetitive and end up being the same name is because it’s a failsafe incase of mutations
Runs from N to C

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bio unit unit 3 Molecular genetics Flashcards

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