Final Flashcards

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1
Q

What are lethal genes?

A

Genes which result in the reduction of viability of an individual or become a cause for death of individuals carrying them - they are usually a result of mutations in genes that are essential for growth or development

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2
Q

Describe study by Lucien Cuenot – how were lethal genes first discovered?

A

The agouti gene in mice is largely responsible for determining coat color
– The wild-type allele (A) produces a blend of yellow and black pigmentation in each hair of the mouse – this yellow and black blend may be referred to as ‘agouti’ in color
– One of the mutant alleles (AY) of the agouti gene results in mice with a much lighter, yellowish color

1) When two homozygous wild-type agouti mice were crossed, only agouti offspring were obtained (which was expected)
• When yellow mice were crossed with homozygous wild-type agouti mice, a 1:1 phenotypic ratio of yellow and agouti offspring were obtained
– This indicated that the yellow mutation is dominant, and all the parental yellow mice were heterozygous for the mutant allele ( AYA)
- Instead, he observed a 2:1 phenotypic ratio of yellow to agouti mice
2) By mating two yellow mice, Cuénot expected to observe a usual 3:1 phenotypic ratio of yellow to agouti
- Instead, he observed a 2:1 phenotypic ratio of yellow to agouti mice
– After many genetic crosses, he was unable to produce any mice that were homozygous for the yellow allele
- As it turned out, this unusual ratio reflected that some of the mouse embryos (homozygous AYAY genotype) died very early in development, long before birth

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3
Q

What are recessive lethals and give examples

A

These are most common type of lethal genes, and their expression occurs in homozygous conditions only, hence survival of heterozygous ones are unaffected
• They are a pair of identical alleles that results in the death of an
entity ultimately
• Despite the fact that recessive lethals can code for recessive or dominant characteristics, they turn fatal only under homozygous conditions
– Examples involving recessive lethals that code for dominant traits = Manx cats, yellow coat color in mice, creeper chickens
– Example involving recessive lethals that code for recessive traits = sickle cell anemia, cystic fibrosis

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4
Q

What are dominant lethal genes and provide an example?

A

Lethal genes which reduce viability even in heterozygotes, are said as dominant lethals
• These are the alleles whose presence is required in one copy in an entity for them to turn fatal
• The dominant lethal inheritance pattern is one in which an allele is lethal both in the homozygote and the heterozygote
– These alleles are not commonly found in populations because they usually result in the death of an organism before it can transmit its lethal allele on to its offspring – This allele can only be transmitted if the lethality phenotype occurs after reproductive age
EXAMPLE: Huntington’s disease
– Huntington’s disease is a rare neurodegenerative disorder that ultimately results in death
– This disease is due to a dominant autosomal allele ( H), where the onset of the disease in heterozygotes (Hh) is delayed, usually well into adulthood

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5
Q

What are conditional lethals and give an example of

A

These alleles turn deadly only when there is an external environmental aspect involved
• EXAMPLE: Favism
– A sex-linked inherited condition that causes the carrier to develop hemolytic anemia when they eat fava beans
o Hemolytic anemia = a condition in which red blood cells break apart and block blood vessels
o Blockage can cause kidney failure and result in death
- Results from deficiency in an enzyme called glucose-6-phosphate dehydrogenase (G6PD)

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6
Q

What does each nucleotide consist of?

A

• Phosphate group
• Deoxyribose sugar group
• Nitrogen-containing base

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7
Q

What are the four kinds of bases in DNA?

A

Pyrimidines: one single ring, with 6 members and 2 nitrogen atoms
- Utracil
- Thymine
- Cytosine
Purines: two-ringed structure with 9 members and 4 nitrogen atoms
- Adenine
- Guanine

Useful mnemonic
PURE As Gold
CUT the PYE

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8
Q

What are the parts of a DNA (structure/bonds, etc)

A

DNA is a double-stranded helix
Strands are antiparallel to each other
– The two strands run side by side to each other but in opposite directions – The 5’ end has a free phosphate group while the 3’ end has a free hydroxyl (OH) group
• One turn of the helix occurs every 10 base pairs
• Negatively charged phosphate groups face out (making DNA hydrophilic overall)
- Important bonds that hold the structure – hydrogen bonds between nucleotide bases and phosphodiester bonds between the 5-carbon sugar backbone

Complementary base pairing (Watson-Crick):
– Thymine and adenine (2 hydrogen bonds)
– Cytosine and guanine (3 hydrogen bonds)

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9
Q

What is an origin of replication?

A

• Origin of replication = DNA sequence where replication is initiated
• A replication bubble is formed (each bubble has two replication forks)
• A eukaryotic chromosome may thousands of replication origins, bacterial chromosomes have a single origin of replication

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10
Q

What is the mechanism and proteins in DNA replication

A

– DNA helicases unwind the double helix by breaking hydrogen bonds between strands
– Single-stranded DNA binding proteins prevent DNA strands from rejoining after helicase separates strands
– The initial nucleotide strand is an RNA primer (RNA primase catalyzes the synthesis of short RNA primers)
– DNA replication is catalyzed by DNA polymerase III which needs an RNA primer
– DNA polymerase I degrades the RNA primer and replaces it with DNA nucleotides (synthesized in 5’-
to-3’ direction, reads this template strand in a 3’-to-5’ direction - has proofreading activity)
– DNA ligase joins the short DNA fragments on lagging strand (Okazaki fragments) into a continuous daughter strand (introduces a phosphodiester bond between two fragments)
– Topoisomerase binds to DNA and relieves torsional stress experienced further upstream along helix that occurs as a result of unwinding due to helicase activity (cuts phosphodiester bond from one or both strands of double helix to relax DNA, and then cut strand(s) is/are reannealed

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11
Q

What is the leading vs lagging strand?

A

Leading strand: continuous synthesis
- the leading strand is synthesized continuously in the 5’-to-3’ direction (new strand) by DNA polymerase, starting from 3’ end of original
Lagging strand: discontinuous synthesis
- synthesized discontinuously starting from 3’ end of original, primase synthesizes a short RNA primer which is extended by DNA polymerase to form an Okazaki fragment
- DNA ligase joins the Okazaki fragments

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12
Q

What was the Fredrick Griffith Experiment?

A
  • Experiment involved the use of two strains of the pneumococcus bacterium: One strain of bacteria had a smooth polysaccharide capsule (also known as the “smooth” strain) that is essential for infection while the other (“rough” strain) did not have a capsule
    • Injection with live encapsulated bacteria (virulent form: S strain) – mice contracted
    pneumonia and died
    • Injection with live naked bacteria (non-virulent form: R strain) – mice lived
    • Injection with heat killed encapsulated bacteria – mice remained healthy
    • Injection with heat killed encapsulated bacteria and live naked bacteria – mice contracted pneumonia and died
  • conclusion: a chemical substance from one cell is capable of genetically transforming another cell
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13
Q

What was Oswald Avery’s Experiment

A

Decided to identify specific molecules that could transform non-encapsulated bacterium into encapsulated form – In a significant departure from Griffith’s procedure, however, Avery’s team employed a method for transforming bacteria in cultures rather than in living mice, which gave them better control of their experiments
1) heat-killed virulent S strain filtrated treated with enzymes that either destroy RNA, proteins, or DNA
2) the treated samples were added to cultures of R strain bacteria.
- cultures treated with RNase or protease contain transformed S strain bacteria.. ..but the culture treated with DNase does not.
CONCLUSION: Because only DNase destroyed the transforming substance, the transforming substance is DNA

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14
Q

What was the Hershey-Chase Experiment?

A

Wanted to confirm if DNA is the carrier of genetic material based on the findings by Avery
– Experiments involved the T2 bacteriophage, a virus that infects E. coli bacterium
o Use radioisotopes to trace the fate of phage’s protein and DNA:
1) Experiments begins with culturing of viruses in two types of medium
– One set of viruses was cultured in medium of radioactive sulfur ( 35S) – which would incorporate
into the phage protein
– Other set of viruses was cultured in medium of radioactive phosphorous (32P) – which would
incorporate into the phage DNA
2) The radioactively labelled phages were allowed into infect E. coli bacterial cells in separate flasks – After attachment of a phage to E. coli the phage DNA will enter cytoplasm of bacterial cell
• The infected E. coli cell cultures are agitated through blending to separate the phage viral coats (or phage “ghosts”) from the bacterial cells
• Centrifugation allows for separation of the pellet (containing the bacterial cells)
from the supernatant (containing the phages)
• 32P is discovered within the bacterial pellet whereas 35S is discovered in the
supernatant where the phage viral coats are found
• CONCLUSION: Injected DNA of the phage provides genetic information

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15
Q

What are the three postulated mechanisms of DNA replication:

A

Conservative: both strands of the daughter double helices contained new material
Semi-conservative: each daughter double helix contained one parent strand and one new strand (so total of two strands after replication)
Dispersive: results in two DNA molecules that are mixtures/hybrids of parental and daughter DNA

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16
Q

What was the Meselson/Stahl Experiment?

A

Question: which model of DNA replication-conservative, dispersive, or semiconservative-applies to E. coli?
• Bacterial ( E. coli) DNA is placed in a medium containing heavy nitrogen ( 15N isotope), which binds to the DNA, making it identifiable
• This DNA is then placed in a medium containing light nitrogen ( 14N isotope) and left to replicate for specific rounds of replication - the new bases will contain 14N while the originals will contain 15N
• The DNA is placed in test tubes containing caesium chloride (heavy compound) and centrifuged at 40,000 revolutions per minute - The caesium chloride molecules sink to the bottom of the test tubes creating a
density gradient (the DNA molecules will position at their corresponding level of density – taking into account that 15N is more dense than 14N)
• These test tubes are observed under ultraviolet rays (DNA appears as a fine layer in the test tubes at different heights according to their density)
Conclusion: DNA replication is semi-conservative

17
Q

What was the Meselson/Stahl Experiment?

A

Question: which model of DNA replication-conservative, dispersive, or semiconservative-applies to E. coli?
• Bacterial ( E. coli) DNA is placed in a medium containing heavy nitrogen ( 15N isotope), which binds to the DNA, making it identifiable
• This DNA is then placed in a medium containing light nitrogen ( 14N isotope) and left to replicate for specific rounds of replication - the new bases will contain 14N while the originals will contain 15N
• The DNA is placed in test tubes containing caesium chloride (heavy compound) and centrifuged at 40,000 revolutions per minute - The caesium chloride molecules sink to the bottom of the test tubes creating a
density gradient (the DNA molecules will position at their corresponding level of density – taking into account that 15N is more dense than 14N)
• These test tubes are observed under ultraviolet rays (DNA appears as a fine layer in the test tubes at different heights according to their density)
Conclusion: DNA replication is semi-conservative

18
Q

What are the three main types of RNA?

A

Messenger RNA: Used as a template to make proteins
- 3 to 5% of total RNA
Transfer RNA: Brings amino acids to mRNA and ribosomes for assembling proteins
- 15% of total RNA
Ribosomal RNA: Makes up structural component of ribosomes
- 80% of total RNA

19
Q

What is the process of RNA transcription?

A

In transcription, a portion of double-stranded DNA template gives rise to a single-stranded RNA molecule
• Occurs in nucleus of eukaryotes and in cytoplasm of prokaryotes (i.e.,bacteria)
1) INITIATION STEP:
– RNA polymerase binds on the gene’s promoter
• Promoter: A short DNA sequence having regulatory function over transcription of an
adjacent gene, and to which RNA polymerase binds prior to transcription
• RNA polymerase binds to promoter with assistance of transcription factors, which are DNA-binding proteins that recognize and accurately initiate transcription at specific promoter sequences
Usually found near the beginning of a gene, the promoter has a binding site for RNA polymerase
Typical promoter for RNA polymerase II consists of a TATA box about 25 bases upstream from the startpoint [startpoint designated as +1]
The typical promoter in prokaryotes consists of the start point (designated as +1), the six-nucleotide -10 sequence, and the six-nucleotide -35 sequence
– -10 sequence is also known as -10 element (located 10 nucleotides upstream from
the startpoint)
– -35 sequence is also known as -35 element (located 35 nucleotides upstream from
the startpoint)
• RNA polymerase recognizes and binds directly to these sequences.
– The two elements position the RNA polymerase in the right spot to start transcription, and they also make sure this polymerase is pointing in the right direction
ELONGATION STEP:
– Synthesis of new RNA strand is 5’ to 3’ (by complementary base pairing) – Reads template strand 3’ to 5’ – As RNA polymerase moves along DNA, it untwists
DNA double helix, ~10 to 20 bases at a time – RNA polymerase has a proofreading mechanism
that can replace incorrectly incorporated bases
TERMINATION STEP:
– End point of transcription that involves disassembly of RNA polymerase from DNA
– RNA polymerase stops when it comes to the termination site on DNA
– “Completed” RNA (also known as the primary RNA transcript or pre-RNA) is then subjected to further processing before it leaves the nucleus of the eukaryotic cell

20
Q

What is the process of RNA processing?

A

• In eukaryotes, primary RNA
transcript (i.e., pre-RNA) of a gene
needs further processing before
translation (“RNA processing”)
• RNA processing involves:
1. Addition of cap to the 5’ end
2. Addition of poly-A tail to the 3’ end
3. Splicing out introns and joining exons
• In prokaryotes, RNA is ready to be translated into protein (RNA DOES NOT undergo RNA processing)

21
Q

What is the addition of cap and Poly-A tail?

A

Cap and poly-A tail
• At 5’ end, a cap is added in the form of 7-methyl guanosine triphosphate
• At 3’ end, ~100-250 adenine nucleotides are attached (the poly-A tail)
• Examples of important functions of cap and tail modifications:
– The tail facilitates the export of RNA out of the nucleus and into the cell cytoplasm – The cap and tail protect RNA from enzymes (namely RNAses) that can break it down – The caphelps ribosomes attach to the 5’ end of mRNA at the beginning of translation

22
Q

What is splicing of introns vs exons?

A

Introns and exons
• INTRONS = DOES NOT code for protein (“intervening sequences”)
– Intron sequences need to be removed, and remaining ends are re-attached so the final RNA consists of exons only, However, not all eukaryotic RNA transcripts have introns: i.e., histone RNA transcripts lack introns
• EXONS= codes for protein (“expressing sequences)
- Introns are removed from the primary RNA transcript (or pre-RNA) while it is still in the nucleus
• Introns are “spliced out” by spliceosomes
Spliceosome = complex of small nuclear RNAs (snRNAs) and proteins (composed of over 300 proteins)
• Intron sequences are removed, and remaining ends are re-attached so the final RNA consists of exons only

23
Q

What is the process of RNA translation?

A

Process of “decoding” mRNA and using its genetic information to assemble a protein
• Major players of translation:
– mRNA (messenger RNA)
– tRNA (transfer RNA)
– Ribosomes
– Amino acids
• Four phases: activation, initiation, elongation,termination

mRNA is “read” according to the genetic code, which relates DNA
sequence to amino acid sequence in proteins
― mRNA is read in a series of triplets (3 nucleotides) called codons
• Each codon specifies a particular amino acid that is to be placed in polypeptide chain for construction of protein

24
Q

Explain why the genetic code is degenerate

A

Code is almost “universal” – used by both prokaryotes and eukaryotes
• Each group of 3 nucleotides on the mRNA is a codon
• 64 possible codons, but code for 20 different amino acids; 61 codons code for amino
acids (5’-AUG-3’ is used as start codon (codes for methionine), 3 codons are stop codons (5’-UAA-3’, 5’-UAG-3’, 5’-UGA-3’) and do not code for amino acids
• More than one codon is used for most amino acids: the genetic code is “degenerate”
– This means that it is not possible to take a protein sequence and deduce exactly the base sequence of the gene it came from
In most cases, third base of the codon (wobble base) can be altered without changing the amino acid

25
Q

Describe what an anticodon is in tRNA?

A

Each tRNA has 3 bases that make up the anticodon
– These bases pair with the 3 bases of the codon on mRNA during translation
• Each tRNA has its corresponding amino acid attached to the 3’end (part of activation step of translation, which is the process of attaching an amino acid to tRNA

26
Q

What are the parts of a ribosome?

A

Ribosomes are “Protein factory” of cells made up of two subunits: large and small
• The P site (peptidyl-tRNA site) holds tRNA with newly forming
polypeptide chain attached
• The A site (aminoacyl-tRNA site) holds tRNA carrying next amino
acid to be added to polypeptide chain
• The E site (exit site) holds free tRNA before it is released from
ribosome/RNA template translation complex

27
Q

What is the initiation stage of mRNA translation?

A

• Initiation stage brings together mRNA, a tRNA carrying the first amino acid (methionine), and two ribosomal subunits
• With assistance of special proteins called initiation factors, the small ribosomal subunit binds near 5’ end of mRNA (more specifically it binds to the ribosomal binding site on mRNA)
– Ribosomal binding site = sequence of nucleotides upstream from mRNA start codon
• Afterwards, tRNA carrying methionine binds to start codon(5’-AUG-3’) on mRNA
• The large subunit is brought in next with help from an energy-rich molecule GTP in order to complete the whole initiation complex

28
Q

What is the elongation step of mRNA translation?

A

1) Elongation begins with binding of tRNA carrying an amino acid to A site (this step is catalyzed by GTP and special proteins called elongation factors)
2) Once incoming tRNA binds to A site, the polypeptide chain is moved from tRNA in P site to amino acid attached to tRNA in A site
3) Ribosome moves downstream along mRNA to next codon (translocation is promoted by elongation factors and GTP)

29
Q

What is the termination step of mRNA translation?

A

• When translocation opens the A site and exposes one of the stop codons, a protein called a release factor fills the A site
• Release factors bind to codon - there are no tRNA molecules with anticodons for stop factors
• Release factor binding triggers polypeptide chain to be cleaved from tRNA
• Polypeptide is released, and then the tRNA is released
• The two ribosomal subunits and mRNA dissociate from each other
(catalyzed by GTP)

30
Q

How does mRNA translation differ in eukaryotic vs prokaryotic cells?

A

eukaryotic transcription and translation are spatially and temporally isolated, but coupled and occur simultaneously in prokaryotes

31
Q

What’s the difference between monocistronic vs polycistronic mRNA

A

• Monocistronic
– One mRNA molecule translates for only one type of protein – Refers to eukaryotic mRNA
• Polycistronic
– One mRNA molecule translates for more than one type of protein – Refers to prokaryotic mRNA

32
Q

What are point mutations and what are the categories?

A

Point mutations (small scale mutations) are changes in just one base pair of a gene, and can be divided into two general categories:
1) Nucleotide-pair substitutions
- A nucleotide-pair substitution replaces one nucleotide and its partner with another pair of nucleotides
• Silent mutations have no effect on the amino acid produced by a codon because of redundancy in the geneticcode
• Missense mutations still code for an amino acid, but not the correct amino acid (i.e., different amino acid produced)
• Nonsense mutations change an amino acid codon into a stop codon, nearly always leading to a non functional protein
• Generates truncated protein (i.e., shortened protein)
2) One or more nucleotide-pair insertions or deletions
- Insertions and deletions are additions or losses of nucleotide pairs in a gene
• These mutations have a disastrous effect on the resulting protein more often than substitutions do
• Insertion or deletion of nucleotides may alter the reading frame, producing a frameshift mutation

33
Q

What is gene regulation?

A

Gene regulation refers to the processes that cells use to turn genes on for the purposes of making gene products (or to turn genes off)
• This functional gene product may be RNA or protein
• Gene regulation is essential for viruses, prokaryotes and eukaryotes as it increases the versatility and adaptability of an organism by allowing the cell to express protein when needed