Lecture Exam 2 Flashcards

1
Q

Mendel’s Conclusion; aka Particulate Theory of Inheritance (5)

A
  1. Alternative versions of genes (alleles) account for variations in inherited characters
  2. For each character, an organism inherits two alleles, one from each parent; paired condition (of alleles) are restored by random fusion of gametes at fertilization
  3. If two alleles differ, then one (dominant) is fully expressed in the organism’s appearance; the other (recessive) has no noticeable effect on the organism’s appearance
  4. Law of Segregation: two alleles for each character segregate during gamete production
    predicts that in a monohybrid (one factor) cross, the genotype of the F2 generation will approximate a 3:1 ratio (dominant to recessive) or a 1:2:1 ratio (homo-dom, hetero, homo-rec)
  5. Law of Independent Assortment: each allele pair segregates independently during gamete formation via separate processes
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2
Q

Types of Crossing (3) + How

A

TEST CROSSING: lets you know what trait is dominant since you’re crossing the partially unknown (which requires at least ONE dominant allele) with a homo-rec (which can only pass on recessive alleles). THEREFORE, the phenotype of the offspring is dependent on the P/? (aka the partially unknown)

    • Ie. If P/P then offspring will be P/p and all purple == OR == If P/p then offspring will be 1:1 ratio of purple to white.
    • Can also use the Punnett square as a better visualization.

DIHYBRID CROSS: lets you hybridize for two traits.
– Consider:
P = Yellow; Round seeds x Green, Wrinkled seeds
F1 = All Yellow; Round
– Therefore, from the F1 generation, you know that Yellow and Round are the dominant alleles and that the offspring of this generation are all heterozygous.
– F2 generation is as follows → 9:3:3:1 ratio BUT still follows 3:1 ratio if genes are considered individually
== 9/16 Yellow; Round
== 1/16 Green; Wrinkled
== 3/16 Yellow; Wrinkled Novel combinations
== 3/16 Green; Round
** first two are the same combination as in the P generation; last two are novel combinations, with one gene dominant and the other recessive

TRIHYBRID CROSS:
Consider this:
– P = Yellow seed ; Round seed ; Long stem x Green seed ; Wrinkled seed ; Short stem
– F1 = Y/y ; R/r ; L/l
– F2 = UUUUUH → actually, don’t freak out. Just do your Punnett squares (separately, then multiply the ratios of what you want together).

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3
Q

Meiosis

  • Interphase 1
  • PMAT I
  • Yield
  • Interphase II
  • PMAT II

Be sure to note all the things that happen here that DO NOT HAPPEN in mitosis.

A

Interphase I: chromosome replicates in S phase; centrioles in animals also replicate.

Prophase I: chromosomes condense; centrosomes move apart; tetrads form via synapsis; crossing over at chiasmata occurs; nuclear membrane and nucleoli disappear
– THINGS THAT HAPPEN HERE THAT DON’T HAPPEN IN MITOSIS: synapsis and crossing over

Metaphase I: tetrads align on metaphase plate; each homologue is attached to a kinetochore microtubule from the pole it faces
– THINGS THAT HAPPEN HERE THAT DON’T HAPPEN IN MITOSIS: kinetochore microtubules are no longer connected to both homologue; instead now only one per pole

Anaphase I: homologues move toward opposite poles by motor proteins; chromosomes are still doubled and the sister chromatids remain attached but are no longer identical
– THINGS THAT HAPPEN HERE THAT DON’T HAPPEN IN MITOSIS: sister chromatids are no longer alike due to crossing over

Telophase I: chromosomes arrive at spindle poles; (sometimes) the nuclear membrane will reappear; cytokinesis
– THINGS THAT HAPPEN HERE THAT DON’T HAPPEN IN MITOSIS: no reappearance of the nuclear membrane (usually)

At the end of Meiosis I, each daughter cell (there’s 4) is HAPLOID; chromosomes are still duplicated (therefore, still exist in the form of paired chromatids).

Interphase II: short; no DNA replication before Meiosis II
– no S phase as all the DNA was already duplicated in Interphase I.

Meiosis II: just like mitosis

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4
Q

Benefits of Meiosis

A

Meiosis REDUCES the chromosome SET number (and, therefore, the chromosome number).

GENETIC DIVERSITY

    • Independent assortment (occurs in Metaphase 1), yielding unique combinations of chromosomes. There variations are the set up for Darwin’s “survival of the fittest” theory.
    • Number of possible combos = 2n where n is the number of haploids → ie. for humans, n = 23 yields ~8 million
    • Random fertilization in sexual reproduction → Therefore, number of possible combos is now SQUARED !! ~64 million combinations !!!
    • Crossing over produces unique chromosomes that contain the genes from both parents. → There can be multiple crossovers per chromosome → → to infinity and beyond with genetic variation!
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5
Q

Blending Theory of Heredity

A

basically says that all the mixing of genes will result in a homogenous population

Assumes no new input of genetic material, therefore offsprings of subsequent generations will slowly become genetically similar

Once traits are blended, they cannot be separated again → think: mixed paint

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6
Q

Particulate Behavior of Genes

A

Inheritance is due to discrete factors (genes) that are passed on from generation to generation

Segregation and Assortment are RANDOM events that obey simple laws of probability

If F2 seed is planted, we will not be able to apriori predict the phenotype (basically, can’t theorize / calculate the outcome of the offspring) BUT can state that there will be a ¼ chance that it will have white flowers.
– Among a large sample size, about ¼ or 25% will have white flowers based on the 3:1 ratio. (Because small systems will have microcosms that will deviate the result from the ratio.)

** The larger the sample size, the closer results will fit the prediction.

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7
Q

Chromosome Theory of Inheritance + evidence

A

transmission of chromosomes paralleled transmission of genes in Mendel’s Theory → Mendel was rediscovered by scientists in the early 1900s and given credit posthumously

Genes are located on chromosomes

Chromosomes segregate and assort independently during meiosis


Evidence: working with Drosophila melanogaster (fruit fly)
- Easily cultured
- Breed like flies, therefore prolific – lots of offspring
- Short generation time, therefore have quick generational turnover (yields F2 in a few weeks)
- Have only 4 pairs of chromosomes, which are easily seen in a microscope
- 3 pair autosomes + 1 pair sex chromosomes
- Females X/X and males X/Y (heterogametic)

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8
Q

Discoveries:

  • mutant genes
  • linked genes
  • parentals and recombinations
A

MUTANT GENES DISCOVERED
Mutation: the “not normal”; deviated from the regular phenotype; not necessarily dominant though

LINKED GENES DISCOVERED
Genes that are on the same chromosome
Therefore, do not segregate independently (ie. assort independently) and are INHERITED TOGETHER
Linkage: tendency for genes that are closer together to be inherited together
Note: dihybrid cross does not produce a dihybrid cross ratio in F2

PARENTALS AND RECOMBINANTS DISCOVERED
Parental: highest in number
Recombinants: aka novel combinations; arise due to crossing over in meiosis
Double crossing over yields double recombinants

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9
Q

Fruit Fly Denotation

A

Genes are named on the basis of the mutant phenotype

  • If dominant, then capital; if recessive, then lowercase
  • Wild type (aka “standard” strait) denoted by at +
  • Ie. recessive white eye mutation = w ;; versus dominant red eye wild type = w+
  • Remember! Mutant is the phenotype that is less present in the population

Homozygous if the alleles are the same → ie. sn+ // sn+
Heterozygous if the alleles are different
Hemizygous if the animal is a male and the allele maps to the X chromosome → ie. Xsn+ // Y or Xsn // Y → denotes situation based on X-linked or Y-linked genes in the male

Linked genes: found on same autosome; genetically linked, allowing for crossover → heterozygous = w sn // w+ sn+
Unlinked genes: found on different, non-homologous chromosomes; denotation has a semicolon separating them; allows for independent assortment → heterozygous = w // w+ ; sn // sn+

NOTE: DOUBLE LINED FRACTIONS IDENTIFY ONE FOR EACH STRAND → IE. TRAITS FOR ONE STRAND ARE ON TOP; TRAITS FOR THE OTHER STRAND ARE ON THE BOTTOM

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10
Q

Approaching three point crossover problems
Mapping genes: basically, look for the middle
Determining the map units in between genes

A

EXPECTING A DATA COLLECTION OF 8 VALUES
Top 2 = parentals
Lowest 2 = double recombinants → can also be zero!! Be careful!
Everything else = recombinants
Compare and contrast pairs of traits.
If the numbers match up then genetically UNLINKED
If the numbers are different from expected, then GENETICALLY LINKED

Recombination data can be used to map genes on a chromosome; constant and will never change within the species
Compare and contrast the double recombinant and the parent → Middle allele is the one that differs from the parent
Rewrite the given data to match this gene mapping

Recombination frequency is proportional to the “distance” between genes

Double crossover will yield an UNDERESTIMATE (meaning that the actual distance between the genes will be larger than the calculated) because they look similar to the parentals

Map units = number of recombinants over total number of progeny * 100

    • Can substitute total number of progeny for (distance given)
    • Aka centimorgans (cM); Related to probability that crossing over will occur between the two genes; larger distance = more likely to cross over
    • Remember! Crossing over occurs at the distal ends of homologous chromosomes → contrast to linkage: tendency for genes close together on a chromosome to be inherited together
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11
Q

PHYSICALLY LINKED versus GENETICALLY LINKED

A

If the distance between two genes exceeds 50 map units (=50%), then they are considered to be UNLINKED (either bc they are v far apart on the same chromosome, therefore allowing for lots of crossover [thus act as unlinked], or on diff chromosomes)

Physically linked: on same chromosome
Genetically linked: can be detected by crossing

Ie. Mendel’s seed and flower color were PHYSICALLY LINKED but were too far apart and appeared to assort independently, thus was not considered genetically linked

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12
Q

SEX DETERMINATION

A

Chromosomal systems: common but not universal

Y chromosome: much smaller than the X; determines maleness
– Has SRY gene on Y: when activated, will create testosterone
– DAX1 on X inactivated → X/Y still male but sterile; extra on X/X then will develop testis
– Extra X chromosomes tolerated: X/X/X and X/X/Y
X Centered sex determination (fruit flies)

Ratio of X to autosomes determines femaleness
– X/X is female; X/Y is male; X/O is sterile male; X/X/X is dead

Environment systems: not a chromosomal system

– Temperature Dependent: temperature at which egg develops determines sex; some reptiles / turtle / alligators → concerns the temperature sensitive Dmrt1 gene that is expressed at lower temperatures, yielding a male; at higher temperatures, histones at Dmrt1 methylate, thus shutting down expression and yielding a female

– Location dependent: environmental sex determination in certain marine worms → male if the larva comes into contact with a female; becomes female if larva is alone on the seafloor

Note: the larva initially do not have a sex; their sex determination arises from their environment via contact with other larva

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13
Q

Sex Determination Animal Examples

A

Platypus: monotremes (basal mammal that lays yolky eggs); no SRY gene; sex chromosomes are not homologous to those of Eutherians

Many insects / few mammals → females are X/X, males are X/O (one less chromosome)

Birds / reptiles / moths / butterflies → Z/Z and Z/W system

    • Females are heterogametic Z/W ; males are homogametic Z/Z
    • No genes in common between mammalian X/Y and avian Z/W

Bees / ants / wasps / order hymenoptera → Haplodiploidy: specific type of chromosomal

    • Males are haploid, called drone → Parthenogenesis: developed from unfertilized eggs
    • Females are diploid → queen bee mates with drone (male); daughters share ¾ of their genes with each other (not ½ as in XY or ZW genes)
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14
Q

Streptococcus pneumoniae experiment

A

accidentally answered the question “what is genetic material”; injected S/R virus into mice

Initially, DNA (2 basic building blocks) was not accepted as genetic material because was too simple; proteins (4 basic building blocks) were more likely.

Premise: S killed mice (ie. caused disease); R did not; heat-killed S did not → combined heat killed S + living R, which killed mice BUT the blood revealed living S = concept of transformation.

    • Transformation: uptake and expression of genetic material from surroundings; common in bacteria / prokaryotes but not in eukaryotes (sorry humans)
    • Horizontal transformation of R into S as R took in heat-killed S

Due to transfer of genetic material from dead S to living R, the living R began expressing the genetic material from S, thus changing into the S phenotype
– Purified components from the heat-killed S revealed that only DNA transformed (not protein!)

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15
Q

T2 Bacteriophage

+ experiment

A

virus that affects E. coli

Programs host cell to make new virus, which lyse from the cell, thus destroying the host
Viruses don’t reproduce! They use the host cells to make more viruses.
Consists of a DNA encapsulated in a protein coat (called “capsule”) with a tail piece that can change shape in order to pierce the membrane of a target host cell, allowing for injection of the virus DNA

Experiment: how do you know that DNA is inserted (versus proteins)? ** Remember that viruses are made of both DNA and proteins.

    • Grew E. coli and T2 together with 35S (radioactive blue stain for proteins) → incorporated into the phage protein (capsule)
    • Grew E. coli and T2 together with 32P (radioactive stain for DNA, which has P groups)

Procedure:
1. Harvested the phage
2. Infected one group of bacteria with S-labelled phage and another group with the P-labeled phage
Samples blended to dislodge capsule
3. Centrifuged (capsules float; bacteria in pellet + whatever injected)
4. Check for location of radioactivity

Results:

    • S labelled phage (protein stained) had radioactivity in the SUPERNATANT where the protein capsules were
    • P labelled phage (DNA stained) had radioactivity in the PELLET where the DNA was
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16
Q

Name two important studies + some other important clues that led to the conclusion that DNA really was the genetic material (as opposed to proteins).

A

Streptococcus pneumoniae experiment; from Frederick Griffith and (later) Avery et al.

T2 Bacteriophage and E. coli experiment; from Hershey and Chase

DNA composition (ratios of nitrogen bases A:G:T:C) are species specific (ie. will vary from species to species since no two species will be the same) BUT strange regularity of amount of A = amount of T and amount of G = amount of C → Base Pairs (aka Chargaff’s rules) from Erwin Chargaff

1950s - DNA officially accepted as Genetic Material

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17
Q

Watson-Crick Model of DNA Structure

+ what does it mean to be double stranded

A

X-Ray Crystallography Photo method (from Rosalind Franklin): specimen is crystallized and examined with x ray beams, which result in a scattering of light that can then be caught by photosensitive film → the scatter pattern that results can be inferred

Led to deduction of the DNA Double Helix: 2 nm wide; 2 strands (2 poly nucleotides entwined) → 10 layers of nitrogenous bases (pairs) in one turn of the helix, where one full turn is made every 3.4 nm

Each strand consists of sugar-phosphate backbone
Strands are held together by H-bonds between the nitrogen bases

For two bases to fit in the space, has to be Pyrimidine (one ring) with Purine (double ring) → can’t be pyr-pyr bc too small / doesn’t fill space of 2nm, can’t be pur-pur bc too big / overfills space of 2nm

Explains Chargaff’s Rules since A-T and G-C theory is held up (equal each other and fill up space) → A/G are pyrimidines, T/C are purines

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18
Q

DNA Directionality

A

both strands are antiparallel to each other, thus operate in different chemical directions → because of base pairing rules, if we know the base sequence of one strand, we can predict the base sequence of the opposite strand (aka complementary!)

3’ is considered “longer / business” end because it has the capacity to form long carbon chains due to the HYDROXYL GROUP on the THIRD CARBON of the nucleotide sugar; also the location of phosphodiester linkages → allows for (weak) hydrogen bond attractions between the strands, hence creating the double helix form

5’ is considered the “shorter / hanging out end because the PHOSPHATE GROUPs do not interact with anything in the strand / environment besides being attached to the FIFTH CARBON of the nucleotide sugar

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19
Q

DNA Replication Models

+ types of replication

A

replication occurs during the S phase of the cell cycle

DNA strands separate → each strand serves as a template for constructing new complementary strand; deoxyribonucleoside triphosphates attach to new strand, matching base of template according to the base pairing rules

Types of Replication: the winner and the losers + why they lost

  1. Conservative: strands are replicated separately but come back together, thus yielding the original strand joined back together and a completely new strand made from combining the replicas
  2. Dispersive: strands are more or less randomly distributed as original or synthesized; aka splicing information into each “new” strand
  3. Semiconservative: proven theory; one strand is from the original, the other is the newly synthesized – (from reader) as double helix replicates, daughter molecules will consist of one strand from the original molecule and one newly synthesized strand
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20
Q

Meselson and Stahl experiment

A

Tested which of the three models were correct by double replicating the strands in order to differentiate them from one another + compare

Procedure:
1. Used bacteria cultured in a medium containing 15N (heavy nitrogen; therefore DNA synthesized will be heavier) – remember that nitrogen is incorporated into the bases
2. Bacteria transferred to a medium containing 14N
DNA Sample centrifuged after 20 minutes (which is after the first replication) and again after 40 minutes (second replication)

Prediction and Results: original DNA model is of higher density than its replicas

    • Conservative predicted that there would be heavy and light density DNA present and no intermediates → eliminated because intermediates were present after the first replication
    • Dispersive predicted intermediates after the first replication BUT also predicts that the DNA density should become lighter after the second replication → eliminated because intermediate density does not change
    • Semiconservative predicted that intermediates after the first replication BUT also predicts that the densities will not change after the second replication → proven
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21
Q

DNA Replication

A

Rapid and accurate despite having about 6 billion bases in the human genome
DNA polymerase: synthesizes; super accurate (makes about 1 mistake in a billion bases, therefore only 6 mistakes in the entirety of genome replication); however, cannot initiate
– New DNA is created from 5’ to 3’ at a rate of 50 nucleotides / s (but there’s a lot working at once)

Origins of Replication: lots of different types in eukaryotes but there’s a specific sequence of nucleotides that act as a marker in prokaryotes

  1. 2 strands separate at the origin of replication
    - - Problem: DNA polymerase can add new nucleotides only to an existing 3’-OH therefore CANNOT INITIATE
    - - Solution: RNA primer attaches to the complementary sequence at the origin of replication to which the DNA polymerase can extend; DNA primase can then make a primer out of it (teamwork!)
  2. Replication bubble formed as helicase unwinds
    - - Creation of replication fork
    - - Creation of leading strand (continuously replicated in 5’ to 3’ direction) and lagging strand (discontinuously replicated because need to be in 5’ to 3’ direction)
    - —- Lagging strand requires additional primers and DNA polymerases but still unable to fill in the gaps in between, aka Okazaki fragments (about 100 to 200 nucleotides long)
    - —- DNA polymerase will remove the primers it meets (after the first) but gap still present
    - —- DNA ligase comes in to seal the P-sugar backbone with a covalent bond; basically creates a phosphodiester linkage
  3. Elongation continues via helicase until replication forks from adjacent regions meet
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22
Q

Why is RNA used instead of DNA as a primase??

A

DNA polymerase is very active; can incorporate nucleotide triphosphate; if it makes a mistake, can back up and correct it – super accurate tho

DNA Primase that is put down is prone to errors – if used to replicate genome then sooo many mutations would occur

Use RNA instead because there’s a chemical difference from DNA – can be erased with good DNA → “it’s a matter of conserving the integrity of the DNA” – Jim Baxter 2018
RNA primase makes mistakes too but it doesn’t matter because of amplification + the DNA can get replace it

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23
Q

What is DNA?

A

Contains the information needed to construct the primary structure of proteins.

Genes: encodes information for one polypeptide or structural RNA
– Not all genes are “on” in all cells or at all times

Need a mechanism to selectively express specific genes without expressing all the genes on a chromosome (basically, how to turn something on or off)

Eg. Liver cells are different from neurons because even though the DNA content of the cells are the same, different proteins are present and thus yield a different gene expression

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24
Q

Central Dogma of Biology

A

DNA → RNA → Protein

Overview: DNA can undergo replication or can be transcribed into RNA (intermediate), which can be translated into a different language that codes for proteins!

DNA = Genomic DNA encodes the information needed for life and for differentiating one species from another

RNA = Information from the genome, representing only a small fraction of the total DNA content, is transcribed into mRNA for further processing

Protein = Information in a portion of the mRNA sequence (aka coding sequence) is used to guide the synthesis of a specific protein

Gene expression: production of polypeptide or structural RNA encoded for by a gene → “active” protein or RNA required; usually one gene per one polypeptide or RNA BUT can be extended

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25
Q

DNA versus RNA

A

2 polynucleotides (double stranded) – 1 polynucleotide (which can fold to form double stranded portions if complementary, but typically single stranded)

Deoxyribose sugar – ribose sugar

Thymine – Uracil

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26
Q

Transcription

A

RNA polymerase II used to separate DNA strands and DOES NOT need priming; however, less accurate than DNA polyermase

Adds and links together ribonucleotides in a 5’ to 3’ direction according to the sequence in the template and base pairing rules

Non template strand will be identical (due to complementation to the template) to the RNA being produced (which also complements the template strand)

Note: Eukaryotes have 3 types of RNA polymerase; I and III are mostly transcribed for structural RNAs, ie. ribosomal RNA

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27
Q

Three Stages of Transcription

A
  1. Initiation: general transcription factors (GTF) bind to the specific nucleotide sequence in the promotor
    - - Promoter: aka “switch”; can change the environment only when GTF binds to it, thus allowing RNA polymerase to attach to the initiation site and allowing transcription to begin
    - - TATA box: TA box rich region in the promoter of eukaryotes where TATA transcription factor binds → one of the first to bind, thus allowing others to bind after it because the of the induced environment change
  2. Elongation: RNA polymerase separates the DNA into two strands, the directionality of which determines which will be the template for RNA because the RNA nucleotides are linked in a 5’ to 3’ direction; processes at a rate of 30 to 50 nucleotides / s
    - - Newly made RNA will peel away from the template and DNA strands reanneal
  3. Termination: RNA polymerase proceeds until the terminator sequence is reached/transcribed, which will form a hairpin loop (aka “proles”) to push the DNA polymerase off (ends transcription)
    In eukaryotes: will often be AAUAAA
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28
Q

Translation

A

mRNA → amino acid sequence

Direct correspondence of linear sequences of codons in mRNA and amino acid sequences of polypeptides

Important players:

    • Ribosomes
    • mRNA (product from post transcriptional processes)
    • tRNA: aka transfer RNA; structural
    • Aminoacyl-tRNA synthetases (specific class of enzymes):
    • Lots and lots of enzymes
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29
Q

More on tRNA

A

acts as a special adaptor between mRNA and the AA sequence by transferring AA’s from the cytoplasm to the ribosome

    • Single polynucleotide but forms several double stranded regions; about 80 bases long
    • Transcribed in nucleus from tRNA genes via RNA Polymerase III
    • Has an amino attachment site at the 3’ end

Anticodon: base triplet complementary to codon; can polymerize and thus be antiparallel

45 distinct types of RNA arise bc some tRNAs recognize more than one codon for a specific AA → concept of wobble: relaxation of base pair rules at third position of codon
– Note: in anticodons, their 3rd position can have the modified base INOSINE, which can pair with U/C/A → ie. CCI can pair with GGU / GGC / GGA (all glycine)

30
Q

More on Amino Acid Attachment sites

A

located on the 3’ end of a tRNA → ACC (?) ⇒ How is the correct AA matched to the proper tRNA?

For each of the 20 AAs is specific aminoacyl-tRNA synthetase → each enzyme is specific for the AA and tRNA, thus the enzyme makes the connection

Process:

  1. Activation of AA with AMP – active site binds AA and ATP, followed by the hydrolysis of ATP
  2. Attachment of AA to tRNA – proper tRNA binds to the activate site, displaces AMP, creates a covalent bond, becomes a “charged tRNA”
31
Q

More on ribosomes

A

the site of protein synthesis; consist of 2 subunits (one small and one large) which will separate when not working

About 60% structural rRNA + 40% proteins (about 80ish kinds)

Three binding sites for mRNA called E, P, A sites that each refer to one codon
E = before P
P = area of start codon
A = after P

In eukaryotes, subunits will assemble in the NUCLEOLUS

32
Q

Signal amplification

A

usually, many RNA polymerases transcribe genes simultaneously → basically work in tandem for the same gene in the exact same sequence, thus barring or “drowning” out mutations ;; usually the final product in prokaryotes but not in eukaryotes

33
Q

Inosine @ the third position
+ Clarification about “wobble” base pairing
+ codon
+ anticodon

A

Inosine @ the third position:
– Codon has three bases → third base may have “wobble”
– Modified base that does not usually occur in mRNA or DNA – only occurs on tRNA bc of degeneracy →
Can base pair with A , C , U → thus allowing it to subvert base pairing rules

Clarification about “wobble” base pairing:

    • Occurs in translation, in regards to the degeneracy of codons
    • Last base of a codon (made of three bases) can vary in terms of what base it can be BUT still results in the same amino acid
    • First two bases must strictly follow base pairing rules
    • “Incorrect” base pairing occurs because the tRNA curves away from the third base in the codon, thus introducing more space between the already weak H bonds

Codon: made of three bases for mRNA in a 5’ to 3’ direction
– AUG start codon - methionine

Anticodon: what is on the tRNA that base pairs with the codon; made in 3’ to 5’ direction
– AUG anticodon = UAC (can’t be TAC because no T in RNA)
On exam, can be written as 5’ – CAU – 3’ or 3’ – UAC – 5’ (both are correct)

34
Q

Post transcriptional modifications

A

occurs only in eukaryotes; includes inosine

Primary transcript: initial RNA transcribed from gene; not encoded, just modified due to addition of 5’ cap

    • 5’ cap added, thus modifying G; protects 5’ end from degradation by hydrolytic enzymes
    • The one RNAse in the cytosol is a hydrolytic enzyme that can break RNA down into oligonucleotides / smaller molecules

Poly A tail: found at 3’ end; addition of approximately 30 to 200 nucleotide base A; may protect against degradation or influence the “lifespan” of an RNA molecule
– RNA half life is prokaryotes is minutes BUT is much longer in eukaryotes

Introns: non coding sequences that are removed as they do not contribute to the signal for protein synthesis ;; discussed later in lecture as affecting bacteria’s ability to clone human genes

35
Q

RNA Splicing

A

removal of introns done by spliceosome, which is a large molecular complex (also a type of ribozyme)

Consists of small nuclear ribonucleoproteins (snRNPs), which are made of snRNA (structural RNA with a v specific sequence; also comes from gene) and proteins

snRNPs contain RNA that is complementary to the exon and intron ends, therefore multiple snRNPs can be found and degraded by a spliceosome, thus removing the introns

36
Q

Genetic Code

A

refers to the direct relationship between sequences of bases in mRNA and amino acid sequences of polypeptides; universal bc same code in all organisms (with the exception of codons in the mitochondria)

Problem 1: only 4 different bases but 20 different AA → no 1:1 correspondence otherwise 4 AA would be specified (< 20 therefore not enough); no 2:1 correspondence because does not satisfy requirement for 20 (4^2 = 16 < 20); must be 3:1 correspondence bc satisfies 20 AA (4^3 = 64 > 20)
– Triplet: every three bases is considered a CODON that refers to a specific AA; can overlap multiple codons that specify for a specific AA (aka degenerate)

Problem 2: triplets mean there are three possibilities for reading frames BUT only one is correct / functional → meaning no overlap; contrast to viruses that have overlaps
– Make sure to check for start and stop codons !!

37
Q

Three Stages of Translation

A
  1. Initiation:
    - - Small ribosomal subunit binds mRNA to 5’ end → 5’ cap aides in binding of 5’ UTR (leader sequence)
    - - Initiator tRNA binds start codon AUG (aka methionine) → mRNA + initiator tRNA + small r subunit ;; requires initiation factors (proteins) and GTP
    - - Large ribosomal subunits bind, releasing initiation factors → initiator tRNA in P site = initiator complex
  2. Elongation:
    - - Codon recognition: tRNA with anticodon complementary to codon in A site binds, which requires the hydrolysis of GTP
    - - Peptide bond forms between AA in P-site and AA in A-site, which is catalyzed by peptidyl transferase → yields rRNA in large ribosomal subunit (aka ribozyme)
    - - Ribozymes: enzymes usually made of proteins BUT this is an RNA molecule that functions as a catalyst / enzyme (ie. ribosomes, snRNP)
    - - snRNP: ribonucleic protein; complex that has both RNA and proteins on it – make up most of the ribozymes today
    - - Translocation: tRNA and mRNA in A site move as a unit to the P site; ribosomes move along the mRNA in a 5’ to 3’ direction by hydrolysing GTP (provides energy) → uncharged tRNA in P site moves to the R site then exits
    - —– Alternate explanation: EPA unit moves down the mRNA, translating it by forming peptide bonds between the AA in the P site and the AA in the A site BUT breaks the covalent bonds from the mRNA and the AA in the P site, thus allowing for elongation.
  3. Termination: stop codons UAA / UAG / UGA enter the A-site and DO NOT CODE for AA → release factor (protein) will bind to stop codon and hydrolyze the bond between the polypeptides and tRNA in the P site, thus allowing the polypeptide and tRNA leave the ribosome while the small / large subunits and mRNA dissociate
38
Q

Genes determine primary structure of proteins

A

Allow for spontaneous folding into 3D conformation with the help of chaperone proteins

Post translational modifications

    • Addition of sugars / lipis / phosphate / etc
    • One or more AA removed from amino end → ie. begin with methionine BUT can be removed
    • Enzymatic cleavage into 2 or more pieces
    • Joining together of subunits, thus creating a tertiary structure

Translation begins on free ribosomes in cytosol

If polypeptides are to function in the membrane / outside the cell / etc. then will need:

    • First few AAs to be SIGNAL PEPTIDES (aka zip code), made of protein
    • rspRNA (another structural RNA); requires SRP (signal recognition protein) to bind receptor into the ER membrane, thus becoming a bound ribosome
39
Q

Eukaryotes versus Prokaryotes

+ alleles

A

Eukaryotes versus Prokaryotes
– In eukaryotes, transcription in nucleus and translation in cytoplasm can happen concurrently.
– In prokaryotes, no such compartmentalization, therefore translation may begin before transcription complete
Post transcriptional modification in eukaryotes – NONE in prokaryotes

Alleles:
Differ slightly in DNA sequence; may lead to slightly different AA sequence in polypeptide
New alleles arise by mutation, which is a change in genetic material → point mutations: limited to changes in a few bases in a single gene

40
Q

Point mutations

A

Substitutions: replace one base with another; can be a silent mutation (one codon changed but codes for the same AA)

    • In intron start sequence: intron not removed, thus will be in mRNA and will add AA and may shift reading frame
    • In intron end sequence: subsequent exon “regarded” as an intron, therefore removed which is bad

Deletions / insertions: shift reading frame downstream; usually results in nonfunctional protein (unless insert / delete near end of gene)
– Frameshift: shifts reading frame downstream from insertion / deletion, resulting in massive missense bc introduces immediate nonsense

Missense: base substitution leads to substitution of different AA

Nonsense: base substitution changes codon into stop codon, resulting in premature termination of translation

Sense: changes “stop” codon into codon for AA, thus allowing for continuation of adding AA to polypeptide (can
maybe affect folding)

Considerations: for all point mutations, have different consequences regarding heredity

    • Immortal in DNA, therefore has larger consequences
    • Versus RNA – RNA destroyed with some not good proteins BUT they are usually drowned out due to amplification

In somatic cells, will not be passed to kids BUT can lead to cancer (specifically in genes that work in the cell cycle by stopping meiosis, therefore have uncontrollable division)
– Versus germ cells – if gamete then every cell in offspring will have mutations

41
Q

Genes, Proteins, and Phenotypes

A

Organisms is what it is by virtue of the interactions of its proteins

Gene (molecular definition): unit of DNA that encodes information for the primary structure of a polypeptide

Flower color: presence or absence of protein (pigment)
Seed texture: enzyme converts sugar to starch → if defective, no conversion
– Starch does not accumulate in seed but sugar does; sugar takes in a lot of water and therefore will swell and stretch the skin; after maturation, the cell will shrink but since the skin is already stretched, resulting in WRINKLES

42
Q

Viruses

+ viral genomes

A

Viruses: acellular; do not metabolize energy (bc no cell resp / photosynthesis) therefore do not make ATP; never arise directly from pre existing viruses; are obligate intracellular parasites
– Can also have phospholipid bilayers that they obtain from past cells they’ve ruptured from

Viral genomes: enclosed in protein coat (aka capsid)

    • Also double stranded DNA (basically a universal for cellular life) BUT can have ss/ds DNA/RNA
    • Single or many molecules
    • Linear or circular
    • As few as 4 genes, or as many as several hundred (depending on the type of virus and its strategy) → More genes = more complex virus / strategy
43
Q

Host Range + cell/tissue specificity

A

limited: often single species or a few closely related species; often limited to one or a few types of cell / tissues in the host

Recognize host cells by fitting between the viral protein and specific molecules on cell surface of host

Eg. influenza A – sialic acid
Rabies: acetylcholine receptor → broad host range because receptor is at a nerve (releases acetylcholine) and muscle junction (receives and contracts) via conformational changes; can affect the nervous system

Cell / tissue specificity: defines the damage to the host
Ie. HIV attacks the t-cells that helps you fight off foreign bacteria, thus you lose your entire immune system

44
Q

Reproductive Cycle of a Virus – Overview:

A

Main point: viral genes use host’s enzymes, ribosomes, tRNAs, amino acids, ATP, and other resources to reproduce

  1. Infect host with viral genome → attachment, penetration, uncoating
  2. Replicate genome
  3. Manufacture capsid proteins
  4. Assemble viral particles
  5. Exit cell
45
Q

RNA Viruses

A

genome is RNA but therefore has a special problem due to mismatch of cellular machinery

Polycistronic: mRNA that encodes for several proteins; consists of a leader sequence that precedes the first gene

Positive Strand: genome serves as mRNA where one gene encodes for the replicase (aka RNA dependent polymerase) → genome is positive strand

Infect cell, translate into replicase, replicase binds to positive strands and transcribes into multiple negative strands, transcribe back into even more positive strands (some of which are translated into capsids), packaging of positive strands into capsid, lyse host cell

Can be retroviruses: use a special enzyme called REVERSE TRANSCRIPTASE (special DNA polymerase that can use RNA/DNA as a template to yield DNA); can also degrade RNA to leave only DNA thus yielding complementary DNA

Negative Strand: genome is the complement of RNA; replicase already packaged in virion (comes from host cell that packaged it) → genome is negative strand

Infect cell, use existing replicase to transcribe original negative strand to positive strand, translate into replicase, replicase binds to positive strand (some of which are translated into capsids) and transcribes into more negative strands, packaging of positive strands into capsid, lyse host cell

Note: putting a negative strand alone into the cell won’t work as it cannot code for the replicase (unlike positive strand, which has the information to code)

46
Q

Reproductive Cycles of Bacteriophages

A

Bacteriophages: attack bacteria

Lytic Reproductive Cycle: viral production cycle that leads to death of the cell; takes about 20 to 30 min at 37 degrees C between infection and cell death; immediate; ie. virulent phages (T4)

  1. Phage attaches to cell surface
  2. Phage injects DNA (ATP stored in phage tail, which can change shape in order to puncture the cell)
  3. E. coli transcribes and translates viral DNA → one of the first proteins produced from viral DNA destroys the host cell’s DNA, good strategy because everything will follow viral DNA
  4. Host makes many copies of phage components
  5. Phage components self assemble into virions
  6. Enzymes (lysozymes) specified by viral gene digest cell wall of bacterium, releasing the phages

Lysogenic Cycle: temperate phages that integrate their genome into that of their host and can wait before going into the lytic cycle

  1. Binds to surface of E. coli
  2. Injects DNA into host
  3. DNA forms a circle
  4. DNA inserts itself into the host cell’s chromosome at specific site
  5. When the host cell divides, prophage copied along with the rest of bacterial genome can be passed onto the next (few) generation(s)
    - - Prophage: genes inactive EXCEPT for one that codes for the repressor protein, which can shut off other prophage genes → in eukaryotes, will be provirus
  6. With appropriate environmental factors, prophage will leave bacterial chromosome
  7. GOES LYTIC → enters lytic cycle, resulting in cell death
47
Q

Bacterial protection against viruses

+ How do bacteria protect itself from their own reaction enzymes?

A

Bacterial protection against viruses:

    • Mutations: change receptor size used by viruses to gain entry into bacterial cell
    • Restriction enzymes: endonucleases that cut up foreign DNA and recognize short, specific sequences at which a cut will occur; will often cut phosphodiester bonds of two strands in a staggered manner, thus yielding short single stranded ends called STICKY ENDS
    • Recognition sites are often palindromic sequences (sequence who complement reads the same when read in the same chemical direction) → ie. GAATTC // GAATTC

How do bacteria protect itself from their own reaction enzymes?
– Modification of reaction enzymes recognition site via methylation → methyl groups on nucleotides blocks access to the site for reaction enzymes

48
Q

Bacterial Genetics

A

Reproduce by binary fission → semi conservative replication of bacterial chromosome

Single origin of replication; 2 replication forks move until meet

No mitosis / meiosis, therefore genome is HAPLOID → generates genetic variation via mutations

49
Q

Generation of Genetic variation

A
  1. Mutations: rare but immediately expressed (since genome is haploid)
  2. Genetic Recombination: three methods

– Transformation: discussed in Streptococcus pneumoniae experiment; DNA taken up from surroundings, usually via surface receptors that recognize DNA from CLOSELY RELATED SPECIES; integrated into chromosome by “crossing over”

    • Transduction: transfer of DNA from one bacterium to another via bacteriophage; has homologous pairings and recombinants (splicing of DNA molecules into a new combination) due to crossing over → yields a complex change of phenotype and genotype
  • —- Initial phage will insert DNA into host cell and lyse after reproduction; resulting phage has donor DNA (called transducing phage) and will insert donor DNA into new host cell, which will be incorporated into recipient’s chromosome by recombination (yields complex change of phenotype and genotype)

– Conjugation: direct transfer of genes by “mating”; DNA donating cell grows sex pilus (cytoplasmic bridge for transfer of DNA) via F plasmid, attaches to DNA-receiving cell

50
Q

Plasmids
+ F plasmids
+ R plasmids

A

small circular, extrachromosomal double stranded DNA; can be beneficial under stressful conditions; replicate independently of bacterial chromosome; few genes are not necessary for growth and survival of bacteria

F plasmid: 25 genes; creation of sex pilus; replicates in synchrony with chromosomal DNA → F+ = bacterial cells without F plasmid; will donate H
– During conjugation: F plasmid replicates by ROLLING CIRCLE REPLICATION (once nicked, replication fork occurs in both F+ / F-); 5’ end inserted first into F- cell while 3’ exits F+ last); F- cell becomes F+, thus yielding two F+
– Hrf Cell: high frequency of recombination; occurs between recipient chromosome and transferred Hfr fragment
F plasmid can become integrated into chromosome; new copy inserted through sex pilus to F- cell, carrying with it adjacent genes from the bacterial chromosome
– BUT not all of F plasmids are transferred so the recipient remains F- (partially diploid) → amount of F+ chromosomes transferred proportional to conjugation time

R plasmids: carry gene for resistance to antibiotics; some may carry up to 10 genes; can be transferred between species via conjugation

51
Q

Control of Gene Expression

A

in prokaryotes

Bacteria adjust to changes in the environment; not all genes active at same time; insufficient resources to support expression of all genes simultaneously
– Need to economize: turn off and on metabolic pathways / enzymes when needed

Regulate activity of enzymes present → feedback inhibition: good for short term responses
Regulate transcription → only transcribe / translate as needed (ie. no storing)

52
Q

Operons + example

A

Operons: organization of functionally related genes in prokaryotes
– consists of control region (promoter and operator, switches that are NOT transcribed) and structural genes (codes for polypeptide products)

Operator: usually located between promoter and structural genes, controls access of RNA polymerase to structural genes
Advantageous bc only need one switch to turn on expression of genes that need to be coordinated

Example: E. coli prefers glucose for metabolic needs BUT can use other sugars (ie. lactose) if glucose not available → enzymes to utilize other sugars usually not present (or in low concentration) until needed
In lactose-free / glucose-saturated environment, only few molecules of beta-galactosidase
In high lactose / low glucose: mRNA for b-gal goes up BUT lactose permease (transport protein for lactose) will also go up → presence of lactose induces/turns on genes involved in lactose metabolism ⇒ controlled via OPERONS

53
Q

Lac operons and tryptophan operons

A

Lac Operons: example of inducible operon

    • RNA Polymerase: translates structural genes LacZ/Y/A into leader sequences b-gal/permease/transacetylase → high levels of transcription require presence of allolactose and promoter activation
    • High lactose = high levels of allolactose induces conformational change; repressor not bound to operator
    • Low glucose = cAMP levels increase, allowing cAMP-CAP to bind to promoter (cAMP - CAP = transcription factor that can recognize sequences in the promoter and thus can bind to the promoter)
    • Regulatory gene (aka Lac repressor): not part of operon but ALWAYS ON; transcribes monocistronic RNA that binds to the operator, thus RNA polymerase cannot bind and the structural genes are not transcribed → aka HARD SHUTDOWN
    • Allolactose: isomer of lactose that binds to Lac repressor, causing conformational change such that the repressor can no longer bind to the operator; RNA polymerase cannot bind and the structural genes are transcribed
    • CAP: catabolic activator protein; present in cell BUT cannot bind to promoter unless it is activated by cAMP; levels proportional to levels of glucose (low glucose = high AMP)

What about…
– High glucose AND high lactose levels → Repressor removed but cAMP-CAP not bound; low amount of transcription
– Low glucose and low lactose levels → repressor bound and cAMP-CAP also bound BUT remember – HARD STOP !!
Presence of tryptophan with lac operons: explanation of process of how it’s controlled; example of repressible operon

Tryptophan: repressible; similar to Lac operon; made in INACTIVE conformation; has a binding site for tryptophan, thus yielding a conformational change into ACTIVE → can now recognize sequence at the operator and shuts down transcription
Glucose can make tryptophan based on a specific pathway, usually always on (thus repressor is inactive) BUT need to maintain homeostasis
Too much tryptophan activates the repressor, which binds to the operator and stops transcription
As levels drop, the trytophan on the repressor will drop off and transcription begins again
Simpler than Lac operon because solely controlled by levels of tryptophan

54
Q

Cellular differentiation in eukaryotes

+ “loss of DNA” theory

A

Cellular differentiation in eukaryotes:

    • Human has ~200 different cell types → look and behave different
    • All arose from zygote by mitotic cell division, therefore should have the same genes
    • BUT different proteins mean that different genes in the cell are turned on, aka differentiation

“Loss of DNA” Theory: disproved by cloning experiments that recreated an entire organism

Procedure: take DNA from donor and fuse with de-nucleated cell via electric shock; fused cell will divide normally; place blastula into uterus of a foster mother; yields an organism that is genetically the same as the donor nucleus

55
Q

Gene Expression in Multicellular Eukaryotes

+ Gene Regulation

A

only expresses 20 - 30% of their genes

Housekeeping genes: proteins common to all cells
Structural proteins of chromosome, ie. histones
– RNA polymerase
– Cytoskeleton
– Metabolic processes: cell respiration enzymes
– DNA repair enzymes + ribosomal proteins
BUT some proteins are only in specialized cells
Ie. hemoglobin in red blood cell; insulin in pancreatic cells

Many genes are “switched off” while other genes (or their protein products) are “regulated”

Due to complexity of chromosome structure, gene organization, and cell structure,, there are many opportunities for control of gene expression IN EUKARYOTES

56
Q

Pre transcriptional controls

A

regarding chromatin structure

Chromatin organization: condensed heterochromatin therefore not expressed
– Euchromatin: less densely packed; unravelled chromatin is more accessible → necessary condition BUT not the only condition for expression

Chemical modification of chromatin:

– Histone acetylation: placing acetyl groups on histone tails so that they are not wrapped so tightly
COCH3 attached to AA (aka histone acetylases)
—– Histones bind to DNA less tightly
—– Histone deacetylation = less available
Histone methylation: effect dependent on lysine; can be conducive (ie. arginine) to transcription but mostly repressive (ie H3K27)

DNA methylation: epigenetic inheritance; aka inheritance within a cell lineage → shutting down genes in order to specialize and thus turn on other genes

    • Epigenetic: heritable patterns due to DNA modification BUT doesn’t change DNA sequence
    • Usually promoter in a certain gene within a cell is methylated, therefore rendering it inactive
    • Makes it inheritable because when gene A is methylated in a parent cell, then all the progeny from that parent will also have gene A methylated
57
Q

Pattern of Methylation

+ genomic imprinting

A

can be inherited

Eg. X inactivation in Mammals

    • Female have 2 X chromosomes but males have 1.
    • Dosage compensation: female embryo has 1000 to 3000 cell stages and will deactivate a single X chromosome at random
    • Xist = x inactivation specific XIC (X chromosome inactivation center) that will bind to one X chromosome; attracts protein complex Polycomb Repressor Complex , which is an enzymatic system that will methylate the chromosome, thus creating a BARR BODY

Eg. female calico cats (hetero) and their patches of color as a result of methylated / unmethylated skin cells (males are homo and are single colored) → similarly, patterns of DNA methylation inherited by progeny cells of individually methylated genes, yielding SILENCED GENES

Genomic Imprinting:

    • Maternal: one gene is expressed in parent and not expressed (therefore, methylated) in the offspring → Methylation retained during crossing
    • Paternal: methylated gene is not expressed in parent but is expressed in offspring → Methylation is stripped off during crossing
58
Q

What are barr bodies

A

X inactivation center (Xic): not in DNA; will be transcribed into X inactivation specific transcript (Xist), therefore composed of RNA → excess amounts of this RNA will randomly coat one of the X in a X/X

This excess will attract a polycombRC2 which is a methylation machine, therefore becomes very condensed in its bar body THUS not expressed
– The X in X/X that becomes methylated is totally random → 50% covered will be from Mom; 50% covered will be from dad

ALSO EPIGENETIC WITHIN CELL LINEAGE because offspring will have the exact same X chromosome methylated
Ie. skin cell will retain methylation, thus creating a patch due to the same bar bodies from the ancestral cell → is this the calico cat example

59
Q

Transcriptional Controls

+ specific transcription factors mode of action

A

transcription requires specific interactions, such as transcription factors and gene control regions (non-coding sections of DNA)

    • Promoter usually binds to general TF in upstream direction
    • Enhancer and Silencers: within the DNA sequence; consist of control elements (short sequences of DNA; recognizes proteins); can be located far from the promoter BUT can fold to meet / bind to the promoter via direct interactions
  • —- Activators: stimulate, can bind to enhancers ONLY
  • —- Repressors: inhibit; can bind to BOTH enhancers and silencers

Specific transcription factors mode of action:
– TF direct control – bind enhancer and silencers

Repressors:

    • Comparative DNA binding by overlapping the site for activation
    • Masking the activation surface by binding next to each other (ie mashed together)
    • Direction interaction with the general transcription factors

TF indirect control can affect chromatin structure

    • Activator: attract histone acetyltransferases → looser therefore more favorable for transcription
    • Repressor: recruit histone deacetylases, which remove acetyl groups from histone trails and promote tail interactions → righter, therefore less favorable for transcription
60
Q

Coordinately controlled genes

A

no operons in eukaryotes

Genes related to same metabolic pathway can be on different chromosomes; has its own promoter; transcribed separately

Control regions of functionally related genes often have the same DNA sequences (enhancers); are recognized by the same TFs (activators / repressors); and are often activated by external signals

    • Steroid hormones: operate by changing gene expression via conformational change into either activation or inhibition
    • Growth factors and non steroid hormones
  • -MAPK: mitogen activated protein kinase; phosphorylated into nucleus by transduction → binds to inactive to activate

** Same signal but different responses by cell type

61
Q

Post transcriptional control

+ general procedure for transcription

A

gene transcribed as a unit and codes for set of closely related polypeptides

  1. Transport to cytosol delayed
  2. Alternative RNA splicing
  3. Longevity of mRNA: affected by PT modification
    - - Poly A tail: if longer, then longer time of expression
    - - Sequence in 3’ – UTR (untranslated region; uncapped and rapidly broken down)
  4. Localization in cytosol: 3’UTR signal is more spread out, thus needs more time
  5. miRNA (micro) and siRNA (small interfering): basically disrupt or inhibit mRNA transcription
    - - miRNA = endogenous; encoded by the genome; transcribed RNA polymerase II
    - - siRNA = almost exactly the same process BUT exogenous; something other than your cell is making it and inserting it into your cell, therefore mostly done by virus

General Procedure for transcription:

    • Genome goes through RNA polymerase II → creates pri-miRNA (primary) that can fold together to form hairpin loops → some sort of protein will cleave it and allow it to go through the nuclear membrane (specifically use Drosha), called a pre-miRNA (precursor) – other parts that are not the severed hair pin loops are degraded
    • Dicer will cut off the not double stranded part of the hairpin loop, will yield only the dsRNA segment (aka microRNA duplex) – argongonate 2 (ago2) will combine will the single stranded portion that was cut from the hairpin loop and ss mature microRNA and create RISC complex (RNA induced silencing complex) → silences by finding mRNA that it can base pair to and then makes a cut, thus making it nonfunctional → will be found by the cell and degraded
62
Q

Difference btwn RNA Polymerase II versus I and III

A

II = all protein encoding genes are transcribed here

I or III = structural RNA → ie. rRNA

    • rRNA: Allows for folding up, thus yielding secondary structures with double stranded stems (due to base complementary) with single stranded hairpin loops → Can complex with proteins and create large subunits or small subunits within ribosomes
    • tRNA: transcribed by RNA polymerase III; not translated into protein, the transcript is the end product once it folds up into its secondary structure
    • srp RNA: transcribed by either I or III ; signal recognition particle (allows for binding of specific amino acid sequence – basically how a free ribosome becomes a bound ribosome); made of both protein and RNA (RNA in it = snRNA) ⇒ reverse transcriptase inserts transposons everywhere BUT can also recognize srpRNA and thus stick it elsewhere lol fuck→ not talked about so not going to be on exam
63
Q

Transposable Elements

A

aka “jumping genes”; consequence dependent on ending location

DNA Transposons: cut and paste

    • Needs: ligase (integrase), enzyme to cut it out → enzyme will cut out transposon while ligase will take the ends of the transposon and put it anywhere ⇒ done currently
    • Gap formed is called a DOUBLE STRANDED BREAK → can be repaired but you won’t be tested for that
    • Can be dangerous because can be pasted anywhere and disrupt a thing
    • Transposase: removes transposons by recognizing the ends
    • Integrase: randomly insert transposons into genome

Retrotransposon: retroviral-like; copy and paste
similar to RNA virus except already integrated into genome
– Needs: reverse transcriptase (to copy it) and also need ligase (integrase) (to paste it)
– Done for mRNA and without genome replication – also no splicing so the genome remains intact

  • If inserted into intron, then will be cut out during transcription
    • If inserted into promoter, then will not work / will not allow for transcription
    • Promoters are not random DNA but rather have a specific sequence that allow for transcription
  • ** If inserted into exon, then will be transcribed while actual exon is removed → acts as an “exon imposter”; can also shift the reading frame
    • Will create a protein with a bunch of amino acids that are not supposed to be there, thus will not fold properly (aka nonfunctional, needs to be degraded)

Ie. inserts into exon, creating 1000bp long transcription which is NOT divisible by 3, thus shifts reading frame

Introns are useful because make up a lot more than exons / promoters (which make up 21.5% total), thus when a transposon is inserted then it won’t matter.

Different in prokaryotes, which have a lower number of introns, thus less spacers; any transposons can wreak major havoc

64
Q

Gene cloning (3)

A

What: one gene enters the host; host expresses the gene

Why: complexity of eukaryotic studied in simply systems; difficult to study gene in eukaryotes, therefore easier to study in bacteria; produce human gene product

    • Ie. insulin: place human insulin into bacteria and harvest for use
    • Ie. insertion of useful genes into agricultural plants → increase starch yield / resistance to pathogens

Tools: bacteria, plasmids, restriction enzymes, phages, DNA ligase, reverse transcriptase

65
Q

Cloning vectors (6)

A

vehicle for transferring genes from one organism to another (ie. human genes into bacterium) → either phage or plasmid

Plasmid: specifically pBLU plasmid

AMPR: gene for ampicillin resistance

LacZ: gene coding for b-galactosidase

MCS: multiple cloning site, aka “polylinker”; contains restriction sites for large numbers of different restriction enzymes

Restriction enzymes: recognizes short sequences that will cut → if palindromic, will cut in staggered fashion thus yielding sticky ends

Promoters: gene expression BUT different from previous mentions; only one copy per restriction site

66
Q

Steps in Gene Cloning

A
  1. Isolate source genomic DNA and vectors
  2. Cut both with the same restriction enzyme
  3. Mix, allow to anneal → weak H bonds come together and fall apart between the sticky ends
  4. Seal with DNA ligase by making phosphodiester linkage
  5. Transform LacZ (-) DNA bacteria with plasmid mixture → LacZ (-) cannot make b-gal gene
  6. Plate transformed bacteria onto agar plate containing ampicillin (only allows plasmid growth) and x-gal (colorless)
    - - Three types of bacteria yield: no plasmid; plasmid but no DNA insert; plasmid w foreign insert (desired pdt)
    - - Lactose analog turns blue when cleaved by b-gal from foreign insert
    - - Colony consists of millions of bacterial cells that have all descended from one cell → blue if cleaved, white if not
  7. Overlay petri plate with nylon membrane, thus making a replica plate to allow for growth → be gentle! Only want a touch with no excessive spreading.
  8. Treat replica, lyse cells, denature DNA (make single strands) BUT in the same spot via high alkaline conditions (which create unstable RNA which are removed to leave single stranded DNA); bake to fix the single strands
  9. Wash with radioactive probe:
    - - Probe: single stranded oligonucleotide with base sequence complementary to some part of gene of interest; must be long enough to be UNIQUE and stick only to the gene of interest
  10. Wash away excess probe; radioactivity will remain hybridized to the target sequence
  11. Overlay photographic film, develop it, locate the radioactivity by comparing to ORIGINAL template

Problem: gene cloning doesn’t work because the introns are not removed → eukaryotic genes in a prokaryote will contain introns as prokaryotes don’t have spliceosomes, therefore the introns will also be translated
– Another approach: construct a complementary DNA library (cDNA)

67
Q
Compare
LacZ (-) w/o insert
LacZ (-) w insert 
pBLU w/o insert 
pBLU w/o insert
A

LacZ (-) w/o insert = non functional b-gal gene = white
LacZ (-) w insert = functional b-gal = blue bc x-gal cleaved
pBLU w/o insert = functional b-gal = blue bc x-gal cleaved
pBLU w/o insert = non functional b-gal gene = white

68
Q

cDNA Library cultivation procedure

A
  1. Isolate mRNA from tissue where gene of interest is being expressed
  2. Use reverse transcriptase to convert mRNA to double stranded DNA (aka cDNA)
  3. Use same steps as previously outlined for genomic library to insert cDNA into the plasmid
    - - Problem: no sticky ends for plasmid insertion but also need to protect cDNA by methylation at EcoR1 sites; will later use EcoR1 linker to ligate cDNA; then cleave with EcoR1, creating cDNA with sticky ends!
  4. Plate on plate with ampicillin and x-gal bc will still get 3 different yields
69
Q

Genomic Library

versus cDNA library

A

Genomic Library: all genes of source DNA are represented somewhere on petri plate
– To find gene of interest, you COULD sample each white colony, isolate the plasmids, cut with restriction enzymes, use electrophoresis on it BUT takes a lot of time to test all the colonies

Versus cDNA library: not every gene in the genome is represented; only those genes are actively being expressed in the tissue from which the mRNA was isolated (no introns tho; only exons) → search using radioactive probe to find the gene of interest ⇒ should work now!!

70
Q

Restriction Fragment Length Polymorphisms

A

one way of doing DNA fingerprinting (slightly outdated tho lol)

Basic idea: different individuals will have different locations of restriction enzyme sites that, when cut with restriction enzyme, will yield different results on the electrophoresis gel
– Agarose gel has lower resolution than polyacrylamide gel (which can identify lengths by a difference of ONE base)

Different lengths = polymorphism → restriction fragments are the resultant pieces of DNA that are created by restriction enzymes – poly because different lengths, thus yielding different banding sites

Not really accurate at low number of low locis → the more locis you compare, the more accurate your test becomes