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AP Biology Study > Unit 6 - Gene Expression/Regulation > Flashcards

Unit 6 - Gene Expression/Regulation Flashcards

1
Q

Explain the conclusion drawn from Griffith’s transformation experiment.

A

Pathogenic bacteria are killed by heat and mixed with living nonpathogenic bacteria → leads to pathogenic living cells

Discovered transforming substance = DNA (with mice)

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2
Q

Explain the conclusion drawn from Avery, McCarty, and MacLeod’s transformation experiment.

A

Tested for the transforming factor with three separate vials - DNA, proteins, RNA

Supported the claim that transforming substance = DNA

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3
Q

Explain the meaning of semiconservative.

A

DNA replication is semiconservative because each of the daughter molecules produces have one parent strand and one newly synthesized strand

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4
Q

Describe the steps of DNA replication.

A
  1. Begins at sites called origins of replication that have short sequences of DNA that initiation proteins recognize and attach to
  2. These proteins then separate the two strands and open up a replication bubble. These bubbles are spaced out on the strand
  3. Helicase untwists the double helix at replication forks (Y shaped regions at the end of each bubble) by breaking hydrogen bonds between the two strands
  4. Untwisting causes strain and tighter twisting ahead of the replication fork. Topoisomerase relieves this by breaking, swiveling, and rejoining DNA strands
  5. DNA polymerase comes to add nucleotides for the synthesis of a new DNA strand. However, it can only add to existing strands, not create a new one from scratch
  6. Primase creates a short complementary RNA chain called a primer, using the parental DNA strand as a template
  7. Now DNA replication proceeds in both directions along the strand. DNA polymerase can only add to the 3’ end of a primer, elongating the DNA in a 5’ –> 3’ direction.

For the leading strand, replication is continuous, into the replication fork, and only one primer is required

For the lagging strand, which is antiparallel, replication works in segments in the direction away from the replication fork. These segments are called Okazaki fragments - each of them must have a separate primer. DNA polymerase adds nucleotides to connect the primers. At the end, the RNA nucleotides from the primers are replaced by DNA nucleotides

  1. Finally, DNA ligase joins the sugar phosphate backbones of all the Okazaki fragments into a continuous DNA strand
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5
Q

What 3 factors contribute to the accuracy of DNA replication?

A
  1. Specificity of base pairing (A=T, C=G)
  2. Mismatch repair - special repair enzymes fix incorrectly paired nucleotides
  3. Nucleotide excision repair - incorrectly placed nucleotides are excised/removed by enzymes called nucleases, and the gap left over is filled in with the correct nucleotides
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6
Q

What are telomeres and why are they useful?

A

The fact that DNA polymerase can only add nucleotides to the 3’ end of a molecule means that it cannot replicate the 5’ end of the DNA molecule at the end of the chromosome. As a result, repeated rounds of replication (EX: of chromosomes for mitosis) produce shorter and shorter DNA molecules with uneven ends.

In order to prevent essential genes from being lost, the linear ends of eukaryotic chromosomes are capped with telomeres - short, repetitive nucleotide sequences that don’t contain genes. Telomeres get eroded in place of essential genes.

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7
Q

Euchromatin vs Heterochromatin

A

Euchromatin is DNA that is uncoiled and loose (non methylated and histones are acetylated) so it’s available for transcription [found in interphase cells]

Heterochromatin is DNA that is fully condensed into coils so it’s inaccessible to transcription and thus generally does not undergo it [EX: Barr bodies - permanently condensed X chromosomes in female cells]

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8
Q

What is the difference between replication, transcription, and translation in terms of the starting material and the product?

A

Replication: DNA –> DNA
Transcription: DNA –> RNA
Translation: RNA –> protein

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9
Q

What is the difference between replication, transcription, RNA processing, and translation in terms of location?

A

Replication: Nucleus in eukaryotic cells (during S phase of the cell cycle)
Transcription: Nucleus in eukaryotic cells
RNA processing: Nucleus in eukaryotic cells –> exits nucleus through nuclear pores in the membrane
Translation: Cytoplasm at ribosomes

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10
Q

What is the difference in gene expression between prokaryotes and eukaryotes?

A

In prokaryotic cells, mRNA does not undergo RNA processing and transcription & translation can occur simultaneously. In eukaryotic cells, transcription yields pre-RNA or primary transcript that undergoes processing to yield the final mRNA

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11
Q

Central Dogma

A

DNA –> RNA –> Protein

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12
Q

Describe the steps of transcription.

A
  1. RNA polymerase recognizes and binds to the promoter.

In eukaryotic cells, transcription factors recognize the TATA box and bind first, then help RNA polymerase bind in the correct position/orientation. The transcription initiation complex consists of transcription factors + RNA polymerase + promoter

  1. RNA polymerase untwists the two strands of DNA and attaches complementary RNA nucleotides to the template strand. It can only add nucleotides to the 3’ end so RNA elongates in the 5’ –> 3’ direction
  2. The new RNA molecule peels away from its DNA template and the double helix forms again
  3. After RNA polymerase transcribes a terminator sequence in the DNA (which signals the end of transcription), the RNA transcript is released and polymerase detaches. In eukaryotes, a polyadenylation signal is transcribed
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13
Q

Template/Noncoding strand

A

Provides the template for transcription; the mRNA produced is complementary to this strand

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14
Q

Coding strand

A

The sequence is identical to the mRNA produced only with T instead of U

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15
Q

Describe what occurs during RNA processing.

A

Addition of a 5’ cap (modified guanine) and a poly-A tail to the 3’ end (string of adenines). These facilitate the export of mRNA from the nucleus and protect it from degradation by hydrolytic enzymes. In the cytoplasm, they also help ribosomes attach to the 5’ end of the mRNA to begin translation

The length of the poly A tail also determines the lifespan of the mRNA fragment

RNA splicing - introns (noncoding segments) are removed and exons (coding segments) are joined together by a spliceosome. This forms a continuous coding sequence rather than segments

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16
Q

There are 25,000 genes but only 100,000 proteins in an organism. Propose an explanation.

A

In alternative RNA splicing, a single gene can encode more than one kind of polypeptide depending on which introns are moved. So processing can create multiple mRNA sequences which then produce multiple polypeptides and proteins

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17
Q

What did Hershey and Chase’s experiment discover?

A

They labeled DNA with radioactive phosphorous and protein with radioactive sulfur. Then they analyzed what happened when phages infected bacteria.

The radioactive sulfur did not get in while the radioactive phosphorous was found inside the bacteria

Conclusion: DNA is what allowed the virus to infiltrate and replicate its DNA.

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18
Q

Describe the structure of a tRNA molecule.

A

The protruding end at the top (3’) is the specific amino acid attachment site.

The bottom three bases make up the anticodon (3’ –> 5’) that base pairs with the complementary codon on mRNA

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19
Q

Describe the function of tRNA.

A

They transfer amino acids from the cytoplasm (taken from other compounds or from the surrounding solution) to a growing polypeptide chain in a ribosome

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20
Q

Describe the structure of a ribosome.

A

Made up of rRNA and protein

Consists of a small subunit that binds to the mRNA strand and a large subunit that has 3 binding sites for tRNA molecules (E, P, A sites)

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21
Q

Describe the three binding sites on the large subunit of the ribosome.

A

E site - where empty tRNA molecules exit
P site - holds the tRNA that carries the growing polypeptide chain
A site - holds the tRNA that carries the amino acid that will be added to the chain next

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22
Q

Describe the steps of translation.

A
  1. The small ribosomal subunit binds to the mRNA strand in such a way that the first codon (AUG) is placed in the proper position
  2. A special initiator tRNA, which carries the amino acid “Met” and has the anticodon UAC, hydrogen bonds to the first codon
  3. Then the large subunit attaches, allowing the tRNA to attach to the P site. The A site is now available to the tRNA that will bring the second amino acid.

Initiation factors are proteins that helped bring these components together into a complex

  1. The anticodon of the next tRNA molecule base pairs with the complementary mRNA codon in the A site
  2. The polypeptide chain on the tRNA in the P site is transferred to the amino acid on the tRNA in the A site
  3. The empty tRNA is then moved to the E site and released while the polypeptide tRNA in the A site is moved to the P site

mRNA kinda moves like a conveyer belt in the left direction from 3’ –> 5’

  1. This continues happening until the ribosome reaches a stop codon on mRNA. A release factor binds to the stop codon at the A site and promotes hydrolysis of the bond between the tRNA in the P site and the polypeptide (basically H2O is added instead of another amino acid)
  2. The complete polypeptide is released through the E site. The signal peptide (a sequence of amino acids at the leading end of the polypeptide chain) determines the destination of the protein by serving as a sort of luggage tag
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23
Q

What are the two main types of mutations?

A
  1. Point mutations (aka nucleotide pair substitution) - changes in a single nucleotide pair of a gene
  2. Frameshift mutations (aka nucleotide pair insertion/deletion) - changes in the reading frame or triplet grouping of the genetic message when it’s not a multiple of 3
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24
Q

What are the types of point mutations? What are their effects on phenotype?

A
  1. Silent mutation: the amino acid sequence does not change so there’s no effect on phenotype
    EX: Due to the redundancy of the genetic code, CCG and CCA produce the same amino acid
  2. Missense mutation: the amino acid sequence changes but it could have little effect if the mutation occurred in a region not essential to the function of the protein or if the new amino acid has similar chemical properties to the original one
  3. Nonsense mutation: the amino acid codon is changed to a stop codon which terminates translation prematurely, resulting in a shorter polypeptide and thus a nonfunctional protein; very large, harmful effect
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25
Q

What are the types of frameshift mutations? What are their effects on phenotype?

A
  1. Insertion: adding a nucleotide; shifting the reading frame to the right
  2. Deletion: removing a nucleotide; shifting the reading frame to the left

Both these mutations could have very disastrous effects because they affect multiple nucleotide pairs compared to substitutions which only affect one pair

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26
Q

What is a gene and what is its final product(s)?

A

A gene is a region of DNA that can be expressed to produce a final functional product - either a polypeptide OR an RNA molecule (rRNA, tRNA, mRNA)

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27
Q

What is epigenetic inheritance? What are two examples of it?

A

Epigenetic inheritance is when the nucleotide sequence is not changed yet traits are still inherited; DIFFERENT FROM MUTATIONS

EX: Histone acetylation and DNA methylation (both reversible)

28
Q

Histone acetylation

A

Promotes transcription by adding acetyl groups to histones which opens up chromatin structure (unpack)

29
Q

DNA methylation

A

Reduces transcription by adding methyl groups and thus condensing chromatin

Long stretches of inactive DNA (inactivated X chromosomes) are usually more methylated

Inherited in successive cell divisions and DNA replication

30
Q

List and describe the components of an operon.

A

Promoter – Operator – Coding Genes

Operator - acts as an on/off switch that controls RNA polymerase’s access to genes; where the repressor attaches

Promoter - where RNA polymerase attaches

Coding genes - the entire stretch of DNA that produces enzymes

31
Q

Repressor

A

Binds to operator and blocks attachment of RNA polymerase to the promoter, preventing transcription of the genes

32
Q

Explain how the Lac operon functions and its benefits.

A
  1. The repressor is active by itself and binds to the operator, switching the operon off (no breakdown of lactose molecules)
  2. When there are high levels of lactose in the cell, allolactose acts as an inducer that binds to the repressor and alters its shape, inactivating it so that it can no longer bind to the operator.
  3. Without the repressor, RNA polymerase can attach to the promoter and transcribe the genes in the operon into mRNA

This enables the cell to conserve energy and resources, producing enzymes only when the nutrient is available.

33
Q

Tryptophan Operon vs Lac Operon

A

Tryptophan Operon

  • synthesis, anabolic pathway
  • operon is normally ON (but can be repressed), repressor is inactive
  • tryptophan acts as a corepressor that binds to the repressor and activates it in order to inhibit transcription
  • negative control/regulation
  • repressible operon

Lac Operon

  • breakdown of lactose, catabolic pathway
  • operon is normally OFF (but can be induced), repressor is active
  • allolactose acts as an inducer that binds to the repressor and inactivates it in order to stimulate transcription (take away stop sign)
  • negative control/regulation
  • inducible operon
34
Q

Regulatory genes

A

Produce repressor proteins

35
Q

Explain how the Tryptophan operon functions.

A
  1. The operon is normally on because tryptophan is required by bacteria for survival. The repressor is inactive
  2. When a sufficient amount of tryptophan is synthesized, it acts as a corepressor that binds allosterically to the repressor and changes its shape, activating it.
  3. The repressor binds to the operator, thus blocking RNA polymerase from initiating transcription.

This enables the cell to conserve energy and resources, since when the end product is in sufficient supply, it hinders its own synthesis [feedback inhibition]

36
Q

What is the difference between negative regulation and positive regulation?

A

Repressible (try) and inducible (lac) operons are examples of negative control because they both involve a repressor protein that turns OFF the operon by blocking RNA polymerase. In order to turn the operon on, you must remove the stop sign

Positive control is when a protein turns ON the operon and increases the rate of transcription. There is no repressor involved. In order to turn the operon on, you must add a go sign

37
Q

Describe an example of positive regulation.

A
  1. Bacteria prefer glucose to lactose as an energy source. When glucose is scarce and lactose needs to be broken down, cAMP molecules accumulate
  2. cAMP binds to an activator called CRP (cAMP receptor protein) which changes its shape into the active form
  3. CRP attaches to a site upstream to the promoter and increases the affinity of RNA polymerase for the promoter, thus increasing the rate of transcription for genes that code for enzymes to break down lactose as an energy source
  4. If the amount of glucose increases, the concentration of cAMP falls and CRP detaches from the operon (inactive) —> RNA polymerase binds less effectively —> slower transcription
38
Q

How does gene regulation work in eukaryotic cells?

A
  1. Activator proteins bind to control elements of the enhancer on the DNA
  2. A DNA bending protein brings the bound activators closer to the promoter where other essential proteins like general transcription factors and mediator proteins are found as well
  3. The activators bind to the mediator proteins and general transcription factors. Altogether, they assemble and position the transcription initiation complex (transcription factors + RNA polymerase + promoter) by binding to RNA polymerase
  4. Transcription of genes begins
39
Q

What is an enhancer made up of?

A

A grouping of three distal control elements

Control elements = segments of noncoding DNA that serve as binding sites for activators (specific transcription factors) during eukaryotic transcription

40
Q

General transcription factors vs specific transcription factors

A

General transcription factors work AT the promoter by binding to the TATA box within the promoter or to RNA polymerase. This interaction only leads to a low rate of transcription

Specific transcription factors bind to control elements close to or far away from the promoter and lead to high levels of transcription

41
Q

What are coordinately controlled genes and what are their benefits?

A

By grouping genes of related function into one transcription unit (one operon with a single promoter), they can be coordinately controlled. This means they are transcribed and translated together to produce a single mRNA molecule and a single polypeptide chain —> allows all the enzymes to be synthesized at one time when the need arises.

42
Q

Why do certain eukaryotic cells only produce certain proteins? Relate this to gene regulation. (Hint: activators).

A

Each cell type has different activator proteins present that may or may not bind to the control elements of the coding gene

EX: A gene has three control elements: liver, eye, and mouth. Let’s say an eye cell wants to make an eye protein. It has specific activator proteins that bind to the eye control element and stimulate transcription. A liver cell only has activator proteins that bind to the liver control element, not to the eye control element so it won’t make eye proteins

43
Q

Where in a coding sequence will a mutation be most detrimental?

A

Beginning

44
Q

Cell differentiation

A

Embryonic cells become specialized in structure and function

Differentiation is marked by the appearance of tissue-specific proteins which are found only in a specific cell type and enable it to carry out its function

45
Q

Morphogenesis

A

Structures and shape of organism develop (EX: tadpole → toad)

46
Q

What two factors control differentiation and morphogenesis?

A
  1. Cytoplasmic determinants - maternal substances in the unfertilized egg that influence early development

After fertilization, the zygote undergoes mitotic divisions and distributes the cytoplasm unevenly to daughter cells

These determinants include internal transcription factors, proteins, and organelles in the cell.

They regulate the expression of genes that affect each cell’s development (initially establish body axes)

  1. Induction - signaling molecules from one tissue or embryonic cell affect the development of another

Signals are conveyed to an embryonic cell from neighboring cells (through contact with cell surface receptors or binding of growth factors)

The response is transcriptional changes in target cells that send them down a specific developmental path and causes cells to differentiate

47
Q

Determination

A

The point at which an embryonic cell is irreversibly committed to becoming a particular cell type; differentiation is observed and it will differentiate into this cell type no matter the conditions

48
Q

Pattern formation

A

Development of a spatial organization of tissues + organs in their characteristic places

Relative positions of head + tail, right + left, and back + front (three major body axes) are set up before the organs appear

49
Q

Homeotic genes

A

Genes that control pattern formation during early embryonic development. These genes encode proteins called transcription factors that direct cells to form various parts of the body (EX: Hox genes)

50
Q

Positional information

A

Molecular cues that control pattern formation

They are provided by cytoplasmic determinants and inductive signals, and they tell a cell its location relative to the body’s axes and to neighboring cells

51
Q

Anterior vs posterior

A

Anterior = front and posterior = back

52
Q

Bicoid

A

Concentrated in the anterior end of the embryo

The gradient of a maternal effect gene called a bicoid protein determines the anterior-posterior axis

53
Q

What are the two causes of cancer and uncontrollable cell proliferation?

A

Gradual accumulation of mutations that convert proto-oncogenes to active oncogenes and knock out tumor suppressor genes

54
Q

Proto-oncogenes

A

Normal version of oncogenes; code for proteins that stimulate normal cell growth and division

55
Q

Tumor suppressor genes

A

Code for proteins that prevent uncontrolled cell growth by inhibiting the cell cycle

56
Q

How do proto-oncogenes turn into oncogenes?

A

They become hyperactive (mutation in coding sequence) OR they are produced in excess (multiple copies of the gene could be made through duplication or it could be translocated to a new promoter that is especially active)

57
Q

What are the components of a virus?

A

A virus is a small nucleic acid genome (RNA or DNA) enclosed in a protein capsid (made of capsomeres) and sometimes a membranous viral envelope containing viral proteins that help the virus enter a cell

58
Q

How do retroviruses work?

A

Retroviruses have RNA genes → use an enzyme called reverse transcriptase to make a DNA transcript of their RNA → this DNA then undergoes the lysogenic cycle, entering into the nuclear DNA of its host and being integrated as a provirus

59
Q

Generalized transduction

A

The transfer of DNA between infected bacterial cells using a lysogenic virus

Viruses have the ability to introduce genetic change into organisms as well as to undergo rapid genetic change themselves. Moving from one host to another, viruses may pick up pieces of the first host’s DNA and carry it to the next cell to be infected

60
Q

Which direction is mRNA read by ribosomes?

A

5’ –> 3’

61
Q

Where is a mutation most harmful?

A

Near a start codon

62
Q

What is the function of the heat shock done during the transformation lab?

A

It enables the bacterial membrane to become more fluid so it can take up the plasmid DNA and undergo transformation

63
Q

Function of gel electrophoresis

A

Separate DNA fragments by length

DNA is negatively charged so it will be drawn to the positive end –> pulled down

Long near negative (top), short near positive (bottom)

64
Q

PCR (function, process)

A

Function - make multiple copies of (amplify) a segment of DNA from a mixture of DNA molecules

Process

  1. Denaturing - separates 2 strands of DNA through heat
  2. Annealing - allows RNA primers to bind to the gene of interest for new strand synthesis (cooling)
  3. Extending - DNA polymerase binds to the primer and replicates the fragment

Enzymes come from thermophilic (heat loving) bacteria so they don’t denature at high temperatures

65
Q

What are the body axes determined by?

A

Gradients of morphogens encoded by maternal effect genes

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