BIOL 103 (Comprehensive Lessons 0-24)

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When bacteria divide, they first copy all their DNA and then each daughter cell receives a copy. Which theme of life fits bets? (L0)

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Biology

Cells

culmination of slide info, LC questions, quiz questions, etc.

288 Terms

1

When bacteria divide, they first copy all their DNA and then each daughter cell receives a copy. Which theme of life fits bets? (L0)

Information Storage, Transmission, and Flow

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2

The phospholipid bilayer of the plasma membrane has a hydrophobic core, and thus only materials that are non-polar (hydrophobic) can get through the plasma membrane without help. Which theme of life fits best? (L0)

Structure fits function

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3

The molecule ATP synthase is nearly identical in zebras, banana plants, and mushrooms. Which theme of life fits best? (L0)

Evolution

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4

When carbon dioxide and water are combined with energy from the sun, sugar molecules are made.

Which theme of life best describes this? (L0)

Transformation of matter and energy

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5

While bacteria are single celled organisms, they can sometimes form communities called biofilms. Bacteria in biofilms work together to secrete various proteins and sugars and they use chemical signals to communicate with each other.

Which theme of life best fits this? (L0)

Interactions between and within systems

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6

You hypothesize that a novel drug (BIDIX), that has recently been approved for the treatment of S. aureus infections of the skin in humans, increases number of neutrophils in the blood of humans over time. Interestingly, previous studies have also demonstrated that active S. aureus infections also increase blood neutrophil levels in human patients. You choose to use a mouse model for your experiment and set up the following groups, each consisting of 20 male mice:

Group 1: 20 male mice with an active S. aureus infection of the skin that receive a saline injection (0.5 mL) twice a day for 10 days.
Group 2: 20 male mice with an active S. aureus infection of the skin that are given 100 mg of BIDIX dissolved in saline (0.5 mL) twice a day for 10 days.
Group 3: 20 male mice with no bacterial skin infection that receive that receive a saline injection (0.5 mL) twice a day for 10 days.
Group 4: 20 male mice with no bacterial skin infection that are given 100 mg of BIDIX dissolved in saline (0.5 mL) twice a day for 10 days.

The use of male mice in this experiment is supported by which of the following themes of life? (L1)

Evolution

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7

You hypothesize that a novel drug (BIDIX), that has recently been approved for the treatment of S. aureus infections of the skin in humans, increases number of neutrophils in the blood of humans over time. Interestingly, previous studies have also demonstrated that active S. aureus infections also increase blood neutrophil levels in human patients. You choose to use a mouse model for your experiment and set up the following groups, each consisting of 20 male mice:

Group 1: 20 male mice with an active S. aureus infection of the skin that receive a saline injection (0.5 mL) twice a day for 10 days.
Group 2: 20 male mice with an active S. aureus infection of the skin that are given 100 mg of BIDIX dissolved in saline (0.5 mL) twice a day for 10 days.
Group 3: 20 male mice with no bacterial skin infection that receive that receive a saline injection (0.5 mL) twice a day for 10 days.
Group 4: 20 male mice with no bacterial skin infection that are given 100 mg of BIDIX dissolved in saline (0.5 mL) twice a day for 10 days.

Which group serves as the negative control for this experiment? (L1)

Group 3

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8

You hypothesize that a novel drug (BIDIX), that has recently been approved for the treatment of S. aureus infections of the skin in humans, increases number of neutrophils in the blood of humans over time. Interestingly, previous studies have also demonstrated that active S. aureus infections also increase blood neutrophil levels in human patients. You choose to use a mouse model for your experiment and set up the following groups, each consisting of 20 male mice:

Group 1: 20 male mice with an active S. aureus infection of the skin that receive a saline injection (0.5 mL) twice a day for 10 days.
Group 2: 20 male mice with an active S. aureus infection of the skin that are given 100 mg of BIDIX dissolved in saline (0.5 mL) twice a day for 10 days.
Group 3: 20 male mice with no bacterial skin infection that receive that receive a saline injection (0.5 mL) twice a day for 10 days.
Group 4: 20 male mice with no bacterial skin infection that are given 100 mg of BIDIX dissolved in saline (0.5 mL) twice a day for 10 days.

What is the greatest limitation of the experimental approach that you used to test your hypothesis? (L1)

Your experimental design only included male mice

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9

What is a polypeptide? (L2)

a string of amino acids in the order dictated by the DNA and mRNA nucleotide sequences

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10

What makes a protein mature? (L2)

Folded properly, some amino acids might be added or removed, sugars or other chemical groups may be added, and other polypeptides may be joined

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11

How might the function of a protein change if the extracellular region which binds to the ligand is mutated/deleted? (L2)

The ligand may not be able to initiate signal. Alternatively, ligand binding site could “activate” without ligand excessively.

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12

How might the function of a protein change if the anchored portion in the membrane is mutated/deleted? (L2)

The protein might not localize to the plasma membrane

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13

How might the function of a protein change if the region that binds to a signal protein inside of the cell is mutated/deleted? (L2)

Signal may either not be propagated intracellularly or there may be excess signal even without ligand.

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14

Proteins have directionality. The N terminus is made ____ by the ribosome and the C is made ____. (L2)

First (5’) and Last (3’)

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15

Proteins made by “free ribosomes” may end up: (L2)

remaining in the cytosol, to the nucleus via a nuclear pore, the mitochondria, the chloroplast, or peroxisome.

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16

Proteins made by ribosomes embedded in the rough ER may end up: (L2)

(After Golgi), a secretory vesicle to exit the cell, a lysosome, or the plasma membrane.

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17

What is the signal sequence of a growing polypeptide? (L2)

The signal sequence in many proteins consists of the first portion of the elongating polypeptide chain (N-terminus) and typically contains about 15 amino acids.W

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18

What does the signal sequence of a polypeptide chain do? (L2)

Entering the ER is dictated by the presence of a signal sequence on a growing polypeptide.

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19

How does the body switch from fetal to adult hemoglobin? (L2)

The Bcl11A protein blocks transcription of Gamma or fetal hemoglobin polypeptide subunits after birth.

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20

What theme of biology best explains why mice make pretty good models for learning about human hemoglobin and hemoglobin based diseases? (L2)

Evolution

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21

How can RNA interference be used as a genetic medicine? (L2)

RNA interference utilizes artificial small interfering RNA types to destroy or degrade certain segments of mRNA, silencing specific genes and their expression.

In the context of sickle cell, the short RNA was complementary to Bcl11A mRNA so they hybridized (bound together.) This resulted into the destruction of Bcl11A mRNA, prevented the translation of the Bcl11A protein, and resulted in the gamma gene continuing to be transcribed and translated into hemoglobin subunits. Although there was still mutant beta globin, enough hemoglobin was made with gamma that overcame the sickle defect of the red blood cells.

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22

How can small molecule drugs perform as a genetic medicine? (L2)

These drugs are easily taken up by cells and spread throughout the body, where they can help temporarily inhibit certain expression or restore proper function.

In the context of sickle cell, where mutated B-globin was aggregating due to “sticky” subunits which created the chains responsible for sickled shape, a drug could attach to the beta globins and prevent that sticking.

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23

How can gene therapy perform as a genetic medicine? (L2)

Gene therapy utilizes viral vectors to deliver and replace mutated/diseased genes with correct, theraputic ones so that they may be replicated and expressed. This works because viruses are effective at entering cells and not only delivering their genetic material, but using cellular machinery to replicate it via entering the genome.

Using the example of sickle cell, blood-forming stem cells from a patient were corrected by lentiviruses delivering a copy of corrected B-globin gene. While cells have an original mutated B-globin gene, they also have an extra, “normal” copy. These cells take up residence in the bone marrow and are able to start making healthy new red blood cells.

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24

How can CRISPR perform as a genetic medicine? (L2)

CRISPR is gene editing technology, a genetic scalpel, that can remove or edit specific sequences of DNA from the sequence.

For example, blood stem cells can be removed from a patient where they are edited with CRISPR to fix the point mutation in the DNA, and reintroduced to produce normal B-globin subunits.

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25

What are gene switches, and how do they perform as a genetic medicine? (L2)

Gene switches regulate protein expression by physically binding to proteins, to either promote or restrict transcription.

This is a more mechanical manipulation of the DNA which doesn’t involve changing the sequence at all, but rather in one study allowed researchers to just bring the enhancers necessary for transcription initiation physically closer to the gamma gene to activate them, in the DNA loop.

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26

The genetic therapy called CRISPR would affect which molecule directly? (L2)

DNA

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27

If a newly synthesized protein is found in the Golgi apparatus, where might it normally be found later? (L2)

A) Plasma membrane

B) Mitochondrion

C) Nucleus

D) Cytosol

A) Plasma membrane

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28

If the DNA sequence for the nuclear localization signal is missing, what might the most likely consequence of this change be for a nuclear protein? (L2)

A) The ribosome won’t be able to translate, so no protein will be made.

B) The mRNA will not leave the nucleus for translation, so no protein will be made.

C) The synthesized protein will remain in the cytoplasm until the cell recycles it.

D) There will be no consequence.

C) The synthesized protein will remain in the cytoplasm until the cell recycles it.

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29

What are the traits of autosomal dominant inheritance? (L3)

Rare (particularly homozygous genotypes if severe phenotype)

Male and female equally affected

Affected individuals have an affected parent

Doesn’t skip generations

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30

What are the traits for X-linked recessive inheritance? (L3)

Rare (assuming females marrying in are not carriers nor affected)

Males affected more frequently

Skips generations

Unaffected individuals have affected children

Never have father to son transmission

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31

What are the traits of autosomal recessive inheritance? (L3)

Rare (assume people entering into family are not carriers)

Males and females equally affected

Unaffected individuals have affected children

Skips generations

Becomes more common with inbreeding

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32

What is the law of segregation? (L3)

The two alleles for each character separate from each other during gamete formation.

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33

What mode of inheritance does Werner Syndrome follow? (L3)

Autosomal recessive

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34

What mode of inheritance does Huntington’s disease follow? (L3)

Autosomal dominant

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35

Jack just learned his father has HD. What is the probability that Jack will also develop Huntingtons, assuming HD is not in Jack’s mothers family? (L3)

1/2

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36

Compounding off of Jack’s probability of having HD, if he wants to start a family, what is the probability the firstborn child will eventually develop the disease? (L3)

¼ (1/2 chance Jack HD * ½ chance transmission)

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37

Describe the different definitions of dominance using the Tay Sach’s allele and the normal allele. (L3)

At the organismal level, biochemical function level, and the molecular level

At the organismal level, the normal allele is dominant and the TS allele is recessive. This is because TT and Tt individuals have no symptoms of lipid accumulation, while tt individuals do.

At the biochemical functioning level, the alleles are incompletely dominant, where one copy of the allele coding for proper enzyme is enough to prevent lipid accumulation.

At the molecular level, the alleles are codominant, meaning that heterozygotes or Tt individuals actually produce equal expression of normal and misshapen enzymes which would prevent that lipid accumulation.

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38

What is in-situ hybridization? What does it detect, how is it done? (L3)

in situ allows us to see mRNA in place. A nucleic acid probe complementary to target mRNA is fluorescently labeled and hybridizes together.

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39

What is immunohistochemistry? What does it detect, what are the basic steps? What are some positive or negative controls needed in an experiment utilizing IHC? (L3)

Immunohistochemistry allows for detection of location of proteins in tissues. The tissue is collected and cut into microneedled sections, where primary antibodies are added to bind to antigens on the tissue directly. The secondary antibodies are fluorescently dyed and bind to the primary antibodies at the sites of antigen in the tissue.

A common positive control are tissues known to express gene of interest. Common negative controls are samples from tissues known not to express gene of interest, or only the secondary antibody which should not bind to any protein in the tissue.

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40

If Alex has a gene that can be transcribed but not translated, what elements of the gene might be absent? (L4) <All that apply>

A) UTR

B) Enhancer Region

C) Promoter

D) Start and Stop Codon

A & D

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41

In a eukaryote, what is the TATA box? (L4)

The TATA box is a segment of DNA that is part of the promoter, upstream from the transcriptional start point.

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42

Why might a gene be expressed in one cell type but not another cell type, from the same organism? (L4)

The activator/transcription factors present in the cell are different, even though the DNA is the same.

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43

What stays the same between all cells regarding differential gene expression? (L4)

The DNA instructions, including the genes and their control or cis-elements are the same in every cell.

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44

How might it be possible to tell if a control element within a regulatory region is an enhancer or silencer? (L4)

Isolating removal and measuring relative levels of reporter mRNA as a percentage of a control. If an element is singularly removed and expression goes up, it was likely a silencer. If an element is removed and expression plummets, it was likely an enhancer.

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45

The difference between a specific gene being “on” in a heart cell and “off” in a liver cell would be: (L4)

A) A different combination of enhancers and activators

B) The enhancers that are present or not present

C) The activators that are present or not present

C) The activators that are present or not present

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46

If mRNA processing did not occur, what parts of a gene would be included in mRNA? (L4)

Options: Introns, Exons, 5’ cap, 3’ poly-A tail, promoter

Only introns and exons

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47

Which of the following statements is true about the location of the transcriptional start/stop sites and translational start and stop sites within a gene? (L4)

The translational stop and start sites must be between the start and stop sites for transcription because it happens after.

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48

What side of the template strand of DNA contains the promoter? (L4)

3’ side

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49

If Olivia has two mutations of a gene, one in the intron and one in the promoter region, how might it affect the transcribed mature mRNA? (L4)

The mutations of the intron would have no effect. Mutations on the promoter likely alter the level of expression rather than the type of expression.

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50

What are control elements and their function? (L4)

Control elements are non coding parts of the DNA which serve as binding sites for transcription factors which initiate and regulate transcription of a gene. This includes enhancers, which are distal control elements upstream of the promoter, and the proximal control elements, which occur just before the promoter.

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51

What is meant when we say a “general transcription factor”? (L4)

General transcription factors are essential for the transcription of all protein coding genes. They bind to the TATA box, to proteins, other transcription factors, and RNA polymerase. They are essential to the initiation of eukaryotic transcription.

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52

What is a point mutation? (L4)

A point mutation is the change in a singular nucleotide pairing in a gene.

Ex. UUG → UUU

Depending on how this changes the amino acid, the severity of mutation varies.

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53

What is a silent mutation? (L4)

Silent mutations take advantage of the wobble characteristic of codons and only change the nucleotide sequence but not the amino acid sequence. They typically have no phenotypic effect.

Ex. GGC → GGU

(Gly → Gly)

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54

What is a missense mutation? (L4)

The point mutation in a missense mutation codes for a different amino acid than in the wild type. Typically does not change length of gene but can change shape and function, depending on how different the amino acid replacement is.

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55

What is a nonsense mutation? (L4)

Nonsense mutations prematurely code for stop codons to cut transcription short and result in short and nonfunctional proteins. These are often catastrophic to the structural integrity of the polypeptide in comparison to the wild type and have severe phenotypical consequences.

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56

What are nucleotide pair insertions/deletions ? (L4)

Single or two point insertions and deletions are often disastrous on hte reading frame and are called frameshift mutations. They change the 3 codon reading pace entirely and often result in immediate nonsense or extensive missense. If 3 nucleotides are inserted or deleted this impacts the length of the protein as well but not as catastrophically as a 1-2 point deletion.

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57

Why is DNA more stable than RNA? (L5)

On the 2’ carbon, Ribose carries an OH where Deoxyribose carries an H. This makes it more chemically reactive and leads to nucleic acid fragmentation and degradation.

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58

What is the purpose of the 1’ carbon? (L5)

Nitrogenous base attachment

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59

What is important of about the 2’ carbon? (L5)

Identification (DNA vs RNA)

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60

What is important about the 3’ carbon (L5)?

Polymerases add onto the 3’ OH to build sugar backbone

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61

What is important about the 5’ carbon? (L5)

The phosphate group attaches here and allows for attachment to previous 3’ carbon of ribose.

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62

What are the properties of the phosphate group in DNA? (L5)

It is attached to the 5’ end of the pentose sugar and possesses a negative charge at physiological pH.

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63

What is gel electrophoresis? (L5)

A technique to separate DNA based on size with electric current. The initial well where sample is loaded is negatively charged, while the other end is positively charged and the negatively charged phosphate backbone is attracted towards it. Shorter strands move farther than longer strands.

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64

How many hydrogen bonds exist between A & T nitrogenous bases? (L5)

2

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65

How many hydrogen bonds exist between C & G nitrogenous bases? (L5)

3

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66

Why does the cell use an RNA primer for DNA replication? (L5)

Because DNA polymerase cannot synthesize an entirely new strand complementary to a DNA strand… it needs to continue building off of a chain.

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67

What are okazaki fragments? (L5)

Okazaki fragments are segments of the lagging strand of DNA because this strand cannot be synthesized continuously and requires a series of primers to keep directionality, which must later be removed by DNA polymerase I and fused together by DNA ligase.

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68

What are major differences between eukaryotic and prokaryotic replication? (L5)

Eukaryotic DNA is linear and has mulitple origin bubbles of replication. Prokaryotic DNA is circular and has one origin bubble.

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69

If you are designing a drug to block DNA polymerization, which carbon should you focus on to block the growing chain? (L5)

3’

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70

What is true about the nitrogenous base composition of dsDNA? (L5)

A+T = G+C

A+G = T+C

A+G = T+C

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71

Heat disrupts hydrogen bonds. What would be the consequence of heating DNA? (L5)

A) The double stranded DNA would become single stranded

B) the nucleotides would separate from each other leading to a pool of free nucleotides

C) the nitrogenous bases would separate from the DNA backbone, leaving free nitrogenous bases in solution

A) The double stranded DNA would become single stranded

This is because the only hydrogen bonds in the DNA structure are holding the nitrogenous bases of different strands together in the middle. The bonds between nucleotides in the backbone and the nitrogenous bases to the backbone are both covalent and wouldn’t be disrupted.

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72

What separates the two strands of DNA in vivo vs PCR? (L6)

Helicase separates DNA strands in a cell, while heat separates the DNA strands in PCR

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73

What enzyme elongates the new strand of DNA in vivo vs PCR? (L6)

In vivo, DNA polymerase adds onto the newly synthesized strand while Taq or some other thermophilic DNA polymerase performs this function in PCR.

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74

What are the primers made out of in vivo vs PCR? (L6)

In vivo: RNA

PCR: DNA

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75

Is Ligase present/necessary in DNA replication in vivo vs PCR? (L6)

In vivo: yes, ligase is needed to link okazaki fragments of the lagging strand.

PCR: There is no lagging strand so there is no need for ligase.

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76

Reverse primer vs Forward primers for PCR? (L6)

Forward primers have the SAME sequence as the 5’ to 3’ strand and the reverse primers have the REVERSE COMPLEMENT to the 5’ to 3’ strand.

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77

What are the steps of PCR? (L6)

Step 1: Denaturation, where the reaction is heated so dsDNA can be separated into single strands as hydrogen bonds break.

Step 2: Hybridization/Annealing, where reaction temperature is reduces to allow primer annealing.

Step 3: Extension/Elongation, where the temperature is raised again so Taq can synthesize DNA off the primer.

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78

Why use Taq polymerase for PCR? (L6)

Taq polymerase is derived from a hot springs thermophilic bacteria. It doesn’t denature at high temperatures, where DNA polymerase would.

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79

Why do we see only a short band on a PCR gel when the original strands of DNA are so long? (L6)

The original longer strands are still present, but after so many cycles of PCR you aren’t likely to see them.

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80

What happens to the DNA primers in PCR? (L6)

They are incorporated into the DNA.

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81

Below is a sequence of one strand of DNA (coding strand). The bold portion is a region you want to amplify. (L6)

5' TTGGCCGTCGGCTGCCTTCTCCTAGGAG 3'

What would be a REVERSE primer could be used to amplify the area of interest? Be sure to pay attention to 5' and 3' directionality.

Note that primers should always be written 5’ → 3’

5’ CTCCT 3’

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82

Below is a sequence of one strand of DNA (coding strand). The bold portion is a region you want to amplify. (L6)

5' TTGGCCGTCGGCTGCCTTCTCCTAGGAG 3'

Which would be a FORWARD primer could be used to amplify the area of interest? Be sure to pay attention to 5' and 3' directionality.

5’ GGCCAA3’

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83

What enzyme is used in RT-PCR? (L6)

Reverse transcriptase, which can create the cDNA from the mRNA.

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84

What are the characteristics of GOF mutations? (L7)

Assume dominant trait inheritance

not always good, can lead to inappropriate/overactivity

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85

Characteristics of LOF mutations? (L7)

Assume recessive trait inheritance

(One copy allele good enough to produce for the phenotype)

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86

Which model of conservation does DNA replication follow? (L5)

Semiconservative model

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87

How do nitrogenous bases pair regarding pyrimidine and purine structures? (L5)

Pyrimidines (One ring, Cytosine and Thymine) pair with Purines (Adenine and Guanine, two rings) to form the proper width of the helix as calculated/imaged.

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88

Is DNA a left-handed or right handed helix? (L5)

Right-handed

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89

Out of 3 PCR cycles, how many molecules are matching target sequence and are the right length? (L5)

2/8

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90

LOF or GOF: Huntington’s Disease? (L7)

Gain of function mutations because the proteins in hte brain take on new functions of clumping together to form aggregates.

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91

Describe the process of a western blot and what it detects. (L7)

Western Blots detect proteins and separate them by size, which can help identify.

SDS-PAGE → transfer to nitrocellulose → block membrane (milk wash) → primary antibody which binds to protein of interest and wash excess → secondary antibody which binds to only primary antibody and contains dye and wash excess→ develop

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92

Mutations that change the size of a gene (DNA) (L7)

Big insertions and deletions

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93

Mutations that affect mRNA expression (L7)

Changes in promoter or enhancer/repressor control elements

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94

Mutations that affect protein size and likely function or location of protein (L7)

Nucleotide insertions and deletions that lead to frameshifts and nonsense mutations.

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95

Mutations that affect protein sequence and maybe function or location of protein (L7)

Nucleotide substitutions that lead to missense and nonsense mutations.

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96

Steps of RT-PCR (L7)

  1. collect cell lysates containing the total cell mRNAs

  2. Reverse transcription reaction to create cDNA from mRNA, which contains the coding region without introns

  3. Perform conventional PCR with the cDNA, with complementary DNA primers and cycles.

    1. Examine with gel electrophoresis

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97

Which of the following would be true if I had a missense mutation in a gene? (L7)

A) The missense mutate allele would be a different size compared to wt by PCR

B) The missense mutant protein would be a different size by Western compared to wt

C) The missense mutant protein would be the same size by western at wt

D) The missense mutant allele would be the same size as wt by PCR.

C & D

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98

Which of the following would be true if I had a nonsense mutation in a gene? (L7)

A) The nonsense allele would be the same size at wt by PCR

B) The nonsense allele would be a different size compared to wt by PCR.

C) The nonsense protein would be the same size as compared to Western as wt

D_ The nonsense protein would be a different size by Western compared to wt.

A & D

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99

Which of the following elements of a eukaryotic gene would be closest to the 5' end of the non-template strand of DNA? (U1Q)

A) Promoter

B) Stop Codon

C) Start Codon

D) Exon

E) Intron

Promoter

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100

A bacterial cell has a mutation in its ligase gene, making the gene product non-functional. As a result, which of the following would you expect to occur? (U1Q)

DNA replication will be unsuccessful and E. coli will not replicate or produce daughter cells.

Ligase is necessary to bind okazaki fragments and a strand of DNA cannot be formulated without it, leading or lagging regardless.

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