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Metabolism > Final problems pt 2 > Flashcards

Final problems pt 2 Flashcards

1
Q

Citric acid cycle intermediates that are alpha-keto acids

A

Oxaloacetate and alpha-ketoglutarate

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2
Q

Role of alpha-keto acid in a transamination reaction and goal

A

The general role of an alpha-keto acid in a transamination reaction is to accept a nitrogen atom which is converted into free ammonium

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3
Q

How does low energy charge affect oxidative deamination?

A

Low energy charges activates deamination

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4
Q

What is the benefit of the reduced product?

A

the alpha-keto acid can be regenerated

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5
Q

Which enzyme in ammonium ion removal is regulated? what makes this step an ideal one to regulate?

A

Glutamate dehydrogenase is regulated because glutamate us an acceptor of the ammonium ion and can use NADH or NADPH as reducing power.

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6
Q

What happens to the carbon skeletons of amino acids after the alpha-amino group is removed?

A

The carbon skeletons of amino acids are transformed into major metabolic intermediates that can be converted into glucose or oxidized by the citric acid cycle.

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7
Q

Differences between glucogenic and ketogenic amino acids

A

Ketogenic: degraded to acetyl CoA or acetoacetyl CoA

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8
Q

Which amino acids are solely ketogenic?

A

leucine and lysine

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9
Q

Name of process that converts atmospheric nitrogen to biologically useful form?

A

nitrogen fixation

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10
Q

One type of organism that can fix nitrogen…

A

bacteria

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11
Q

Nitrogen fixation produces _____ from atmospheric nitrogen

A

NH3 (ammonium)

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12
Q

What happens to ammonium?

A

It is reduced and becomes NH4+

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13
Q

What enzyme catalyzes the addition of that nitrogen source onto a carbon skeleton

A

Glutamate dehydrogenase

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14
Q

What is the name of the carbon skeleton that serves as a substrate for glutamate dehydrogenase?

A

alpha-ketoglutarate

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15
Q

What amino acid os produced?

A

glutamine

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16
Q

Define cumulative feedback inhibition

A

Cumulative feedback inhibition is when the final products act as inhibitors and reduce the reactivity of the enzyme

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17
Q

end of case 11

A

yo

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18
Q

Two classifications of nucleic acid bases

A

DNA and RNA

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19
Q

What makes up a nucleoside?

A

A nucleoside consists of a purine or pyrimidine base linked to a sugar

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20
Q

What makes up a nucleotide?

A

A nucleotide is a phosphate ester of a nucleoside

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21
Q

Why does A not pair with C?

A

A and T share two hydrogen bonds. Adenine has 2 bonding sites. Cytosine has 3 bonding sites. The difference in the number of bonding sites prevents adenine from bonding to cytosine.

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22
Q

Why is DNA more chemically stable than RNA?

A

DNA is more stable because it lacks the hydroxyl group on the 2’ carbon. In RNA, there are two possible OH groups that the molecules can form a phosphodiester bond between, which means that RNA has a less rigid structure.

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23
Q

Why does DNA replication proceed in the 5’ to 3’ direction?***

A

DNA polymerases require a primer to begin synthesis. A primer strand having a

24
Q

Explain telomeres

A

Telomeres are located at the ends of chromosomes. They contain hundreds of tandem repeats of a hexanucleotide sequence. One of the strands is G rich (AGGGTT) at the 3’ end and is longer than the other strand, The repeat decreases the likelihood of encoding any information but facilitates the formation of large duplex loops. These strands loop back to firm a DNA duplex with another part of the repeated sequence. This loop structure is formed and stabilized by telomere-binding proteins and protects the end of the chromosome from degrading.

25
Q

Proofreading mechanism of DNA polymerase

A

When a newly added base is added incorrectly, the incorrect base will not pair correctly with the template stand and will be unlikely to be linked with the new strand. Even if an incorrect base is inserted into the new strand, it will most likely be deleted. After the addition of a new nucleotide, the DNA is pulled by one base pair into the enzyme. If an incorrect base pair is there, the enzyme stalls which provides additional time for the strand to wander into the exonuclease site.

26
Q

Polymerase chain reaction (PCR)

A

PCR is a reaction that regenerates DNA by separating the strands, hybridizing the primers, and synthesizing DNA. It is used to regenerate DNA in crime scenes to declare the person on trial guilty or not guilty. PCR can also reveal the presence of HIV in a person who has no immune response.

27
Q

Why is a C to T point mutation so common?

A

5-methylcytosine can spontaneously deaminate to generate thymine, which leads to a T-G base pair.

28
Q

T-G is recognized by which repair machinery?

A

The T in T-G pairs is repaired by base-excision repair mechanism (which recognizes distortions in the DNA double helix caused by the presence of a damaged base), but some T-G pairs can escape and are repaired as a T-A pair resulting in a C to T mutation.

29
Q

Which repair process corrects the C -T mutation?

A

base-excision repair

30
Q

Which enzyme starts the repair process for C-T mutation?

A

DNA glycoslyase

31
Q

Why is a C to U point mutation is so common?

A

Cytosine spontaneously deaminates to form uracil.

32
Q

Which base pair is recognized by repair machinery?

A

U-G

33
Q

Which repair process corrects the above mutation?

A

base-excision repair

34
Q

Which enzyme starts the repair process?

A

Uracil DNA glycolase

35
Q

Ionizing radiation can induce double-strand breaks in DNA. Explain how the DNA breaks can be repaired.

A

The repair begins with digestion at the 5’ end, which generates single stranded regions of DNA that are bound by multiple copies at RAD51. The single stranded DNA displaces one of the strands of undamaged double helixes in a process called invasion. It results in a displacement loop. DNA synthesis occurs using the undamaged helix as a template. A second strand invasion occurs to repair the damaged strands and forms two cross-like structures called Holliday junctions. The junctions are cleaved and ligated which yields two DNA double helixes.

36
Q

Why does a person with XP exhibit sensitivity to sunlight and have a predisposition to skin cancer?

A

XP is a rare human skin disorder that makes the skin sensitive to sunlight and UV light. Changes in the skin worsen with time and cause the skin to become dry and atrophy. Keratoses appear and cause skin cancer to develop. Mutations occur in genes for many different proteins which are components of the human nucleotide-excision-repair pathway. These mutations cause an increase in frequency of tumors.

37
Q

How does cisplatin work?

A

When cisplatin reacts with DNA, the chloride ligands are displaced by purine nitrogen atoms, most commonly in guanine, on two adjacent bases on the same strand. The formation of these bonds causes a kink in the structure of DNA which prevents replication and transcription. Certain nuclear proteins bind of the cisplatin-damaged DNA and prevent access to DNA-repair enzymes. The net effect of cisplatin is that the cell undergoes apoptosis (programmed cell death) which kills the cancer cell.

38
Q

Insulin from ecoli

A

The human insulin gene is built in the lab and then a plasmid is removed. The human insulin gene is inserted into the plasmid. The plasmid is returned to the bacteria and the bacteria is put into a large fermentation tank. Then, the bacteria begins producing human insulin. The insulin is harvested and purified.

39
Q

Why cDNA is better for cloning?

A

cDNA is a collection of DNA sequences representing all of the mRNA expressed by a cell. Mammalian genes contain introns and exons. These interpreted genes cannot be expressed by bacteria because it lacks the machinery to splice the introns out of the primary sequence. cDNA had no intons which is why cDNA should be used.

40
Q

Why are restriction enzymes such vital tools for recombinant DNA technology?

A

Restriction enzymes, aka restriction endonucleases, recognize specific base sequences in a double-helical DNA and cleave specific spot on the duplex. They are important because they are needed for analyzing chromosome structure, sequencing long DNA molecules, isolating genes, and creating new DNA molecules that can be cloned.

41
Q

end of case 12

A

yo

42
Q

Goal of study

A

To determine if induced pluripotent stem cells could be created from adult fibroblasts to avoid the use of embryonic cells.

43
Q

Why did Takashai and colleagues believe it would be possible to achieve their goal?

A

In 2006, the cited that they generated iPS cells from mouse embryonic fibroblasts successfully and were able to fully generate an entire mouse from stem cells. In addition, iPS can give rise to adult chimeras which were competent for germline transmission. In 1962, John Gurdon showed that nuclei from differentiated frog cells could be reprogrammed by transplantation into oocytes

44
Q

Why is iPS better than ES?

A

The use of embryonic stem cells has many ethical issues that may inhibit the use of these cells because it requires the destruction of in vitro fertilized embryos. They are also hard to make disease or patient specific. The use of iPS cells has no ethical controversies and can be made disease and patient specific.

45
Q

What type of biomolecule are transcription factors?

A

proteins

46
Q

What are two regions of DNA to which transcription factors can bind?

A

enhancers/promoter regions or RNA polymerase

47
Q

Which transcription factors were of interest to the authors of the Cell article? Incidentally, these are now known as Yamanaka factors

A

OCT3/4, SOX2. KLF4, c-MYC

48
Q

How does gene expression relate to cellular identity? In other words, why would it be useful to measure gene expression?

A

Cellular identity differentiates between two cells, so gene expression must be measured in order to determine the identity of the cell. Each cell has the same DNA. Genes encode from different proteins which have different functions which give the cells its identity. Gene expression must be measured in order to determine the function and identity of the cell.

49
Q

What were the results of the luciferase assay?

A

The luciferase assay indicated that human OCT3/4 and REX2 promoters have high levels of transcriptional activity in human iPS and ES cells but do not have high levels of transcriptional activity in HDF.

50
Q

Explain the need to include a ubiquitously expressed gene (RNA polymerase II) in the luciferase assay

A

RNA polymerase II is needed as a type of control for the luciferase assay to ensure that the gene was correctly introduced.

51
Q

Epigenetics play a crucial role in regulating gene expression. One mechanism, histone acetylation, is detailed in the textbook (Section 37.4). The authors of the Cell article tested for another mechanism, histone methylation. In the discussion section, the Takahashi study speculates about a link between the two mechanisms. List the sequence of epigenetic events that would occur if the authors are correct.

A

The Cell paper suggests that DNA methylation and histone modification a prevent binding. They could not decipher the mechanism of the four factors that induce pluripotency in somatic cells. DNA methylation controls gene expression. Histone acetylation/methylation controls transcriptional activation and deactivation. They together can determine which histones to transcriptionally activate and deactivate.

52
Q

While this research successfully achieved its goal, the authors end on a note of caution. In your own words, what point do they make clear?

A

They make clear that iPS cells are not identical to hES cells and further research must be done to determine the full functions of iPS cells.

53
Q

List the two techniques for examining gene-expression level

A

mRNA quantities/levels measured by a quantitative PCR (qPCR). Transcriptome found by a DNA microarray or gene chip.

54
Q

Briefly describe how DNA microarray works

A

Oligonucleotides or cDNAs are affixed to a solid support such as a microscope slide which creates a DNA microarray or gene chip. Fluorescently labeled cDNA is then hybridized to the chip to reveal the expression level for each gene, identifiable by its known location on the chip. The intensity of the fluorescent spot on the chip reveals the extent of the transcription of a particular gene.

55
Q

What is the advantage of DNA microarray (or gene chip) technique over quantitative PCR (qPCR)?

A

DNA microarray can investigate an entire transcriptome, the pattern and level of expression of all genes in a particular cell or tissue. PCR can only quantify a small number of transcripts in any given experiment.

Metabolism (14 decks)

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