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Bio1A03 Midterm 2 > applied lectures > Flashcards

applied lectures Flashcards

1
Q

What is byssus in muscles

A

underwater adhesive

-fibres holding muscles to rock are proteins secreted by the muscular foot of the muscle
-these proteins have keratin and resinous protein which forms byssus

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2
Q

application of muscle bysas to real life

A

genetic engineers insert segments of muscle DNA into yeast cells (cause yeast has DNA), serve as factories that translate mussel genes into bysuss

-surgeons want better ways to hold repaired body parts together (baby surgery), dentists, adhesive manufacturers

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3
Q

Kermode “Spirit” bear

A

white subspecies of the American black bear
-not albino since skin and eyes are pigmented

a single nucleotide mutation results on modified protein product from the mc1r gene

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4
Q

Kermit Ritland and coworkers

A

discovered single nucleotide mutation in the melanocortin 1 receptor gene (mc1r) - which caused a turosine-to-cysteine replacement at codon 298

mc1r gene encodes for a protein responsible for regulating skin and hair colour in mammals

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5
Q

Androgen Sensitivity (Eden Atwood)

A

mutation that changes the nucleotide sequence of a single gene, causing a single type defective protein to be produced

-person who is genotypically (46,XY). No functional androgen receptor proteins, so cells are unable to respond to testosterone

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6
Q

which of the following would be least likely to be implemented (used) during the process of translation?
a)sigma factors
b) release factors
c)initiation factors
d)elongation factors

A

a) sigma factors (used in transcription not translation)

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7
Q

describe and explain the function of the TATA box

A

it is a sequence found in the eukaryotic promoter that is recognized by transcription factors. These bind to the sequence then link the appropriate polymerase.

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8
Q

real world example of why translation is so important

A

antibiotics - used to treat bacterial infections
-many operate by disrupting translation at the site of the ribosome

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9
Q

process of translation

A

codon-anticodon pairing contributes to primary sequence of amino acids

  1. assembly: initiation factors
  2. Delivery of charged tRNAs: elongation factors
  3. Termination: release factors

driven by GTP hydrolysis

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10
Q

proteome complexity

A

1.genome (20-25000 genes)
-alternative promoters, alternative splicing mRNA editing

2.transcritome (100,000 transcripts)
-post transcriptional modification

  1. proteome (>1,000,000 proteins)
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11
Q

functional insulin protein is composed of 2 chains:

A

alpha - 21 aa’s
beta - 30 aa’s

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12
Q

translated polypeptide coded by insulin gene is _ amino acids long

A

110

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13
Q

post-tranlational modifications of insulin

A

preproinsulin (detaches signal sequence) , proinsulin (detaches chain c), insulin

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14
Q

George Beadle and Edward Tatum

A

studied relationship between genes and enzymes

-examined neurospora crassa mutants, which were created by UV or X-ray treatments

Grows on: minimal medium, able to make all the amino acids and other substances that it requires to survive

It is a support medium: combined only with molecules that are necessary for the growth of the wild-type individuals

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15
Q

Srb and Horwitz hypothesis

A

certain genes were involved in making each of the 3 types of enzymes -one gene one enzyme hypothesis

found: for a mutant to grow on either ornithine or citrulline, it must be able to make arginine from either of these compounds
-if a mutant is unable to grow with either ornithine or citrulline added to the medium (ex: arg 3), then it must have a defect in the enzyme that converts citrulline to arginine

conclusions:
-different mutants possess defects in different genes
-the metabolic pathway, which leads to the synthesis of arginine can be determined from the examination of mutants
-orthinine is a precursor to citrulline
-citrulline is a precursor to arginine

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16
Q

3 enzymes needed for extracting energy from lactose

A
  1. galactokinase (adds phosphate)
  2. gal-1-P uridyltransferase
  3. UDP-Gal epimerase
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17
Q

lactose production decreases after weaning

A

reflection of environmental factors
-lactose gene expression is high for early life /leading up to birth

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18
Q

mutations that keep the lactose gene permanently ON

A

-lactose gene does not drop as these ppl age!

mutations arose randomly in adjacent/nearby regulatory gene (MCM6) and once is arose it gave advantages of enhancing lactase gene transcription through lifetime

(low in asian/african populations, high in arab and european populations, not swedes or danes)

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19
Q

lactose non-persistance

A

as you age, can retain some lactase activity and can include varying amounts of lactose in their diets w/o experiencing any difficulties

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20
Q

lactose intolerance /congenital lactase deficiency

A

born w/o ability to express the lactase gene that codes for lactase protein

results:
-expressive lactose in intestine attracts water molecules, which prevents water from being properly absorbed into the bloodstream
-intestinal bacteria acts on lactose (via fermentation)

LCT gene on chromosome 2 codes for lactase enzyme
-LCT gene mutations in the coding region of the gene have been found to cause congenital lactase deficiency

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21
Q

galactosemia

A

meaning: galactose in blood

-individuals that are born w/o enzyme needed for galactose processing
-same symptoms as intolerance

-brought about by inherited mutations in the GALT, GALE, GALK1 and genes on chromosome 9
-mutations alter single amino acids or lead to polypeptides that are too short
-almost completely eliminate enzyme activity

results: toxic accumulation of galactose which can lead to organ and tissue damage

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22
Q

Ellie Metchnikoff

A

-hypothesis of the positive role played by certain bacteria

-Known at the time that milk fermented with lactic acid
producing bacteria inhibits the growth of proteolytic bacteria
due to low pH produced by fermentation of lactose

  • Observed populations in rural Europe that lived on fermented
    milk were exceptionally long lived

Consumption of fermented milk would “seed” the intestine with harmless lactic-acid producing bacteria and suppress growth of harmful bacteria

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23
Q

Fermented yogurt/milk as treatment of lactose intolerance

A

-Microbial B-galactosidase in yogurt is present in active forms in duodenum
-B-galactosidase enhances lactose metabolism in lactase-deficient individuals

-yogurt enzyme has protection by bacterial cell membrane which is why it is not affected by acidity of stomach

-yogurt enzyme has buffering capacity, as a result B-galactosidase is not denatured

Supplementation w/probiotics modified amount of bacteria and increased B-galactosidase activity in faeces from lactose intolerant subjects (modifies microbiome of colon)
BUT
Probiotics cannot stimulate endogenous (intestinal) lactase enzyme activity in the small intestine

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24
Q

When glucose “flatlines” on graph:

A

the “lag” phase: the period when the expression of beta-galactosidase and lactose permease is being inducedWh

no longer have glucose available, no increase in growth because bacteria strop growing - however, they do not stop dividing

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25
Q

lactose is an inducer, it turns on:

A

gene that produces B-galactosidase

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26
Q

Where can an operon be inhibited:

A

at the operator.

27
Q

genes within an operon are regulated by:

A

one single promoter and are expressed at the same time

-one promoter turns on all genes

lac operon encodes a polycistronic mRNA (mRNA molecule that contains more than one protein encoding segment)

28
Q

How is the lactose operon regulated in the presence of glucose?

A

negatively regulated in the presence of glucose and no lactose, kept in a mostly off state.

lactose removes negative regulation of lac operon .. so lactose is an inducer of this operon

29
Q

Positive regulation of lac operon (cAMP)

A

influenced by glucose levels
-involves signalling molecule called cyclic AMP (cAMP)

-cAMP is low when glucose is high
-cAMP is high when glucose is low

with the presence of lactose is the environment, CRP- cyclic AMP complex facilitates the ability of RNA polymerase to bind to the lac promoter

30
Q

Is the lac operon ever fully off?

A

no!

the repressor protein is not permanently bound, there will be periods where is dissociates from the operator allowing lac operon transcription
-when it reattaches, it can negatively regulate operon again

31
Q

Are eukaryotic genes clustered into operons?

A

no, only prokaryotic genes are

-instead each gene is regulated by its own promoter and regulatory regions

32
Q

How is eukaryotic transcription regulated?

A
  1. General transcription factors bind to the promoter, and transcriptional activator proteins bind to enhancers
  2. Through looping of DNA, transcriptional activator proteins, mediator complex, RNA Pol II, and general transcription factors are brought into close proximity, allowing transcription to proceed.

^(interruption of any of these interactions can be a means of transcriptional regulation of gene expression)

-regulatory regions are often quite far from the core promoter regions on eukaryotes

33
Q

Transcriptional repressors can halt transcription

A

Silencer: binds repressor proteins, can stop gene expression @ transcriptional level

34
Q

transcriptional regulators determine cell types

A

you can change adult (stem) cells by turning on/off genes to change gene expression

-all cells have the same DNA blueprint!
but, in some cells, certain genes are activated , in others, certain genes are kept off (makes them different)

35
Q

In situ hybridization

A

cells are alive
-can be RNA or DNA

target mRNA (5’ to 3’): ACUUACCCGGGACUA
fluorescent complementary probe (3’ to 5’) : TGAATGGGCCCTGAT

can do it to small cells and embryos, but not people. so we must use microarray analysis

36
Q

Microarray analysis procedure

A

1.take biopsy of normal breast cells and cancerous breast cells, mush them up in dishes
2.isolate mRNA
3.reverse transcription to cDNA w/fluorescent nucleotides
4. use microarray chip, each well represents a gene of interest - is pre-seeded with hundreds of DNA molecules
5.combine equal amounts of both breast cells and apply cDNA mixture to the prepared microarray chip
6.wash and measure fluorescence

37
Q

what does florecensence indicate

A

well become fluorescent when genes match the ones in well.
-empty: genes being used as controls, gene that you know is not being expressed

green: gene is more expressed in normal epithelial cells (on in normal, off in cancer)
red: gene is more expressed in breast carcinoma cells
yellow: gene is equally expressed in both cell types

38
Q

Regulating eukaryotic gene expression:

A
  1. Histone Modifications (regulating amount of mRNA that is transcribed, at DNA level)
  2. DNA methylation (regulating amount of mRNA that is transcribed, at DNA level)
  3. mRNA stability (regulating amount of mRNA that is translated into proteins, post transcriptional level)
  4. protein stability (regulating the amount of time a protein can function in a cell, post transcriptional level)
39
Q
  1. Histone Modifications
A
  1. positively charged tails of nucleosomal histone proteins probably interact with negatively charged phosphates of DNA

2.Acetylation of tails weakens their interaction w/DNA and may permit some transcription factors to bind to DNA

40
Q
  1. DNA methylation
A

happens @ promoter

-in many organisms, some cytosine bases in DNA are changed by enzymes into 5-methyl cytosines

-Heavy methylation of the CpG island inhibits transcription

Heavily methylated promoters also recruit histone deacetylase which is able to deacetylate or remove acetyl groups from the histone protein tails.. inhibiting gene expression at the transcriptional level

41
Q
  1. mRNA stability
A

-miRNAs result in inhibition of translation (blocks ribosome)

-siRNAs (cute mRNA you want to get rid of) lead to mRNA degradation

42
Q
  1. Protein stability
A

-proteasomes degrade unneeded or damaged proteins
-proteins are removed from the cell by marking them for enzymatic degradation

UB - ubiquitin , monitors age of proteins and destroys them, also destroy eachother

43
Q

Epigenetic mechanisms

A

-Modifications of histone tails (Alters wrapping of DNA around histones)

-DNA methylation (alters ability of transcription factors and RNA polymerase to interact with promoter)

Overall result: Chromatin remodeling

44
Q

cancer cells have

A
  • lower levels of methylation
    -higher levels of methylation
45
Q

Brown mice:

A

high DNA methylation
low agouti gene expression
high maternal methyl source

46
Q

yellow mice:

A

low DNA methylation
high agouti gene expression
low maternal methyl source

47
Q

Agouti gene

A

encodes protein that signals to melanocytes to switch from producing black to yellow pigment

The epigenome leads to these differences (specifically: CpG methylation)

48
Q

The term gene expression refers to
a) transcription, translation and post-translational modifications of the functional gene product.
b) the process of transcribing DNA into mRNA.
c) the post-translational modifications of a protein.
d) the transcription and subsequent translation of a protein coding gene.

A

a

49
Q

When growing bacterial cells in an environment containing only a limited amount of glucose as a nutrient source

a) bacterial growth is arrested once all the glucose is depleted.
b) bacterial switch to utilizing lactose as a source of energy.
c) there is no change in bacterial growth once glucose is depleted.
d) bacterial growth is accelerated once all the glucose is depleted.

A

a)

50
Q

_______________________ is the enzyme that can metabolize lactose in bacteria.

a) Lactose permease
b) Amylase
c) Galactose
d) Beta-galactosidase

A

d

51
Q

Both miRNA and siRNA work in conjunction with

a) RNA polymerase to bind RNA transcripts in the nucleus.
b) a protein complex called RISC to bind RNA transcripts in the cytoplasm.
c) hairpin structures to bind RNA transcripts in the cytoplasm.
d) a protein complex called RISC to bind RNA transcripts in the nucleus.

A

b)

52
Q

Small interfering RNA (siRNA) regulates gene expression by

a) inhibiting translation by binding to RNA transcripts.
b) inhibiting ribosomal RNA synthesis.
c) inhibiting RNA splicing.
d) inhibiting transcription.
e) inhibiting translation by degrading RNA transcripts.

A

e)

53
Q

DNA sites bound by regulatory transcription factors

a) are trans acting factors.
b) are promoters.
c) are cis acting factors.
d) have identical sequences in all cells.

A

c

54
Q

In general, when cytosine bases in CpG islands are methylated

a) translation is active and rapid.
b) transcription is active, but slow.
c) transcription is active and rapid.
d) transcription is repressed.

A

d

55
Q

A CpG island is a cluster of alternating C and G bases

a) found anywhere along a DNA strand.
b) found in a small region near or in the promoter site of a gene.
c) most often found near the silencer region of a gene.
d) found near bacterial operons.

A

b

56
Q
  1. When comparing the effects of regulatory transcription factors that affect TFIID and those that affect adaptor or mediator proteins

a) TFIID binding prevents transcription while mediator binding allows transcription.
b) mediator proteins allow for the binding of RNA polymerase but transcription can be prevented by transcription factors that bind to silencer regions.
c) transcription factors which bind to silencers prevent binding of both proteins.
d) transcription factors which bind to enhancers stimulate binding of both proteins.

A

b

57
Q

The human body contains approximately 200 major cell types. They look and function differently from one another because

a) each has a slightly different genome and each expresses a different set of genes.
b) each expresses the same set of genes, but in different orders at different times.
c) each expresses a different set of genes.
d) each has a slightly different genome.

A

c

58
Q

Which statement about the elongation of the lagging strand during DNA replication is correct?

a) It is synthesized in a 3’ to 5’ direction.
b) It requires a short RNA primer to proceed.
c) It progresses (grows) toward the replication fork.
d) It is synthesized by DNA ligase.
e) It is synthesized continuously.

A

b

59
Q

A new nucleotide can only be added to the ________________ of a growing DNA strand. DNA therefore always grows in the ____________________direction.

a) 3’ end; 5’ to 3’
b) 5’ end; 3’ to 5’
c) 5’ end; 5’ to 3’
d) 3’ end; 3’ to 5’

A

a

60
Q

Describe the structure of a ribosome. How does the structure of the ribosome relate to its function?

A

structure for translation to take place. ribosomes make proteins. made up of large and small subunits.

3 spots for tRNA: APE
A - bind to incoming aminoacyl tRNA, carries to a polypeptide chain
P - hold tRNA with growing polypeptide chain
E - exit, threshold without its amino acid which is then released by the ribosome

61
Q

Explain why the genetic code is redundant and unambiguous.

A

-several codons code for one amino acid
-one codon does not code for several amino acids

62
Q

significance of reading frame during translation

A

If a deletion or insertion occurs in the RNA sequence, the entire reading frame will be thrown off, which can lead to a nonsense mutation (premature stop) or missense mutations ( change in amino acid)

63
Q

Describe the steps in the process of translation. What enzymes, protein factors, and energy sources are required?

A

Initiation: brings together mRNA, a tRNA with the first amino acid, and the two ribosomal subunits

Elongation: tRNA is base paired with mRNA, another tRNA binds to A site, ribosome creates covalent bond between amino acids, mRNA is moved the length of a codon, GTP is converted to GDP and energy is used

Termination: once an stop codon is reached, a release factor enters the A-site, a water is in place of an amino acid and the polypeptide is detached

64
Q
A

Bio1A03 Midterm 2 (5 decks)

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