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Bio Mocks > Macromolecules 2 > Flashcards

Macromolecules 2 Flashcards

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Biology 101 flashcards
1
Q

Competitive/non-competitive and reversible/irreversible inhibition examples

A

reversible competitive (methanol, dehydrogenase (formic acid, formaldehyde), ethanol)
reversible non-competitive (fructose-6-phosphate, phosphofructokinase, xylitol-5-phosphate)
irreversible competitive (peptidoglycan (bacteria cell wall), transpeptidase which catalyzes peptidoglycan production, penicillin)
irreversible non-competitive (heavy metals like Hg and Pb)

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2
Q

Name the monomer of nucleic acids and draw and annotate its structure

A

nucleotide (deoxyribonucleotide or ribonucleotide), …, pentose sugar (deoxyribose, ribose), a phosphate group, and one nitrogenous (N) base (adenine (A), guanine (G), cytosine (C) and thymine (T) in DNA or uracil (U) in RNA

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3
Q

The difference in nitrogenous bases’ structures

A

adenine and guanine are purines (two rings), and cytosine, thymine, and uracil pyrimidines (one ring)

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4
Q

Full names of DNA nucleotides (for each N-base)

A

deoxyriboadenosine, deoxyriboguanosine, deoxyribocytosine, and deoxyribothymosine

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5
Q

Draw and annotate the dehydration reaction between two nucleotides, and name the bond between them

A

…, phosphodiester bond (3’ to 5’ linkage) – form a polynucleotide chain

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6
Q

What is the shape of two polynucleotide strands, in what relation are the two strand

A

coiled around each other in a double helix, they are antiparallel which means that they run alongside each other but in opposite directions (each is oriented in the direction from 5’ to 3’) – the two connected strands need to have complementary N-base pairs (A-T, C-G) – A and T form two H-bonds, and C and G three

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7
Q

Three types of RNA

A

messenger, transfer and ribosomal

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8
Q

Describe the Hershey-Chase experiment – phage (radioactive S (protein)/P (DNA)) infects bacteria

A

only phages with radioactive DNA produced radioactive E. coli (which was turned into a phage-producing factory) – reflected as a radioactive pellet (bacteria are denser so they fall as a pellet (instead of supernatant) after being separated from phages in a centrifuge). It was hypothesized that proteins carry genes because they have more capacity for variation 20^n vs 4^n, but DNA can be any length, unlike proteins which adds to the potential diversity of sequences

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9
Q

DNA replication, semiconservative meaning

A

the process of cloning the DNA to produce two identical copies which happens prior to cell division so that each daughter cell receives one copy of the original DNA – newly synthesized DNA is made of one old and one new polynucleotide chain

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10
Q

Summarize the process of DNA replication

A

DNA helicase attaches to one site on the parent DNA and unwinds and unzips the two strands. It creates a replication fork (where the DNA is opened) – primers (ribonucleoside triphosphates) get added to both template strands to start the polymerization process – there is only one on the leading and multiple on the lagging strand (creating Okazaki fragments) – then free DNA nucleotides (deoxyribonucleoside triphosphates) bases get linked with their complementary bases on the template strand by the DNA polymerase III (forming phosphodiester bonds) – DNA synthesis may occur bi-directionally from the origin – primers are later removed by DNA polymerase I, and enzyme ligase seals the gaps in the chains – gyrase (topoisomerase) moves in advance of helicase and coils the DNA strand to relieve the pressure made by supercoiling of the DNA (because of helicase’s unwinding) which could cause a block in replication – SSB (single-strand binding) proteins attach to template strands to prevent their spontaneous rejoining

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11
Q

Origin of replication in prokaryotic vs eukaryotic DNA

A

there is more than one origin of replication on eukaryotic DNA because eukaryotic chromosomes are much larger and more numerous than prokaryotic – speeding up the process

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12
Q

How are deoxyribonucleoside triphosphates different from bound nucleotides

A

have three phosphate groups: ATP (3) -> ADP (2) -> AMP (1) – as the phosphodiester bond is formed, the two phosphate groups are removed from the free nucleotides, providing E for the bonding

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13
Q

What is the limitation of DNA polymerase III

A

its specificity; it can only add new nucleotides onto a preexisting chain of nucleotides so it cannot initiate polymerization

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14
Q

What is the central dogma of molecular biology, are there any exceptions to it

A

the process of gene expression, retroviruses (their RNA gets turned into DNA and then into mRNA) – reverse transcription

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15
Q

Gene

A

a part of the DNA with a specific base sequence that codes for the a-a sequence of one polypeptide chain

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16
Q

Describe the transcription process

A

only one (anti-sense) strand is used for transcription, the one complementary to the one carrying genetic info – initiation: RNA polymerase binds to the promotor region (short DNA seq. before the gene), unzips the gene and lets ribonucleoside triphosphates (ATP, GTP, CTP, UTP) pair up with the anti-sense strand free nitrogenous bases – elongation: RNA polymerase forms bonds between ribonucleoside triphosphates (in 5’ to 3’ direction) – termination: RNA polymerase reaches the terminator region (end of the gene) and the polymerization is stopped (mRNA is complete, it gets released from the template strand and travels to the cytoplasm (ribosomes))

17
Q

Which two processes in gene expression can occur simultaneously?

A

DNA replication and transcription

18
Q

Repair mechanisms purpose

A

maintaining the stability of DNA templates during transcription when the sense strand is more exposed and subjected to the impact of mutagens – they minimize mutations but deteriorate as we age so mutations accumulate and cause diseases

19
Q

Describe post-transcriptional modification

A

introns (the non-coding regions of the DNA) get spliced out of the pre mRNA by spliceosome because they should not get translated (only in eukaryotic because prokaryotic only have exons – transcription and translation occur almost simultaneously) – also, poly-A-tail gets added to the 3’ end of pre mRNA because it aids in the translation, it transports mRNA to the cytoplasm and protects it from enzymatic digestion in the cytoplasm

20
Q

What is mRNA organized in

A

in codons, that is, groups of three nitrogenous bases that code for one amino acid – some codons signal for the end of translation (stop codons)

21
Q

Genetic code function

A

enables cells to convert base sequences into amino acids – it is universal (used by all species) and degenerate which means that more than one codon will code for the same a-a (reduces the effect of base pair substitution mutations)

22
Q

Ribosome structure

A

proteins and ribosomal RNA – composed of two subunits, one big and one small which are usually separated but join during translation – there are three tRNA binding sites (E-exit, P-peptidyl, A-acceptor)

23
Q

tRNA function, structure

A

transports amino acids towards ribosomes and attaches to specific binding sites on the ribosome – made out of base pairs and has a shape like a clover as a result of attraction forces between complementary bases – in its middle, there is an anticodon which is complementary to the mRNA codon (they later pair up) – the 3’ end is the a-a attachment site where the a-a specific for that tRNA joins to be transported to a ribosome

24
Q

Describe the translation process

A

polypeptide synthesis using mRNA as a guide – initiation: mRNA binds to the mRNA binding site on the small subunit, initiator t RNA binds to the start codon, big subunit joins and initiator fits into the P-site, the next t RNA bind onto A-site and a peptide bond is formed between the two a-a – the first a-a detaches (dipeptide carried by the one in P) – elongation: ribosome moves for one codon towards 3’ of m RNA (P->E), new t RNA attaches to A site and the chain grows – termination: when a stop codon attaches to the A-site and the release factor (protein) is added instead of an a-a – it releases the polypeptide chain and ribosome disassembles

25
Q

Prokaryotic vs eukaryotic DNA organization

A

location (nucleoid vs nucleus, mitochondrion, chloroplast), number and shape of molecules (one circular vs two and more linear), types of sequences (mostly coding vs mostly non-coding), association with histones (no vs yes)

26
Q

Histones

A

globular proteins that control the gene expression by controlling the level of DNA condensation – DNA supercoils around them and they form chromosomes which prevents entangling of DNA (and mutations) – genes in condensed DNA are not approachable to transcription enzymes

27
Q

Nucleosome structure

A

eight histones and two turns of DNA around them – two nucleosomes are linked by linker DNA – N-terminal histone tails protrude from each histone

28
Q

N-terminal histone tail acetylation (explain the effect)

A

without the acetyl group, histone tails attach and pull nucleosomes closer (increasing condensation) – without the Ac group, the possibility of gene expression is increased

29
Q

What are the types of non-coding sequences

A

1| regulatory (promotors and terminators, they don’t get transcribed)
2| introns (non-coding sequences within a gene, they get transcribed and then spliced out)
3| genes for tRNA and rRNA (get transcribed and then spliced out)
4| short tandem repeat sequences or telomerases (unique for each organism, basis for DNA recognition, it can afford to mutate excessively because it doesn’t code for anything)

30
Q

Role of STR in DN replication

A

they are mostly found in centromeres and telomeres – after each DNA replication, the terminal part of the DNA gets lost (DNA polymerase cannot add on the 5’ end) – once STR is used up, the DNA can no longer replicate, otherwise it would lose genes (the cell can only die) – enzyme telomerase can recover telomerases (cancer cells)

31
Q

Alternative splicing

A

one gene can encode for more than one protein

32
Q

Epigenetics

A

a process that influences gene expression or function of genes without changing the DNA sequence, using chemical markers/environment – epigenome is the set of small chemical markers (methyl, acetyl group) that are associated with the DNA – e.g. histone tail acetylation and DNA methylation

33
Q

How do acetylation and methylation affect gene expression

A

acetylation promotes gene transcription and methylation inhibits it

34
Q

Explain the difference between genotype and phenotype

A

all cells of the same organism have the same genotype but a different phenotype (some genes switched on/off) – gene expression is the process that turns a genotype into a phenotype (has three stages: transcription, translation, and protein function)

35
Q

Gene expression can be controlled:

A

at the level of transcription, by controlling the mRNA lifespan, or by external factors

36
Q

Controlling gene expression at the level of transcription

A

TATA box is a part of the (eukaryotic) promotor region where DNA binding protein specifically bind, indicating to RNA polymerase where transcription should start – promotors are similar for all genes – upstream from the promotor lie two regulatory sequences, silencer, and enhancer (unique for each gene) – they are the binding sites for specific transcription factors (activators or repressors of transcription)

37
Q

Controlling gene expression by controlling the mRNA lifespan

A

poly-A-tail protects the mRNA so the same one can be transcribed multiple times (it shortens with time – the rate of poly-A-tail shortening determines how long mRNA persists and various factors can impact that variable – e.g. in egg-laying animals, protein vitellogenin is a precursor for egg yolk, high estrogen levels increase the longevity of mRNA that encodes for vitellogenin and results in a high-quality egg (more yolk)

38
Q

Controlling gene expression by external factors

A

E. coli can utilize lactose from the medium but needs enzyme lactase – synthesis of lactase is inducable and is regulated by the presence/absence of lactose (it will only be produced if there is a need for it, that is if lactose is present). Gene regulator produces the repressor protein and it is constantly getting transcribed. If lactose is present, lactose molecules will join with the repressor protein, and change its conformation so that it can no longer join with the operator – nothing is preventing the mRNA from transcription of the lactase gene. After lactase digests all lactose, it can no longer be produced – the cell doesn’t want to waste E by producing an enzyme when there is no substrate specific to it present

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