Test #1 Flashcards

Question

What is required for DNA replication?

Answer 1

DNA polymerase -adds nucleotides to the growing DNA strand using the existing template strand as a guide DNA template DNTPs

Answer 2

DnaA inititates -Start at the origin and proceed in both directions -Occurs in cytosol -Topoisomerase and Helicase unwind

Answer 3

-DNA Polymerase a/Primase Complex initiates -During S-phase in nucleus -Multiple replication origins

Answer 4

-there must be RNA primers -Primase produces these primers

Answer 5

Remove RNA primers with RnaseH and replace with DNA: synthesis of the lagging strand -Seal with DNA ligase -Repair Damaged DNA -t 1/10000bp errors

Answer 6

major replicase- required for synthesis -Does the majority of elongation -Leading strand -three subunits: a (the polymerase); ε (the 3'– 5' exonuclease); and θ (the stimulator of the 3'–5' exonuclease). The τ subunit is responsible for the dimerization of the core DNA polymerase.

Answer 7

-DNA polymerase-α is mainly involved in the initiation of the synthesis of the leading strand and the lagging strand -Associated with primase -DNA polymerase-α has 5’ to 3’ polymerase activity but lacks 3’ to 5’ exonuclease activity -synthesize a 7 to 10 bp RNA primer and then extend it with an additional 15 or so dNTP, which is called initiator DNA.

Answer 8

-DNA polymerase-δ is mainly involved in the lagging strand synthesis DNA polymerase-δ has 5’ to 3’ polymerase activity and 3’ to 5’ exonuclease activity, so it proofreads -does not associate with primase -requires a sliding-clamp protein called PCNA (Proliferating Cell Nuclear Antigen) and RFC (replication factor C, a clamp loader).

Answer 9

-DNA polymerase-ε is mainly involved in leading strand synthesis -has 5’ to 3’ polymerase activity and 3’ to 5’ exonuclease activity, so it proofreads -does not associate with primase and does not require PCNA

Answer 10

Terminates at specific sites which are polar and arrest replication forks

Answer 11

-Primer removed:RNase H1 & flap endonuclease-1 -Nucleosome repackaged -Telomere capping -Protection of the end of the chromosomes. The telomeres are consumed during cell division. -Separation of one chromosome from another to prevent chromosome fusion and massive genomic instability

Answer 12

-Telomeres protect the ends of chromosomes from degradation, fusion, and recognition as DNA damage -Only found in eukaryotes -Telomerase is a specialized enzyme that can elongate telomeres by adding repetitive DNA sequences to chromosome ends - telomerase is active in germ cells, stem cells, and the vast majority of cancer cells ( Not active in normal healthy adults)

Answer 13

- Prokaryotes:DNA polymerase -Eukaryotes: RNase H and other RNase

Answer 14

Plants have photolyase that can directly repair, but humans do not -Endo/Exo nucleases can remove damaged DNA in humans

Answer 15

Glycosylate glycosylase cleaves the glycosidic bond between the altered base and ribose. - In nucleotide excision repair, the entire nucleotide is removed at once. The gap formed by the incision (cut) and excision (removal) of endonucleases is usually several nucleotides wide

Answer 16

Telomerase reverse transcriptase (TERT) uses RNA as a template to elongate and maintain eukaryotic chromosome ends

Answer 17

In retrotransposition, an RNA intermediate is reverse-transcribed to insert DNA copies into other areas of the genome

Answer 18

Similarities: * Requires template * Form phosphodiester bond Differences: Transcription * Does not require primer * Lack 3’ to 5’ exonuclease activity * More errors

Answer 19

-Promoter regoin The start point for the transcription of each gene (promoter region) * Upstream of coding region * Each gene has own promoter

Answer 20

Promoter is a DNA sequence that specifies the transcription initiation site and contains consensus sequences needed for transcription

Answer 21

* DNA with promoters * A single RNA polymerase (RNAP) holoenzyme to generate all of the different types of RNA : α2ββ′ωσ * Nucleoside triphosphates (NTPs) -σ (sigma) factor binds the core enzyme and directs binding of RNA polymerase to specific promoter regions of the DNA template.

Answer 22

Found in prokaryotes Initiation requires the formation of a transcription bubble * Binding of RNA polymerase with a σ factor to the promoter region of DNA causes the two DNA strands to unwind and separate and form transcription bubble.

Answer 23

* DNA with promoters * RNA polymerase (3 RNA Polymerases in Eukaryotes) * Nucleoside triphosphates (NTPs) * Transcription factors (replace σ)

Answer 24

recognize TATA box: recruit TFIIA and TFIIB: pos and neg regulatoy function

Answer 25

* Requirements: * dsDNA (single DNA strand template) * RNAP core enzyme - α2ββ′ω * Nucleoside triphosphates (NTPs) * Topoisomerases * Acts like a gyrase (removing H-bonds) * NO Helicase (RNA Polymerase has this activity) * The elongation reactions continue until the RNA polymerase encounters a transcription termination signal. * Higher error rate but tolerable

Answer 26

* The TATA-binding protein (TBP), a component of TFIID, binds to the TATA box. * Transcription factors TFIIA and -B bind to TBP. * RNA polymerase binds, and then TFIIE, -F, and -H bind. * This complex can transcribe at a basal level * Higher error rate but tolerable

Answer 27

Prokaryotes * Requirements * Rho (ρ) dependent -p binds to C-rich regoins: contact bw Rho and RNA polymerase terminates transcription * Rho (ρ) independent (intrinsic) * Hairpin structure forms + U’s to cause release Copyright © Wolters Kluwer

Answer 28

Eukaryotes * Transcription continues beyond the end of the coding sequence * Not well-defined termination signals * Polyadenylation signal (AAUAAA) appears to be important

Answer 29

Prokaryotes * Coupled transcription-translation: mRNA – used immediately in translation * Bacterial Operon * A cistron encodes a single polypeptide chain. In bacteria, a single promoter may control transcription of an operon containing many cistrons. * A single polycistronic messenger RNA (mRNA) is transcribed. Its translation produces several polypeptide chains

Answer 30

Eukaryotes * 5’Capping * Poly-A tail * Splicing

Answer 31

* Protection of mRNAs from digestion by 5’ → 3’ exonucleases * Part of initiation to elongation switch * Enhance translocation of mature mRNA from the nucleus to the cytoplasm * Mark the 5’-end of the mRNA which is important for translation initiation

Answer 32

Function of Polyadenylation * Transcription termination * Stability * Nuclear export * Translation

Answer 33

Splicing * Excision of introns (non- expressed) * Joining of exons (expressed) * A at Intron branch site * Spliceosome: snRNAs - protein complexes * Proceeds from 5’ → 3’ end of transcript * Process is “co-transcriptional

Answer 34

Prokaryotes * Mature rRNAs and tRNAs are processed from larger transcripts *One large precursor is cleaved to produce 16S, 23S,and 5S rRNA and some tRNAs

Answer 35

Eukaryotes * Maturation of the 45S ribosomal RNA (rRNA) precursor. The clear regions are removed, and the red regions become the mature rRNAs -5S rRNA is transcribed in the nucleoplasm and moves into the nucleolus. The other rRNAs are transcribed from DNA and mature in the nucleolus, forming the 40S and 60S ribosomal subunits, which migrate to the cytoplasm

Answer 36

* 4 major steps before transport to cytoplasm * Cut 5’ and 3’ end * Remove introns to form anticodon * Nucleotide base Modification * Add CCA– (A) is attached to amino acid * 2 binding sites * Anticodon:recognizes 3 nucleotides and ensures genetic code is translated into correct sequence * 3’end: (A) wheramino acid is attached

Answer 37

In eukaryotic cells, the process of transcription, which occurs in the nucleus, is separated by the nuclear envelope from the process of translation (protein synthesis from the mRNA template), which occurs in the cytoplasm. Because prokaryotes lack nuclei, the processes of transcription and translation occur simultaneously. * Transcription of bacterial DNA requires only one promoter per operon. In contrast, human DNA requires one promoter for each gene. * Eukaryotes have more elaborate mechanisms for processing the transcripts (pre- mRNA). * Eukaryotes have three polymerases rather than just the one present in prokaryotes. * Generally, the transcripts in prokaryotes do not have introns. * Coupled transcription-translation does not occur in eukaryotes

Answer 38

The third nucleotide in the codon-anticodon interaction can form non-standard or "wobble" base pairs. For instance, the third position of the codon can pair with more than one nucleotide at the corresponding position in the anticodon. This flexibility allows for fewer tRNA molecules to recognize all codons coding for a particular amino acid.

Answer 39

-Prior to attachment to tRNA, the amino acid is activated through the hydrolysis of ATP to AMP and pyrophosphate. This reaction activates the amino acid, forming an aminoacyl-AMP intermediate -he activated amino acid is then transferred to the 3' end of the corresponding tRNA molecule -ensures the accurate and efficient translation of genetic information from mRNA into proteins during protein synthesis

Answer 40

-mRNA that is not capped -30S and 50S ribosomal subunit -IF1,IF2,IF3 -Charged initiator tRNA (fMET)

Answer 41

-mRNA with start codon -5' cap - Eif3(helps keep subunits separated until ready to bind) and Eif4(helps find 5' cap) -40S and 60S ribosomal subunits -Charged tRNA

Answer 42

* 70S Ribosome Complex * Aminoacyl-tRNAs * Elongation Factors o E F -Tu – d e l i v e r s A A - t R N A s ; G T P → G D P + P i ( C H A R G I N G ) o E F -Ts – r e g e n e r a t e s E F -Tu * G T P o E F - G – t r a n s l o c a t i o n ; G T P → G D P + P i * GTP

Answer 43

Eukar yotes * 80S Ribosome Complex * Aminoacyl-tRNAs * Elongation Factors o e E F - 1 A - d e l i v e r s A A - t R N A s ; G T P → G D P + P i ( C H A R G I N G ) o e E F - 1 B - r e g e n e r a t e s e E F - 1 A * G T P o e E F - 2 – t r a n s l o c a t i o n ; G T P → G D P + P i * GTP

Answer 44

Step 1 – decoding: binding of the AA- tRNA dictated by the next codon to the A site Step 2 – transpeptidation: formation of the peptide (amide) bond (P to A site) - Peptidyltransferase is not a protein but t h e r R N A o f t h e l a r g e r i b o s o m a l s u b u n i t Step 3 – translocation (A to P site) - A>P>E site

Answer 45

Stop codons: UAA, UAG, or UGA * Release Factors bind to the stop codons * Peptidyl transferase hydrolyzes the bond between the peptide chain and tRNA. * The newly synthesized polypeptide is released from the ribosome. * Ribosome dissociates into its individual subunits, releasing the mRNA

Answer 46

Prokaryotes * Major point of regulation is at initiation * Shine-Dalgarno consensus sequence * E u k a r y o t e s * m R N A e x p o r t – requires , at a minimum , a 5 ʼ- Cap and a 3 ʼ- poly( A ) - tail :energy-requiring process * mRNA stability: length of poly A tail important - Negative translation control: binding proteins at 5' end inhibit initiation -Initiation factor phosphorylation: Dec protein synthesis

Answer 47

Chaperones: proteins that bind to the nascent polypeptide and mediate the folding process (protect against improper interaction) -assist in the folding and assembly of newly synthesized polypeptide chains into their correct three-dimensional structures Trigger Factor: ribosome-associated chaperone known in bacteria and binds short hydrophobic protein segments

Answer 48

-specific peptide bonds within a protein are cleaved after the protein has been synthesized. This process can generate mature, functional proteins from inactive precursors or proproteins. -Removal of fMet in prokaryotes

Answer 49

DNA Replication is semi-conservative DNA Replication is semi-discontinuous DNA Replication is bi-directional

Answer 50

TATA & Prinbow box are similar in stature/nature

Answer 51

Post-Translational

Answer 52

RNA polymerase does not require a primer to initiate transcription(RNA polymerase binds to a specific region of DNA called the promoter and unwinds the DNA double helix. Then, it synthesizes an RNA molecule complementary to one of the DNA strands, using the other DNA strand as a template. This process initiates at the promoter sequence without the need for a primer.) unlike DNA polymerase, which requires a primer for DNA replication

Answer 53

RNA polymerase lacks 3'->5' exonuclease activity

Answer 54

Operons -units of prokaryotic DNA that consist of a cluster of genes under the control of a single promoter and regulatory sequences -Single promoter, many genes -Polycistronic mRNA

Answer 55

-Regulation of operons by repressors When the repressor protein is bound to the operator RNA polymerase cannot bind, and transcription does not occur. -the inducer binds to the repressor, inactivating it

Answer 56

The lac promoter is weak It requires cyclic AMP (cAMP) and Catabolite Gene Activator Protein (CAP) to Stimulates transcription of the lac operon CAP*cAMP binds 5’ of the promoter (activator site), helps RNA polymerase bind to the promoter and form the closed complex Glucose sensing: glucose ↓, cAMP ↑ (+) Control

Answer 57

Tryptophan is a corepressor that binds to the inactive repressor, causing it to change conformation and bind to the operator Inhibits transcription of the operon

Answer 58

Chromatin-remodeling complex Requires energy from ATP hydrolysis -modification of chromatin structure to regulate access to DNA and control gene -Chromatin remodeling involves altering the position, composition, or organization of nucleosomes expression -Increase/maintain positive charge- methylation -Decrease positive charge- acetylation -Increase negative charge- phosphorylation

Answer 59

RNA polymerase II and six general (basal) transcription factors act cooperatively with each other and form the pre-initiation complex (PIC)

Answer 60

Alternative splicing is a process in eukaryotic gene expression whereby a single pre-mRNA can be spliced in multiple ways to produce different mRNA transcripts, leading to the synthesis of distinct protein isoforms from a single gene. Poly A sites specific sequences in eukaryotic pre-mRNAs where a stretch of adenosine nucleotides (poly-A tail) is added to the 3' end of the RNA molecule during mRNA processing

Answer 61

microRNAs (miRNAs) are small RNA molecules that regulate protein expression at a posttranscriptional level. An miRNA can either induce the degradation of a target mRNA or block translation of the target mRNA

Answer 62

Allele: one of two or more versions of a gene. An individual inherits two alleles for each gene, one from each parent. If the two alleles are the same, the individual is homozygous for that gene. If the alleles are different, the individual is heterozygous.

Answer 63

Non-coding region: includes introns, regulatory sequences, and other intergenic DNA -Promoter- sequence the directs of regulates gene transcription -Enhancer- short DNA sequences that impacts transcription of genes by binding protein factors -UTR (untranslated region)- non-coding transcribed region that affects mRNA stability

Answer 64

INDELs: insertion and/or deletion of nucleotides into genomic DNA - Can lead to reading frameshift

Answer 65

Polymorphisms occur >1% of the time Mutations occur <1% of the time

Answer 66

Somatic Mutation: Genetic alteration acquired by a cell that can be passed to the progeny of the mutated cell in the course of cell division. -Later somatic mutations have less effect than early developed Germline Mutation: A germline mutation is any detectable and heritable variation in the lineage of germ cells. Mutations in these cells are transmitted to offspring

Answer 67

variation in the relative frequency of different genotypes in a small population, owing to the chance disappearance of particular genes as individuals die or do not reproduce

Answer 68

the reduced genetic diversity which results when a population is descended from a small number of colonizing ancestors

Answer 69

non-random association of alleles at different loci in a given population

Answer 70

process by which new advantageous genetic variants sweep a population

Answer 71

Answer for blank # 1: Active Answer for blank # 2: Secondary

Answer 72

CBR1 CYP2E1 CYP3A4

Answer 73

“>” represents a DNA base pair substitution Changes in amino acid (AA) sequence is denoted by the original AA followed by the position and the AA found in the mutant A “p” is placed in front to indicate protein sequence *1 denotes the default reference allele (wild type or fully functional) that is first identified. A unique number (*#) is assigned when a novel variant is identified-a variant with no functional impact “xN”,- Number of copies of gene

Answer 74

Haplotype: Combination of a set of alleles on a chromosome

Answer 75

Tag SNPs: a subset of SNPs that can be used to determine a haplotype Knowing the 3 SNPs is sufficient to determine the haplotype (rather than sequencing it all)