Unit 1 review Flashcards

(51 cards)

1
Q

Draw out nitrogenous bases

A

Picture

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2
Q

Which nitrogenous bases are pyrimidines and purines

A

Pyrimidines: Cytosine, Thymine, Uracil
Purine: Guanine, Adenine

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3
Q

What is the name of the bond that holds nucleotides together in DNA and RNA.

A

Phosophodiester bonds between C and N terminus.
Hydrogen bonds between nitrogenous bases

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4
Q

What is the secondary structure of ds DNA?

A

Double Helix

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5
Q

What is the secondary structure of ssRNA and ssDNA?

A

Hair pin

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6
Q

What is DNA methylation. Why is it important?

A

DNA methylation is the addition of a methyl groups (CH3) to different nucleotides. It is important because it acts as an epigenetic mechanism to regulate gene expression by chemically modifying DNA without altering the underlying DNA sequence.

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7
Q

What would negative or positive supercoiling do in a bacterial chromosome?

A

Negative supercoiling: molecules are underwound: promotes process of DNA replication- makes DNA more accessible
Positive supercoiling: molecules are overrotated, hinders process of DNA replication- makes DNA less accessible

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8
Q

What is the difference between eukaryotic heterochromatin and euchromatin?

A

Heterochromatin: tightly packed, condensed form of chromatin with low transcriptional activity
Euchromatin: loosely packed form of chromatin with high transcriptional acitivity

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9
Q

Why does DNA wrap so well around nucleosomes?

A

Due to large positive charge

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10
Q

What is epigenetics?

A

Epigenetics are stable alterations to the chromatin structure that can be passed on to future generations, that do not alter the DNA sequence

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11
Q

What are telomeres? How are they replicated?

A

Telomeres: ends of chromosomes- rich in G repeats
Replication: During DNA replication, the lagging strands os synthesized in short okazaki fragments, requing RNA primers. When the final primer at the end is removed DNA polymerase can’t fill the gap due to there not being a pre-existing 3”-OH group. This leads to the progessive shortening of chromosome ends.

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12
Q

What is the evolutionary origin of chloroplasts and mitochondria? How can sequence comparisons support the evolutionary origin theory?

A

Evolutionary Origin of Chloroplasts: eukaryote engulfed cyanobacteria and formed an endosymbiotic relationship.
Evolutionary Origin of Mitochondria: eukaryote. engulfed proto-bacterium and formed an endosymbiotic relationship
Evidence:
- mitochondria and chloroplasts DNA resembles bacterial genome sequences (circular DNA)
- chloroplasts have double membrane ( supports engulfment event)
- replicate independently (binary fission)

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13
Q

What are the main 7 enzymes and proteins involved in DNA replication and what do they do. Draw on replication fork

A
  1. Helicase: unwinds double stranded DNA by breakung hydrogen bonds between pairs
  2. SS-Binding Proteins: Bind to single stranded DNA to prevent them from rewinding
  3. Topoisomerase (Gyrase in Prokaryotes): relives supercoiling ahead of replication fork
  4. Primase: synthesize short RNA primers to provide starting point for replication
  5. DNA Polymerase III: main enzyme- adds nuclotides to 3’ end of DNA
  6. DNA Polymerase I: Removes RNA primers and replaces them with DNA
  7. Ligase: links okazaki fragments
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14
Q

Differentiate between the leading and lagging strands in a replication fork.

A

Leading: 5’-3’ no breaks (top strand)
Lagging: 5’-3’ with breaks (okazaki fragments) (bottom strand)

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15
Q

What is an Okazaki fragment? What differentiates it from all other nucleotide strands?

A

Okazaki fragments: short segments of newly synthesized DNA on the lagging strand. Made in short bursts due to the anti-parallel nature of DNA
Differentiates:
- short length
- each okazaki fragments begin with RNA primers
- require DNA ligase for joining

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16
Q

Explain the problem of replicating the ends of linear chromosomes. How do eukaryotes and prokaryotes solve the problem?

A

Problem: progressive shortening of chromosomes
Solution: Telomerase: recognizes tip of repeate sequences- telomerase elongates the strand in the 5’-3’ direction and adds additional repeats. The overhang then has a primer added on the DNA polymerase can fill in the complementary strand

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17
Q

Recognize and describe the differences between
- DNA and RNA,
- DNA replication and RNA transcription,
- and DNA polymerases and RNA polymerase.

A

DNA
-deoxyribose (No -OH on C2 carbon)
- Bases: T:A G:C
- double stranded
- only one type of DNA
- only used to store genetic material
- only in nucleus
- DNA lasts long
- Self replicating
RNA
- Ribose (OH on C2 carbon)
- Bases: T:A G:U
- single stranded
- lots of types of RNA
- used for protein synthesis, gene regulation, and catalyze reactions
- RNA degrades easily
synthesized when needed

DNA Replication
- creates a copy of DNA for cell division
- uses DNA polyermase
- template is both strands of DNA
- produces two identical DNA strands
RNA Transcription
- creates RNA for protein synthesis or regulation
- uses RNA polymerase
- template is only on strand of DNA
- produces tRNA, mRNA, and rRNA

DNA Polymerase
- Function: synthesize new DNA during replication
- template: DNA
- Primer: needs RNA
RNA Polymerase
- Function: synthesizes new RNA during transciption
- template: DNA template
- Primer: doesn’t need primer

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18
Q

Draw a typical gene in prokaryotes with its regulatory elements.

A

Drawing

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19
Q

What is a 5’ or 3’ untranslated region in a prokaryotic gene?

A

its the noncoding region

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20
Q

Given a gene determine which strand is the coding strand and which is the template strand.

A

Coding strand (sense strand): This DNA strand has the same sequence as the mRNA, except it has T instead of U. It is not used directly in transcription.

Template strand (antisense strand): This is the strand that RNA polymerase reads to synthesize mRNA. It is complementary to the mRNA.

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21
Q

Provide an RNA sequence from a DNA template strand.
5’ GACT 3’

A

3’ CUGA 5’

22
Q

Be able to translate an RNA open reading frame with a genetic code. (Give strategy)

A

Non Coding region- Start Codon AUG- Translated Region- Stop codon- None Coding region

23
Q

Differentiate between rho independent and rho dependent transcriptional termination in prokaryotes.

A

Rho Independent terminators: are able to cause the termination only in the presence of an rho factor

Rho Dependent terminatorsL are able to cause the end of transcription in the absence of a rho factor

24
Q

How is transcription and RNA modification different in eukaryotes and prokaryotes.

A

Eukaryotes
- Location: Transcription (Nucleus), Translation (Cytoplasm) Happens separately
- Polymerase: Pol 1, 2, 3
- Modifications: 5’ cap and polyadenylation
- Termination: Polyadenylation signal, and specific termination factors.
Prokaryotes
- Location: Both Transcription and Translation (Cytoplasm) Happens simultaneously
- Polymerase: Pol 1
- Modifications: none
- Terminators: Rho-independent, and dependent specific termination factors

25
Describe the role of the spliceosome.
Splice out introns
26
Describe what differential processing does to the primary structure of resulting proteins.
Differential Processing: the process where a single gene can produce multiple different protein variants by selectively splicing its mRNA transcript
27
How is the Start codon recognized in prokaryotes and eukaryotes?
- Shine del garno sequence- positions ribosome relative to start codon - Start codon - Initiator tRNA carries fMet
28
What forms the peptide bond?
peptide bond: carbonyl group and amino group undergo dehydration reaction
29
Differentiate between the A P and E sites in a ribosome.
A (Aminoacetyl site): where new amino acid, carrying tRNA enters ACCEPTS NEW AMINO ACID P(Peptidyl site): holds tRNA with growing polypeptide chain BUILDS POLYPEPTIDE CHAIN E(exit site): where tRNA exits after releasing its amino acid RELEASES SPENT TRNA
30
Describe the process of translation elongation in prokaryotes.
DRAW 1. IF3 binds SSU 2. GTP + IF1 (goes to SSU)+ IF2 (works with GTP and tRNA) + tRNA charged with fMet 3. IF1, IF2, IF3, GDP, Pi, released LSU binds
31
What is release factor? What does it do in translation?
Release factor: terminates translation by recognizing stop codons and telling the ribosome to release the newly formed polypeptide
32
For prokaryotes, what is DNA supercoiling (positive and negative) and why is it important?
Positive Supercoiling- molecules that are overrotated- hinders transcription by making DNA less accessible Negative Supercoiling- molecules that are under rotated- activate transcription by making DNA more accessible
33
Appreciate the importance of genome comparisons for understanding and asking questions about sequence evolution.
Conserved sequences (stayed the same) — likely essential for survival. Divergent sequences (changed over time) — suggest adaptations or evolutionary changes. For example, comparing human and chimpanzee genomes shows how closely related we are and highlights specific changes that led to differences.
34
What is chromatin, and what is it made of? What is the difference between heterochromatin and euchromatin?
Chromatin: the material from which the chromosomes of an organism are composed of- a complex made of DNA and proteins (histones are most abundant) Heterochromatin: highly condensed, characterized by a lack of transcription Euchromatin: un-condensed form (loose), where most transcription takes place
35
Why and when is it important for chromatin to be “open”. How can this state be detected experimentally?
“Open” chromatin refers to a relaxed, less compacted state of chromatin where DNA is more accessible to proteins like transcription factors, RNA polymerase, and other regulatory machinery. Detected Experimentally: DNA Methylation Analysis - Which genes are turned on or off in different cell types or conditions. - Epigenetic differences between healthy and diseased tissue (e.g., cancer). - How development, aging, or environmental exposures affect gene regulation.
36
Explain why DNA tends to wrap tightly around nucleosomes, and how this tendency can be relaxed.
DNA tends to wrap tightly around nucleosomes due to the large positive charge of histones. This tendency ca be relaxed through 1. histone acetylation (neutralizes the negative charge of DNA) 2. Chromatin remodeling complex (slides nucleosomes to promote gene expression 3. Histone methylation (activates chromatin transcription)
37
Why do eukaryotic cells have telomeres?
They act like "caps," preventing the ends of chromosomes from being recognized as broken DNA by DNA repair systems. This prevents them from being mistakenly joined together or undergoing degradation. Telomeres allow for the complete replication of chromosome ends, which would otherwise be lost during the replication process. This is facilitated by the enzyme telomerase, which adds repetitive DNA sequences to the telomere ends, maintaining their length.
38
Why and how are pedigrees used when diagnosing genetic disorders?
Pedigrees are used in diagnosing genetic disorders to track the inheritance of traits and identify patterns that suggest whether a condition is genetic and how it is passed down through families.
39
define semi-conservative replication and make a cartoon drawing illustrating the concept
Semi-conservative replication: the mode by which DNA replicates- each strand acts as a template for a new double helix
40
interpret Meselson-Stahl’s experimental results that test the semi-conservative replication hypothesis, and show how the results would have been different if it was conservative replication instead
The Meselson-Stahl experiment (1958) was a landmark study that demonstrated DNA replicates semi-conservatively—meaning each daughter DNA molecule consists of one old (parental) strand and one new strand. Experiment: Labeling DNA with Heavy Nitrogen (¹⁵N): E. coli were grown in ¹⁵N (heavy nitrogen) medium so their DNA was "heavy". Switching to Light Nitrogen (¹⁴N): Cells were then moved to ¹⁴N (light nitrogen) medium and allowed to replicate. DNA Density Separation: DNA was extracted after each round of replication and separated by density-gradient centrifugation (heavier DNA migrates lower in the gradient). Meselson-Stahl's data strongly supported the semi-conservative model of DNA replication. Each new DNA molecule keeps one old strand and synthesizes one new strand. The conservative model was ruled out by the absence of a heavy band after one generation.
41
List the three constraints that DNA Polymerase has when synthesizing new DNA: - direction of synthesis - can only extend an existing strand - needs a single stand (& 3’ end) template
1. DNA polymerase can only synthesize DNA in the 5′ → 3′ direction. This means it adds nucleotides to the 3′ end of the growing DNA strand. 2. DNA polymerase cannot start synthesis from scratch. It requires a pre-existing short strand of nucleotides (called a primer) with a free 3′ OH group to add new nucleotides to. 3. DNA polymerase needs a single-stranded DNA template to read and copy. The enzyme uses this strand to determine which complementary nucleotide to add next.
42
draw a polymer of nucleotides and show how a nucleotide triphosphate is then attached to the polymer, all at atomic detail as we saw in class (ie, bases not in atomic detail)
Drawing
43
show, by drawing, how a replication bubble forms at an origin of replication, and show which parts of the bubble are forks, and show which direction the forks travel on the drawing, show leading and lagging strands, continuous and discontinuous replication, and show the order in which the Okazaki fragments are formed
Drawing
44
describe--with drawings and text--the processes by which Okazaki fragments are repaired/resolved to remove RNA primers fill in the gaps between fragments form the final phosphodiester bond between the 3’OH of the last nucleotide of one fragment and the 5’phosphate of the next nucleotide
1. RNA primer removal by DNA Pol I- leaves a gap 2. Gap filling by DNA Pol I- adds nucleotides to fill gap 3. Sealing the Gap by DNA ligase by reforming the phosphodiester bond
45
Explain how cells with identical DNA sequences can have different phenotypes, and give examples
1. Epigenetic Modifications - DNA Methylation - Histone modifications 2. Gene Expression Regulation Different cells express different sets of genes due to the presence or absence of specific transcription factors or regulatory molecules that bind to DNA and activate or repress gene transcription. 3. Alternative Splicing Alternative splicing of mRNA transcripts allows a single gene to give rise to multiple protein isoforms. 4. Environmental Influence The environment can affect gene expression and the resulting phenotype, even in cells with identical genetic material. For example, changes in temperature
46
Diagram a gene by indicating what it is made of and what it does - The anatomy of a gene (or transcription unit) Draw the structure of a typical gene in bacteria, and in eukaryotes, labeling all the relevant parts, including 5’ and 3’ upstream and downstream regions Know all relevant names of the parts
Eukaryotes Promoter: DNA sequence where RNA polymerase binds to begin transcription Transcription Start Site (TSS): The first nucleotide transcribed into RNA 5′ UTR (Untranslated Region): Transcribed but not translated; important for regulation Coding Sequence (CDS): Region that is translated into protein Introns (eukaryotes only): Non-coding regions removed from mRNA during splicing Exons (eukaryotes only): Coding sequences that remain in mature mRNA 3′ UTR: Transcribed but not translated; includes regulatory elements Transcription Terminator: Signals RNA polymerase to stop transcription Enhancers/Silencers (eukaryotes): Regulatory sequences far from the promoter that increase or repress transcription Prokaryotes -35 and -10 (TATA) elements: Core parts of the promoter, recognized by the sigma factor of RNA polymerase. TSS (Transcription Start Site): Where transcription begins. 5′ UTR: Untranslated region before the coding sequence. CDS (Coding Sequence): Translated into protein. Terminator: Hairpin loop or Rho-dependent mechanism signals RNA polymerase to stop.
47
Describe--using drawings and text--how transcription initiation works in bacteria Similar to diagram you drew of a transcription bubble on C12 worksheet Know what the core RNAp and the holoenzyme is and does in prokaryotes Be able to interpret/diagram “Christmas tree” diagrams (worksheet)
1. Closed Complex Formation RNA polymerase holoenzyme (core + σ) binds to the -35 and -10 regions of the promoter. DNA is still double-stranded at this point. 2. Open Complex Formation (Transcription Bubble) RNA polymerase unwinds about 17 base pairs around the -10 region. Forms a transcription bubble. The template strand is exposed and used for RNA synthesis. 3. Initiation of RNA Synthesis RNA polymerase begins adding ribonucleotides at the +1 site. The first few nucleotides are added without the polymerase moving. Sigma factor may still be bound. 4. Promoter Clearance After ~10 nucleotides are synthesized, the sigma factor dissociates. RNA polymerase transitions into elongation mode and moves downstream.
48
Be able to determine which strand is coding which is the template strand - Understand complementary base pairing - Know that RNAp can proofread (occasionally).
Steps in Elongation: RNA polymerase reads the DNA template strand from 3′ to 5′. It builds the RNA transcript in the 5′ to 3′ direction, complementary to the template strand. Uracil (U) replaces thymine (T) in RNA. - The template strand is what RNA polymerase reads (from 3′ to 5′). - The coding strand has the same sequence as the RNA (but T instead of U).
49
Define a gene by indicating what it is made of and what it does
A gene is a segment of DNA that contains the instructions to make a functional product, usually a protein
50
distinguish between transcription and translation, and use the words accurately when writing about gene expression
Transcription: The process of copying a DNA sequence into RNA. Translation: The process of decoding an mRNA transcript to build a protein.
51