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BIOL 107 > Exam 3: Learning Objectives > Flashcards

Exam 3: Learning Objectives Flashcards

1
Q

draw a rough sketch of the structure of a nucleotide

A
  • phosphate group
  • deoxyribose (5 C sugar)
  • nitrogenous base
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2
Q

recognize the 4 nitrogenous bases from a picture of the structure

A
  • adenine: double ringed purine, has NH2 branch
  • thymine: single ringed pyrimidine, has CH3 branch
  • guanine: double ringed purine, has NH2 and O branch
  • cytosine: single ringed pyrimidine, has NH3 branch
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3
Q

identify purine, pyrimidines, and complementary base pairs

A
  • purine: double ringed, adenine and guanine
  • pyrimidine: single ringed, thymine and cytosine
  • complementary base pairs: A and T, G and C
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4
Q

describe the structure of the DNA double helix

A
  • phosphate and sugar backbone held together with covalent phosphodiester bonds
  • antiparallel: strand running in opposite directions
  • 5’ (phosphate) to 3’ (sugar)
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5
Q

understand how DNA is compacted into chromatin

A
  • DNA wraps around histone proteins
  • 8 histone proteins make nucleosome
  • nucleosomes coil into strand which become chromatin
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6
Q

explain what chromosomes are and identify different regions

A
  • DNA molecule containing genes
  • condensed chromatin
  • centromere in the middle
  • telomere on the ends
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7
Q

identify homologous vs non-homologous chromosomes

A
  • homologous: have same genes in same places
  • non-homologous: do not have the same genes in the same places
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8
Q

compare euchromatin and heterochromatin

A
  • euchromatin: loosely packed, genetically active, rich in genes used for transcription process, found in inner nucleus
  • heterochromatin: tightly packed, genetically inactive, found in outer nucleus
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9
Q

complete the sequence of a complementary strand of DNA

A
  • when you see an A, complement with a T
  • when you see a T, complement with an A
  • when you see a G, complement with a C
  • when you see a C, complement with a G
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10
Q

understand the meaning of “semi-conservative” replication

A
  • each of two strands of double helix serves as template for new strands be created
  • one DNA splits and parts are replicated to make two
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11
Q

compare DNA replication in prokaryotes and eukaryotes

A
  • prokaryotes: 1 circular chromosome, 1 origin of replication, 1000 nucleotides/sec replication rate, no telomeres
  • eukaryotes: several linear chromosomes, several origins of replication, 50 to 100 nucleotides/sec replication rate, has telomeres
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12
Q

draw an origin of replication and indicate direction of synthesis for each strand

A
  • prokaryotes: one origin of replication, duplicates in both directions until it goes all the way around the circular chromosome
  • eukaryotes: several bubble origins, duplicated in both directions until it meets another duplicated section across the linear chromosome
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13
Q

sketch and label a replication fork

A
  • replication fork extends from origin of replication in both directions
  • leading strand on 3’ to 5’ end
  • lagging strand on 5’ to 3’ end
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14
Q

explain the process of DNA replication including the role of each enzyme

A
  • DNA unwinded using helicase at the replication fork
  • single stranded binding proteins stabilize DNA and ensure it doesn’t rewind
  • topoisomerase prevent overwinding
  • primase synthesizing RNA primers to help DNA polymerase know where to start syntehsizing
  • leading strand synthesizes continuously from 3’ to 5’ end
  • lagging strand synthesizes discontinuously from 5’ to 3’ end
  • DNA polymerase 3 reads template strand and adds nucleotides
  • DNA polymerase 1 removes RNA primers and replaced them with DNA bases
  • DNA ligase joins okazaki fragments to fix the gap in the phosphodiester backbone
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15
Q

discuss the importance of telomere replication

A
  • ensures parts of DNA at the end of the template strands get replicated so the strand doesn’t continuously get shorter over time
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16
Q

understand the 3 types of DNA repair mechanisms: proofreading, mismatch repair, nucleotide excision repair

A
  • proofreading: DNA polymerase checks for correctness as it is replicating
  • mismatch repair: enzymes recognize mistakes after replication and repair them
  • nucleotide excision repair: DNA is unwound and incorrect bases are removed as well as bases on either side; used for thymine dimers
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17
Q

explain the central dogma

A
  • cellular chain of command that dictates the flow of genetic information
  • DNA transcribed to RNA translated to proteins
18
Q

compare gene expression in prokaryotes and eukaryotes

A
  • prokaryotes: no nucleus so transcription and translation take place in cytoplasm and both can occur at the same time; has operons: clusters of genes coding for proteins with similar functions
  • eukaryotes: transcription and mRNA modification takes places in the nucleus; translation takes place in cytoplasm
19
Q

understand beyond the one gene-one enzyme hypothesis

A
  • many genes encode for proteins other than enzymes
  • some genes only encode for part of a protein
  • some genes encode for non-coding RNAs
  • many genes have more than 1 exon and are processed differently to produce multiple products
20
Q

describe the main steps of transcription

A
  • initiation: transcription factors bind to promoter region of gene and recruit RNA polymerase and bind together; RNA polymerase recognizes start sequence and begins synthesizing RNA transcript
  • elongation: RNA polymerase unwinds DNA and reads it to attach complementary RNA bases; RNA nucleotides are joined by phosphodiester bonds along the backbone
  • termination: RNA polymerase transcribes termination sequence and releases RNA transcript; RNA polymerase detaches from DNA
21
Q

describe how eukaryotic mRNA is processed and how this differs from prokaryotic

A
  • prokaryotic: RNA doesn’t need to be processed; transcription and translation can happen at the same time with multiple RNA polymerases and ribosomes
  • eukaryotic: modifications of 5’ cap with guanine, multiple adenines on 3’ end, and RNA splicing
22
Q

list the components of translation

A
  • ribosome
  • transfer RNA
  • messenger RNA
  • polypeptide
23
Q

describe the structure of tRNAs

A
  • carry a specific amino acid on 1 end
  • anticodon on other end
  • single RNA strand about 80 nucleotides long
  • have specific aminoacyl tRNA-synthetase to attach its amino acid
24
Q

understand the relationship between tRNAs and amino acids

A
  • tRNAs carry a specific amino acid using a specific aminoacyl tRNA-synthetase
  • when mRNA call for that amino acid, the tRNA binds its anticodon with the mRNA codon and its amino acid attaches to the polypeptide chain
25
Q

draw and label a diagram of a ribosome and include all the binding sites

A
  • large subunit on top: E, P, A sites from left to right
  • small subunit on bottom: has mRNA binding site on left
26
Q

explain how ribosomes read and translate mRNA into a protein

A
  • small subunit bind mRNA and decodes it
  • large subunit catalyzes the peptide bond formation using amino acids brought by tRNA
27
Q

explain the genetic code and how nucleotide sequence determines amino acid sequence

A
  • genetic code: sequences of 3 nucleotides coding for amino acids
  • decoded during translation where amino acids are brought to polypeptide chain based on codons (3 nucleotides)
28
Q

explain why the genetic code is described as conserved and redundant

A
  • conserved: bases make the same amino acids across all species
  • redundant: repetitive; multiple codons code for the same amino acids
29
Q

describe the main steps of translation

A
  • initiation: mRNA binds to small ribosomal subunit; start codon (AUG) located, initiator tRNA binds to start codon at p site; energy used to bind large and small ribosomal subunits
  • elongation: tRNA enters a site and binds with codon; growing polypeptide chain in p site binds with amino acid on a site; mRNA shifted; tRNA on p site exits trough e site and tRNA on a site shifts to p site
  • termination: stop codon recognized; release factor binds to stop codon causing hydrolysis of polypeptide chain from tRNA; translation components dissociate
30
Q

know the start and stop codons

A
  • start: AUG
  • stop: UAA, UGA, UAG
31
Q

identify the reading frame of a gene, write the mRNA transcript, identify the start/stop codon, and translate to a peptide sequence

A
  • reading frame: sequence of TAC on DNA strand reading 3’ to 5’ or sequence AUG on RNA strand reading 5’ to 3’
  • mRNA transcript: change all nucleotides from DNA strand starting with start codon and ending with stop codon; A=U, T=A, C=G, G=C
  • start codon: TAC on DNA; AUG on RNA
  • stop codon: ATT, ACT, ATC on DNA; UAA, UGA, UAG on RNA
  • peptide sequence: use chart
32
Q

explain why every cell does not express every gene

A
  • conserves energy and space
  • some cells do not need to products of certain genes
33
Q

describe transcriptional regulation in prokaryotic cells

A
  • regulated at transcriptional level because transcription and translation can happen at the same time
34
Q

list the multiple levels of regulation in eukaryotic cells

A
  • epigenetic
  • transcriptional
  • post-transcriptional
  • translational
  • post-translational
35
Q

compare the complexity of regulation in prokaryotic vs. eukaryotic cells

A
  • prokaryotic: less complex
  • eukaryotic: more complex
36
Q

explain the common view of evolution during Darwin’s time

A
  • Plato: species were static and unchanging
  • 18th century: accepted idea of extinct species and change of planet geology
  • Lamarck: inheritance of acquired characteristics
37
Q

explain Lamarck’s theory of evolution and how it was disproved by Weissman

A
  • inheritance of acquired characteristics
  • organism changes during its life to adapt to the environment and it will pass these traits on to its offspring
  • Weissman cut tails of 22 generations of mice to disprove Lamarck; all mice born with tails
38
Q

describe how Darwin’s theory differed from the current view

A
  • explained that species changed over many generations due to environmental conditions
39
Q

understand how the current theory of evolution developed

A
  • Darwin observed finches in Galapagos islands; beak shape and food source; defined natural selection
  • modern synthesis of evolutionary theory grew from joining Mendel’s ideas of heredity with Darwin’s ideas of evolution
40
Q

define population genetics and explain how it is used to study the evolution of populations

A
  • field of study examining what happens to alleles in a population; evolution in terms of generic change
  • uses Hardy-Weinberg equilibrium to study allelic frequencies of populations
41
Q

explain the four evolutionary forces that disrupt equilibrium

A
  • natural selection: phenotype may be advantage or disadvantage depending on environmental conditions
  • mutation: source of new alleles; ultimate source of variation
  • genetic drift: effect of chance; important in small populations; bottleneck effect and founder effect
  • migration/gene flow: flow of alleles due to migration
42
Q

describe how each of the four evolutionary forces that disrupt equilibrium affect the allele frequencies in a population

A
  • natural selection: if phenotype is advantageous, allelic frequency will likely increase
  • mutation: change in allele frequency is very small
  • genetic drift: more change in small populations; if one individual dies in a population, all of its genes will be lost
  • migration/gene flow: some populations are fairly stable and others experience more flux

BIOL 107 (18 decks)

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