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
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
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
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)
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
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
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
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
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
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
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
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
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
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
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
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
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
draw and label a diagram of a ribosome and include all the binding sites
- large subunit on top: E, P, A sites from left to right
- small subunit on bottom: has mRNA binding site on left
26
explain how ribosomes read and translate mRNA into a protein
- small subunit bind mRNA and decodes it
- large subunit catalyzes the peptide bond formation using amino acids brought by tRNA
27
explain the genetic code and how nucleotide sequence determines amino acid sequence
- 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
explain why the genetic code is described as conserved and redundant
- conserved: bases make the same amino acids across all species
- redundant: repetitive; multiple codons code for the same amino acids
29
describe the main steps of translation
- 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
know the start and stop codons
- start: AUG
- stop: UAA, UGA, UAG
31
identify the reading frame of a gene, write the mRNA transcript, identify the start/stop codon, and translate to a peptide sequence
- 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
explain why every cell does not express every gene
- conserves energy and space
- some cells do not need to products of certain genes
33
describe transcriptional regulation in prokaryotic cells
- regulated at transcriptional level because transcription and translation can happen at the same time
34
list the multiple levels of regulation in eukaryotic cells
- epigenetic
- transcriptional
- post-transcriptional
- translational
- post-translational
35
compare the complexity of regulation in prokaryotic vs. eukaryotic cells
- prokaryotic: less complex
- eukaryotic: more complex
36
explain the common view of evolution during Darwin's time
- Plato: species were static and unchanging
- 18th century: accepted idea of extinct species and change of planet geology
- Lamarck: inheritance of acquired characteristics
37
explain Lamarck's theory of evolution and how it was disproved by Weissman
- 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
describe how Darwin's theory differed from the current view
- explained that species changed over many generations due to environmental conditions
39
understand how the current theory of evolution developed
- 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
define population genetics and explain how it is used to study the evolution of populations
- 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
explain the four evolutionary forces that disrupt equilibrium
- 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
describe how each of the four evolutionary forces that disrupt equilibrium affect the allele frequencies in a population
- 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
