topic 6 - final exam Flashcards

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1
Q

Describe how and why DNA is replicated

A
  • To prepare for cellular division dna replication occurs (in the S phase of cell cycle)
  • Is semiconservative replication
    4 main steps
    INITIATION
    Helicase enzyme unwinding the double heliex
  • Strands separated by spinning at 10 000 rmp
  • Hydrogen bonds broken
  • Molecule unwinds
  • Forms replication fork
  • Each strand then acts as a template for the DNA replication process

PRIMER BINDING
- DNA replication initiaded by the binding of short RNA prmer to ssDNA
- Nucleotides can only be added to the 3’ end of an existin nucleotide chain
- RNA primer synthesized by the enzyme, primase

ELONGATION
- Adding DNA nucleotide to the primer
- By DNA polymerase III synthesizing in 5’ to 3’ direction uses 3’ to 5; direction template DNA
- Discontinuous DNA relication: DNA synthesis in short fragments (Okazaki fragments) on the lagging stramd
- 4 types of nucleotides
 Deoxyribosnucleoside triphosphates (dNTPs: dATP, dGTP, dCTP, dTTP)
- Base pairing error may occur in DNA rplication
 Need for proofreading
- DNA polymerase is able to proofread and correct errors
 Removing incorrect nucleotides
TERMINATION
- DNA polymerase I removes RNA with DNA
-
- DNA ligase connects fragment to form on continuous strand

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2
Q

Define what constitutes a gene structure

A

gene refers to the oranisation of specific sequence elements within a gene. genes are segemnts of DNA that contain the code for a polypeptide chains aa sequence, as well as regulatory sequences that control gene expression.

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3
Q

Describe how gene expression is controlled through the actions of regulatory proteins, using lac operon as an example

A

Negative regulation via lac repressor
- Absence of lactose: repressor bound to operator (negative regulation)
- Lactose absent, repressor bound to operator, operon repressed.

Presence of lactose: allolactose binds to repressorcausing it t become an inactive form
- Lactose present, repressor not bound to operator operon derepressed.
-

Positive regulation via catabolite activator protein (CAP)
- CAP also known as cAMP Receptor protein (CRP)
- CAP-cAMP complex facilitates RNA polymerase binding to promoter (positive regulation)
- Light house

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4
Q

Outline how gene expression is controlled in eukaryotes

A

eukaryotic gene regulation
- occurs at 5 different levels
1. genomic eg., chromatin decondensation (euchromatin) and condensation (heterochromatin)
- euchromatin are dna undergoing transcription
2. transcriptional eg presence of enhancers or silencers
3. post-trancriptional (RNA processing and nucleo export)
- RNA splicing to remove introns forming mature mRNA
4. Transitional – binding of a translational repressor
5. Post- translational eg phosphorylation/dephosporylation

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5
Q

Define and describe DNA mutation

A

Dna mutation is any alteration in dna
 Source of new gene
 Source of gene diversity
 May result in abnormal protein produced (if mutation is in the gene coding region)
Causes small to large scale changes

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6
Q

compare and contrast between spontaneous and induced mutations

A

Non induced (spontaneous) mutations
 Unknown that may occur anytime, anywhere
 Damaged genes (resulted from DNA replication without repair)
 More common in organisms with short generation times (eg viruses and bacteria)
Induced mutations
- Exposure to mutagens
o Natural (uv lights, cosmic rays, radon)
o Induced
 Chemical (nitrosguanidine, colchicine
 Physical (gamma rays, x-rays)
 Biological (virus)
 Man made (asbestose and nuclear fallout)

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7
Q

compare and contrast the difference in somatic vs germinal mutations

A

Somatic mutations are not passed onto offspring where as germanal mutations can be

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8
Q

identify small scale mutations at the DNA level

A

small scale mutations (dna level)

  • INDEL (insertion or deletion)
  • Base substitution (transition or transversion)

purine to pyrimidines vice versa = transversion mutation

purine to purine
pyrimidine to pyrimidine
= transition

purine (A, G)
have 2 circles (wedding rings)

pyrimidne ( C, T, U)
have one circle

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9
Q

describe large scale mutations

A

Large scale mutation
Chromosome mutation abnormalities in chromosome number

can lead to a deletion, of a chromotid, duplication, translocation, inversion

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10
Q

state the different types of mutations in proteins

A

Affects PROTEIN synthesis in many ways
- Frameshift
o Due to insertion or deletion of DNA base
o Addition of T(in example) changed the reading frame
o May result in devastating effect since entire sequence may be changed
- Missense
o Due to substitution of DNA (T -> A)
o Resulted in change in mRNA sequence and therefore, a change in AA
o
- Nonsense
o Due to substitution of DNA base (T -> A)
o Resulted in STOP codon added
- Silent
o Due to substitution of DNA base (C -> T)
o However, protein sequence is not changed
o Redundancy of the egenetic code reduced the change og point mutations altering the specified AA

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11
Q

Compare the two sequences of DNA below:

Sequence 1: AGC TTC AGT GAc TGA GCT GGC GTC GAC

Sequence 2: AGC TTC AGT GAt TGA GCT GGC GTC GAC

The two bases in lowercase are an example of:

A

transition

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12
Q

Compare the two sequences of DNA below:

Sequence 1: AGC TtC AGT GAT TGA GCT GGC TTC GAC

Sequence 2: AGC TaC AGT GAT TGA GCT GGC TTC GAC

The two bases in lowercase are an example of:

A

transversion

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13
Q

The normal sequence of genes along a hypothetical chromosome is ABCDEFGH, where letters represent genes and the asterisk () represents the centromere. A mutation event occurs, resulting in a chromosome with the following sequence of genes: ABC*DFEGH. This mutation is an example of:

A

inversion

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14
Q

define a regulatory gene

A

encodes protein that regulates gene expression

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15
Q

define structural gene

A

encodes protein/rRNA/tRNA that is not a regulatory gene

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16
Q

define premotor

A

RNA polymerase binding site