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MCDB 101B > Old Lectures for the Final > Flashcards

Old Lectures for the Final Flashcards

1
Q

Define incomplete dominance

A

Heterozygote phenotype is intermediate between the two

homozygote phenotypes

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2
Q

Define codominance

A

Heterozygote shows trait of both parents

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3
Q

Define epistasis

A

alleles of one gene mask the effects of the alleles of another gene

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4
Q

What does a complementation test tell you?

A

test for determining whether two mutations associated with a specific phenotype represent two different forms of the same gene (alleles) or are variations of two different genes.

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5
Q

Define haploinsufficiency

A 
It is a loss-of-function mutations that
are dominant to wild type.
AKA One wild-type allele
does not provide
enough of a gene
product
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6
Q

What are homologous chromosomes?

A

Homologous chromosomes have almost identical DNA sequences. One is inherited from the mother and one is inherited from the father.

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7
Q

How do Cyclin-dependent kinases (CDKs) control the cell cycle?

A
  1. Inactive CDKs will bind to cyclin (e.g. S-phase cyclin build up during G1 phase) as cycling levels begin to build up
  2. CDKs bound to cyclin will become active, triggering DNA synthesis
  3. Cyclin will be degraded as cell cycle continues
  4. CDK will become inactive due to low levels of cyclin
  5. process starts over for the next stage of the cell cycle
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8
Q

What are the 4 phases of the cell cycle?

A

G1 -> S -> G2 -> M ->G1 -> cont.

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9
Q

What is the main purpose of the S phase?

A

To replicate the cell’s genome

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10
Q

Define what a sister chromatid is

A

The identical products of

DNA replication

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11
Q

What holds sister chromatids together after DNA replication?

A

Sister chromatids are held together by cohesin

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12
Q

What are telomers and why are they important?

A

They are hundreds of repeat sequences (TTAGG) at the ends of linear DNA. They prevent the ends of DNA strands from being lost during replication.

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13
Q

Which enzyme is responsible for adding telomeres to DNA

A

telomerase

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14
Q

What is Shelterin and why is it important?

A

Shelterin is a complex of several proteins that together bind telomeric repeat DNA. It prevents DNA repair systems from being activated and fusing the ends of linear DNA together.

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15
Q

What does Shelterin do in order to protect the DNA?

A

Shelterin forms t-loops at the ends of chromosomes. A t-loop is formed when the end of the DNA circles back around on itself and a 3’ overhand anneals to the DNA strand, creating a closed loop at the end of the DNA.

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16
Q

What is Condensin?

A

Condensin is a protein complex that binds DNA and forms organized loops using ATP. (Condenses the DNA into organized chromosomes)

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17
Q

What type of microtubule attachment is there during the metaphase-anaphase transition (mitosis or meiosis II)

A

There is bipolar attachment of centromeres of sister chromatids.

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18
Q

During the metaphase-anaphase transition, how are sister chromatids separated?

A

When the conditions are met for the cell to move into anaphase, the inhibitor of the separin protease will be degraded and separin will cleave the cohesin, allowing the microtubules to pull the chromatids apart.

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19
Q

What are the three main requirements for Meiosis I?

A
  1. Sister chromatids held together along entire length
  2. homologs must pair
  3. must be at least 1 breakage and rejoining between NON-sister chromatids of a homologous pair (crossover event)
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20
Q

What is SPO11?

A

SPO11 is the enzyme responsible for creating double stranded breaks in DNA during prophase I (meiosis)

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21
Q

How are homologs paired during prophase I (meiosis)?

A

A synaptomenal complex (SC) forms between paired homologs, creating a latter-like structure between them.

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22
Q

What are the 5 substages of Prophase I (meiosis)?

A

Leptotene, zygotene, pachytene, diplotene, and diakinesis

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23
Q

What happens during the leptotene phase?

A

DSB appear

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24
Q

What happens during the zygotene phase?

A

strand invasion and D-loop

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25
Q

What happens during the pachytene phase?

A

double Holliday structure or

SDSA

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26
Q

What happens during Diplotene phase?

A

resolution of double Holliday

structure in NCO or CO pathways

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27
Q

What happens during the diakinesis phase?

A

Chromosomes are pulled apart

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28
Q

What are chiasmata?

A

Chiasmata are structures formed by homologous chromosomes that have undergone double stranded breaks and rejoining to NON-sister chromatids of a homologous pair.

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28
Q

What are chiasmata?

A

Chiasmata are structures formed by homologous chromosomes that have undergone double stranded breaks and rejoining to NON-sister chromatids of a homologous pair.

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29
Q

What type of microtubule attachment is there during the metaphase I (meiosis)?

A

There is monopolar attachment of the sister chromatids’ centromeres to the spindle fibers.

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30
Q

What happens to cohesin during the transition of Metaphase I to Anaphase I? (meiosis)

A

Cohesin along the arms of the chromosomes are degraded. Marked cohesion around the centromeres of the sister chromatids is not degraded and continues to hold sister chromatids together

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31
Q

What type of microtubule attachment is there during the metaphase II (meiosis)?

A

There is bipolar attachment of sister chromatids’ centromeres.

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32
Q

What is nondisjunction?

A

Nondisjunction is when chromosomes fail to segregate to opposite poles

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33
Q

What is monosomy?

A

Monosomy is an outcome of nondisjunction where there is only 1 copy of a chromosome present in
a zygote when there should be two

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34
Q

What is trisomy?

A

Trisomy is an outcome of nondisjunction where there is 3 copies of a chromosome present in a zygote when there should only be 2

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35
Q

What is an SRY gene?

A

SRY is a gene that determines whether a fetus will develop testes or ovaries. The SRY gene is usually only present on the Y chromosome. However, people who have two X chromosomes but an SRY gene on one of them will still develop testes. On the other hand, people who are born with a Y chromosome that lacks the SRY gene will develop ovaries.

36
Q

What are some features of the Y chromosome?

A
  • Male Specific Y (MYS): sends embryo down male pathways plus a few genes needed for testes development
  • Mostly highly condensed heterochromatin
  • a few non-testes related genes scattered on the rest of the Y
  • PARs at tips homologous with tips of Y (allows pairing and recomb.)
37
Q

How is the imbalance between the X and Y chromosomes solved in humans? (i.e. dosage compensation)

A

One of the X chromosomes in females are deactivated. Deactivation of one X versus the other is completely random in each cell. However daughter cells will have the same deactivation as their parent cell.

38
Q

How does histone acetylase effect DNA?

A

Histone acetylase acetylates the histones causing them to become more negative. Because DNA is negatively charged, this acetylation causes weaker interactions between the histones and DNA. This makes it easier for DNA is be transcribed.

39
Q

How does methylating DNA affect transcription?

A

Methylation of DNA causes the DNA to form tightly packed structures called heterochromatin, preventing any transcription of that DNA

40
Q

Which gene is responsible for the deactivation of the X chromosome?

A

The Xist gene, located in the X inactivation center (XIC). Xist is upregulated on the inactive X chrom. and completely shut off in the active X chrom.

41
Q

What does the Xist gene do to promote X shut down?

A

The Xist gene promotes methylation of the DNA, starting at the X inactivation center (XIC) and spreading outward.

42
Q

What is a patchwork female?

A

“patchwork females” are a result of different X chromosomes being shut down at different places in the body, causing certain sex-linked gene expressions to be different at different places in the body.

43
Q

What are PARs?

A

PARs are pseudo autosomal regions. They are located on both X and Y chromosomes (even inactivated X). These regions are located at the tops of chromosomes and allow X and Y chromosomes to pair and recombine. PARs are not inactivated in Y chroms or Barr bodied X chroms.

44
Q

How is Chi square (X^2) calculated?

A

X^2 = sum of all (obs-exp)^2 / exp

45
Q

How are degrees of freedom calculated?

A

d.f. = (#classes - 1)

46
Q

What is the highest recombination frequency for genes on the same chromosome?

A

can never have a recombination frequency that is equal to or larger than 50%

47
Q

What is the equation for calculating interference?

A

I = (1 - (obs# DCO / exp# DCO))(100)

48
Q

What is the first step in the process of recombination?

A

Double-stranded break formation in one chromatid catalyzed by Spo11

49
Q

What is the second step in the process of recombination?

A

The 5’ ends of each broken strand are degraded by an exonuclease to create 3’ single-stranded tails

50
Q

What is the third step in the process of recombination?

A

Dmc1 helps one single-strand tail invades a non-sister chromatid
and forms stable heteroduplex.
Displacement loop (D-loop) from invaded chromatid is stabilized
by single-strand binding protein

51
Q

What is the fourth step in the process of recombination?

A

D-loop and other 3’ tail form a second heteroduplex.
DNA synthesis replaces the DNA degraded during formation of the 3’ tails.
DNA ligase reseals the DNA backbones.
Formation of double Holliday Junction

52
Q

What is the fifth step of recombination?

A

Branch migration: Heteroduplex region of both DNA molecules is lengthened

53
Q

What is the sixth step in recombination?

A

Holliday junctions are resolved by resolvase and ligase

54
Q

What is the SDSA pathway?

A

Recombination without crossing over: synthesis-dependent
strand annealing
Anticrossover helicase disentangles the invading strand from
the nonsister chromatid preventing Holliday junction
formation.

55
Q

What is a heteroduplex region?

A

The region of DNA between breakpoint in recombination. One strand is maternal and the other is paternal. Strands can have mismatched DNA

56
Q

What are the three tetrad types produced by meiosis of dihybrid yeast?

A
  • parental ditype (PD)
  • nonparental ditype (NPD)
  • tetratype (T) (mix of PD and NPD)
57
Q

What indicates that genes are not linked in tetrad analysis?

A

when #PD = #NPD, it is indicated that genes are not linked

58
Q

How do you calculate the recombination frequency of unlinked genes in tetrads?

A

RF = (NPD + 1/2T)/total tedrads

59
Q

What indicates that genes are linked in tetrad analysis?

A

number of PD&raquo_space; number of NPD

60
Q

How do you calculate the recombination frequency of linked genes in tetrads?

A

RF = (1/2T + 3NPD)/Total

61
Q

What is gene conversion and how does it happen?

A 
Gene conversion – the physical
change of one allele in a
heteroduplex to the other.
Either allele can be converted
by DNA mismatch repair
pathway.
62
Q

How does gene conversion effect tetrad recombination analysis?

A

Gene conversion can make a 3:1 PD to NPD in a single tetrad

63
Q

What is a feature of transcription initiation that eukaryotes have that prokaryotes lack?

A

Eukaryotic genes often have enhancers:
• Can be thousands of base pairs away from the promoter
• Required for efficient transcription

64
Q

What is the 5’ cap?

A

The 5’ cap is a methylated G that is added backwards to a primary transcript.

65
Q

How are mRNA introns spliced?

A
  1. The 5’ splice donor site (GU) is cut by spliceosome
  2. RNA folds over on itself, connecting to the donor site (A) in the middle of the intron (Lariat structure)
  3. The 3’ splice acceptor site (AG) is cut by spliceosome
  4. Mature mRNA is ligated together and intron is degraded
66
Q

Are all pre-mRNAs spliced in the same way?

A

No. Alternative splicing allows multiple mRNAs to be produced from the same primary transcript. Many primary transcripts are spliced differently in different areas of the body

67
Q

What are EJCs?

A

EJCs are exon junction complexes. EJCs bind to junction sites in spliced mRNAs, indicating that the mRNA is ready to leave the nucleus and allowing the cell to do a quality check on the mRNA

68
Q

What are the main Eukaryotic translation initiation factors?

A
  • Cap binding protein
  • eIF-4E
  • eIF-4A
  • eIF-4G
  • PolyA Binding Protein
69
Q

What determines the stability and lifespan of an mRNA?

A

The length of mRNA polyA tails determines how long a mRNA can be translated. 5’ cap complex binds to 3’ PolyA tail binding protein, forming a closed loop. PolyA tail is slowly degraded and when it is gone, RNA will become linear and be digested.

70
Q

How is translation initiated?

A

Small subunit of ribosome binds to 5’
CAP and slides down mRNA to first AUG
Large subunit is recruited and translation
initiates.

71
Q

What happens during the translation elongation phase?

A

Addition of amino acids to C-terminus of polypeptide

Charged tRNAs ushered into A site by proteins called elongation factors

72
Q

What happens during translation termination?

A
  • Ribosome reaches stop codon
  • Release factor comes into the A position rather than a new tRNA
  • Ribosome falls off of mRNA
  • Closed loop mRNA promotes ribosome recycling
73
Q

How are truncated proteins formed?

A

Mutations or splicing mistakes can result in early stop codons, leading to truncated proteins

74
Q

How do EJC’s provide translation quality control?

A

If EJCs are found downstream of a stop codon, additional factors will be recruited, and RNA will be cleaved and degraded (nonsense mediated decay)

75
Q

Why is the pioneer round of translation important?

A
  • quality control
  • all EJCs should be removed
  • pioneer ribosome detects downstream EJCs after a stop codon (Nonsense mediated decay)
76
Q

What does the Cap-binding protein do?

A

The cap-binding protein is only involved in the pioneer round of translation. It recruits the small ribosome subunit to initiate translation

77
Q

What does eIF-4E do?

A

The eIF-4E protein binds the 5’ cap in translation rounds following the pioneer round

78
Q

what does eIF-4A do?

A

RNA helicase activity, at cap

79
Q

what does eIF-4G do?

A

at cap, but binds to PABP

80
Q

How is nonsense mediated decay initiated?

A
  • pioneer ribosome will detect EJC downstream of the stop codon (PTC)
  • UPF1 (SURF complex) is phosphorylated
  • mRNA is cleaved and degraded by exonuclease
81
Q

How is non-stop decay initiated?

A
  • invoked when there is no stop codon due to incorrect splicing or mutation
  • Ribosome will stall eventually when translating the polyA tail
  • This recruits Ski complex which will degrade the mRNA
  • protein is ubiquitinated and degraded
82
Q

what are Null (amorphic) mutations?

A

completely block function of
a gene product (For example
deletion of an entire gene)

83
Q

What are Hypomorphic mutations?

A

produces less of wild-type
protein, or less effective
mutant protein

84
Q

What is Haploinsufficiency?

A

one wildtype allele does not produce
enough gene product to avoid
mutant phenotype

85
Q

What are Hypermorphic mutations?

A

generate more gene product or
the same amount of a more
efficient gene product

86
Q

What are Neomorphic mutations?

A 
generate gene product with
new function or that is
expressed at inappropriate
time or place (ectopic
expression)
87
Q

What are Antimorphic alleles?

A
  • dominant negative
  • prevent the normal
    protein from functioning
    • Usually occurs in genes that
    encode multimeric proteins
    • Mutant subunits block the
    activity of normal subunits

MCDB 101B (3 decks)

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