original exam 2022 Flashcards

1
Q

what is true about Restriction enzymes?
A. Restiction enzymes are Exonucleases
B. They make double stranded cuts at certain sequences
C. found in mammals

A

B. They make double stranded cuts at certain sequences

(A should be endonucleases)
(C should be bakteria)

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2
Q

What is true about spectrophotometry?
A. Proteins absorbs UV light at 260nm
B. The sugar molecules in DNA absorb UV light at 280nm
C. The purine and pyrimidine rings of the bases absorb UV light at 260nm

A

c: . The purine and pyrimidine rings of the bases absorb UV light at 260nm

(A Protein = 280nm
B Sugar = 230nm)

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3
Q

What is true about agarose gel electrophoresis
a. DNA molecules are seperated by size
b. the DNA solution must first be treated with SDS
c. the largest fragments travel more slowly than the smallest
d. agarose denatures the DNA molecules so that they can travel in the gel

A

A and C

A. DNA molecules are seperated by size

C. the largest fragments travel more slowly than the smallest

(B not SDS , D not gel but PCR)

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4
Q

Which is false about detection of nucleic acids? (choose one option)
A.- DNA can be detected using UV light if ethidium bromide has been added to the gel
B. A circular plasmid will migrate as a band on agarose gel electrophoresis
C. Nucleic acids can be detected in cells using fluorescence in situ hybridisation
D. A DNA ladder must be included in a well to be able to estimate the size of the DNA fragments

A

B. A circular plasmid will migrate as a band on agarose gel electrophoresis

( A circular plasmid will migrate as SEVERAL band on agarose gel electrophoresis)

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5
Q

What happens in the hybridization step of a Polymerase Chain Reaction (PCR) cycle?
A. The DNA strands are separated
B.Primer binds to DNA template
C. DNA is synthesized

A

B.Primer binds to DNA template

(A .The DNA strands are separated = DENATURING.
C. DNA is synthesized= ELONGING)

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6
Q

What defines the area that is amplified in the PCR reaction?
A. Promoter
B. DNA polymerase
C. Primer
D. dNTPs

A

C. Primer

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7
Q

What is not found in a PCR tube?
A.DNA polymerase
B. Primer
C. Restriction enzymes

A

C. Restriction enzymes

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8
Q

When designing primers, it is important to;
A.have a
hybridization temperature above 55 degrees to avoid mispriming
B.have 40-60% AT
content to promote hybridization
C. avoid partial
complementarity in the primer sequences for to avoid primer dimers

A

A and C

(have a
hybridization temperature above 55 degrees to avoid mispriming
AND avoid partial
complementarity in the primer sequences for to avoid primer dimers

(B 40-60% GC
content to promote hybridization)

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9
Q

Horizontal gene transfer in prokaryotes can occur by
A. Transformation
B. Meiosis
C. Transduction
D. Transfection

A

A and C
(Transformation and Transduction)

(Transfection= PLASMID IN MAMALS CELL)

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10
Q

During Sanger sequencing, all types of ddNTP (ddATP, ddCTP, ccGTP, ddTTP) are added to a
reaction mixture together with other components and DNA template. Which statement is
correct about Sanger sequencing?
A.uses only ddNTPs,
no dNTPs are required for the sequencing reaction.
B. generates fragments that have one
ddNTP with a fluorophore attached to it.
C.The fluorophore means that the DNA polymerase is not necessary during the sequencing
- generating fragments that have one ddNTP terminal with a specific fluorophore attached to
it.
D.The bases are detected by the fluorophore emitting light of a specific wavelength.

A

C AND D
(C.The fluorophore means that the DNA polymerase is not necessary during the sequencing
- generating fragments that have one ddNTP terminal with a specific fluorophore attached to
it.
D.The bases are detected by the fluorophore emitting light of a specific wavelength)

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11
Q

What is true about reverse transcription?
A.needs the enzyme reverse transcriptase.
B. Uses RNA as a template to form cDNA.
C. DNA viruses use reverse transcription to integrate into the host cell’s genome.
D.Used to analyze genomic DNA

A

A and B
(A.needs the enzyme reverse transcriptase.
B. Uses RNA as a template to form cDNA)

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12
Q

What is false about plasmids?
A.They are small in size
B.The presence of insert cannot be detected
C. they can be grown in large quantities.
D.they are replicated independently of the bacteria’s chromosomal DNA

A

B.The presence of insert cannot be detected

A,C,and D CORRECT!
PLASMID= -SMALL size
- grown in LARGE quantities
-REPLICATED independently of the bacteria’s chromosomal DNA

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13
Q

Which enzyme is used in Polymerase Chain Reaction (PCR)? What properties are
necessary for this enzyme?

A

Enzyme PCR: DNA Polymerase .
Properties: stable at high
temperatures and performs DNA synthesis at 72 degrees

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14
Q

In which natural processes in the cell does DNA ligase have a function?
How is DNA ligase used
in DNA cloning?

A

DNA ligase Function: DNA repair( to glue DNA strands together after the correct bases have been
inserted).
DNA ligase used
in DNA cloning : DNA ligase( to glue vector and insert together)

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15
Q

What are the differences between DNA cloning and CRISPR technology?

A

DNA cloning: give new characteristics to an organism.
Enzyme: restriction enzymes
CRISPR : one edits one’s own genes
Enzyme: Cas9 that makes double-stranded cuts

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16
Q

When carrying out DNA cloning using restriction enzymes, the vector is often opened with two
enzymes. Why?

A

Two restriction enzymes are used so that create sticky ends not complementary to each other, therefore will not stick back to each other

SHORT ANSWER ( will be able to give
direction to the insert)

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17
Q

When you clone eukaryotic genes, you use the cDNA sequence of genes (and not the genomic
sequence). Why?

A

don’t want to carry the
introns ( The cDNA sequence does not
contain the introns, which have been spliced out during transcription)

18
Q

Which is false about thymine dimers?
A.Thymine dimers are recognized as a bulge in the DNA helix by repair enzymes
B.Thymine dimers have covalent bonds between two thymine bases on opposite strands
C.Thymine dimers are formed by exposure to UV

A

B.Thymine dimers have covalent bonds between two thymine bases on opposite strands

(B should be dont have COVALENT)

(A. Thymine dimers a bulge in the DNA helix
C.Thymine dimers exposure to UV)

19
Q
  1. What is false about deamination
    A. loss of an amino group
    B. occur when exposed to chemical agents
    C. spontaneously
    D.Deamination is repaired by mismatch repair
A

D. Deamination is repaired by mismatch repair

(D not mismatch but BER)
(A,B,C TRUE! Deamination: -loss of an amino group
-occur when exposed to chemical
-spontaneously)

20
Q

A missense mutation
A.results in a change in codon but does not change the amino acid
B.results in a change in codon that changes the amino acid
C.results in a new and premature stop codon

A

B.results in a change in codon that changes the amino acid

REMEMBER : SMN

(SILENT :a change in codon but does not change the amino acid
MISSENSE : results in a change in codon that changes the amino acid
NONSENSE:in a new and premature stop codon (NGG) )

21
Q

Re-inactivation can occur by
A.nonsense point mutation
B. silent mutation
C.Insertion due to transposon

A

A and C

(A.nonsense point mutation
C.Insertion due to transposon)

22
Q

Gene regulation in prokaryotes takes place mostly at the level of transcription
A. TRUE
B. FALSE

A

A . TRUE
transcription = prokaryotes in Gene regulation

23
Q

Which is false about the lac operon?
A. the lac operon consists of three genes: lacZ, lacY, lacA
B. the lac operon is off when there is no lactose in the
environment
C. lactose is not broken down when the lac operon is on
D.The repressor LacI binds to lacO and inhibits transcription

A

C. lactose is not broken down when the lac operon is on

A,B,D is CORRECT!

LAC OPERON:
- Three genes: lacZ, lacY, lacA
-Lac operon is OFF = NO Lactose
-Repressor LacI binds to lacO, inhibits Transcription

24
Q

Which is false about araC?
A. - AraC is an activator and promotes gene expression when arabinose is in the
environment
B.AraC undergoes a conformational change when it binds to arabinose
C.AraC
binds to the araBAD promoter (also called pBAD
D.Arabinose acts as an inducer

A

A. - AraC is an activator and promotes gene expression when arabinose is in the
environment

B.C.and D is CORRECT!

ARAC:
-undergoes a conformational change when it binds to arabinose
-binds to the araBAD promoter(pBAD)
-Arabinose acts as an inducer

25
Q

A repressor helps to regulate gene expression of a particular gene by:
A. inhibits access to
RNA polymerase by binding the promoter
B. binds to a signal molecule that
causes it to let go of the DNA and allow the RNA polymerase bind
C.bind to a signal molecule that allows it to interact with DNA and recruit RNA polymerase
D.bind constitutively to the promoter of the gene

A

A,B, and D

Repressor helps to regulate gene expression:
-inhibits access to
RNA polymerase by binding the promoter
- binds to a signal molecule that
causes it to let go of the DNA and allow the RNA polymerase bind
-bind constitutively to the promoter of the gene

C is for Activator!

26
Q

What is alternative splicing? How does it contribute to the non-linear relationship between the
transcriptome and the proteome?

A

Alternative splicing is when different exons are spliced together to give rise to different proteins

27
Q

BRCA1 is to be expressed by transfection in a cell line of mammalian skin cells. You want to isolate your
target protein, BRCA1, from the protein solution (lysate) using affinity chromatography.
How do you proceed? Describe what affinity chromatography is, what the various steps are, and what
preparatory work you must do before BRCA1 is to be transfected

A

no tag, but an antibody is present in the column to capture the protein.

28
Q

What is density-gradient centrifugation?

A

separates cell compounds into different liquid phases and good separation

29
Q

A naked virus is taken up by the host cell by endocytosis.
A. True
B. Wrong

A

A True
Virus is Endocytosis cell

30
Q

Choose the correct order for the various steps in the life cycle of an animal virus (HUSK! AEUBAR): (choose one or
more options)
A. - Adsorption, entry, undressing, biosynthesis, assembly, release
B.- Entry, adsorption, undressing, biosynthesis, assembly, release
C.Undressing, entry, adsorption, assembly, biosynthesis, release
D.Entry, stripping, adsorption, assembly, biosynthesis, release

A

A Adsorption, Entry, Undressing, Biosynthesis, Assembly, Release

31
Q

An envelope virus uses parts of the host cell’s plasma membrane to form a membrane around
itself
A.True
B. False

A

A (TRUE)
Envelope virus use host cells plasma membrane

32
Q

What kind of virus can go from lysogenic to lytic life cycle? (select one or more options)
A.DNA viruses
B. RNA viruses
C.Retroviruses

A

C.Retroviruses

(lysogenic to lytic )

33
Q

What happens at the adsorption step in the life cycle of a virus?
A. Assembly and maturation of virus particles
B.The cape is removed
C.Genome replication
D. Viruses attach to the host cell’s receptors
E.Viruses enter the cell by fusion or endocytosis

A

D. Viruses attach to the host cell’s receptors

adsorption step in the life cycle is virus attach to host cell’s receptors

34
Q

What is true about a bacteriophage?
A. bacteriophage can be surrounded by a membrane
B.A bacteriophage does not go through the undressing step
C.A bacteriophage infects mammalian cells
D.- A bacteriophage injects its nucleic acids through the cell wall

A

B and D
. Bacteriophage Does not go through undressing step and injects nucleic acids through the cell wall

35
Q

A cell that has been infected by a virus can present antigens from the virus on its surface. On which
molecule are these antigens presented?
A.MHC I
B. MHC II
C. Antibody
D. T cell receptor

A

A.MHC I

antigent that present from A cell infected by VIRUS is MHC I

36
Q

A phagocyte is a cell that can engulf pathogens by phagocytosis. Which immune substances promote
phagocytosis? (select one or more options)
A. T cell receptors
B.Macrophages
C.Complement system
D. Antibody

A

C and D

immune substances promote
phagocytosis:
-Complement system -
Antibody

37
Q

You want to make 2L of 1X TAE and have a stock solution of 20X. How many mL of 20X TAE do you
need?

A

100 ml

Stock solution (C1): 20X
Final Concentration (C2): 1X TAE
Total Volume (V2): 2 L

Rumus: C1.V1 /C2.V2

V1: (1X TAE . 2L ) / 20X = 0.1 L
L –> ml = kali 1000 ml
0,1 L . 1000 ml =100 ml

38
Q

You must perform restriction digestion of the vector pBR322 (4.361 bp) with the enzymes BsmI (cuts at 1353)
and BsgI (cuts at 1650). What will be the size of the largest fragment

husk : (besar - besar )+ paling kecil ( untuk menemukan paling besar the fregment.

A

4064 bp

pBR322 (4.361 Bp- Bsgl 1650) + Bsml 1353 = 4064 bp

39
Q

The stock concentration of the enzymes BsmI and BsgI is 2 U/µL. You want to incubate 1 µg of pBR322
(stock concentration 123 ng/µL) together with 2 U of each enzyme. How many microliters of each
enzyme will there be? Total volume of reaction is 20 µL

A

How many microliters of each
enzyme1 µL

since they ASK about the enzyme, so focus just with enzym without rumus C1.V1 / C2.V2

langsung saja enzym / enzym

Enzym Bsml : 2U
Enzym 1 µg of pBR322 : 2 U

2U/ µL / 2U/ µL = 1 µL

40
Q

How many microliters of pBR322 must you pipette so that you incubate 1 µg?

husk: 1µg —> 1000 ng

A

8.13 µL

1 µg = 1000 ng
stock concentration incubate 1 µg is 123 ng/µL

1000 ng / 123 ng/µL = 8.13µL

41
Q

Now you will set up a PCR reaction with genomic DNA as a template. You will amplify the gene agr from
the bacterial strain Staphylococcus aureus. Primers for agr have a stock concentration of 20 µM.
The total volume of the PCR reaction is 20 µL and final concentration of primers in the PCR reaction is 1 µM. How many microliters of primers do you need to add?

A

1µL

Primers stock concentration (C1) : 20 µM
Final Concentration (C2) : 1 µM
Total Volum (V2) : 20 µL

Rumus:
C1.V1 / C2.V2

V1: 1µM. 20 µL / 20 µM = 1 µL

42
Q

what is :
- linearized plasmid
- circular plasmid
- PCR product
- newly constructed plasmid (vector + PCR product), which has been restriction cut.

A
  • linearized plasmid : Only one High size bond )
  • circular plasmid : several bonds, High and Low size
    -PCR Product : Low size with one bond
    -Newly Constructed : paling ujung akhir yang vector (1 paling atas)dan PCR product (1 paling bawah) dalam satu column garis