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WorksheetsREVISION CHAPTER 7 & 8 SB015
Total questions: 70
Worksheet time: 52mins
(CHAPTER 7)
Which of the following statements about sickle cell anemia are FALSE?
It is caused by a single base substitution
It is caused by a base substitution on the sixth tripled code for β polypeptide of haemoglobin
Nucleotide with thymine is substituted by nucleotide with adenine base
Glycine amino acid is substituted by valine
Notice the abnormal 23rd chromosome. This individual suffers from which chromosomal genetic disorder?
Klinefelter Syndrome
Down Syndrome
Cri-du-Chat Syndrome
Andrew Syndrome
A person with Cri-du-Chat Syndrome is
missing chromosome # 5
trisomy 13
trisomy 21
missing part of chromosome # 5
Klinefelter Syndrome is caused by the presence of
an extra Y-chromosome (XYY)
an extra X-chromosome (XXY)
a missing X-chromosome ( _X)
trisomy 21
Plant A (2n=10) mates with plant B (2n=12) produce a sterile hybrid which evolved to a fertile hybrid. What is the chromosome number of a fertile hybrid?
33
44
11
22
What type of gene mutation shown in FIGURE 1.
Normal DNA
AAC TGA GGC CTT CTT
Leu Thr Pro Glu Glu
Abnormal DNA
AAC TGA GGC CAT CTT
Leu Thr Pro Val Glu
FIGURE 1
Base Substitution
Base Deletion
Base Insertion
Base inversion
What type of gene mutation occurred when a pair of nucleotide is REPLACED with another pair of nucleotide in the DNA sequence.
Base deletion
Base substitution
Base insertion
Base inversion
Which of the following statements is NOT TRUE about missense mutation.
mRNA codon is changed
Encoded for different amino acid
Will caused disorder which is sickle cell anemia
Produce non functional protein
Which type of mutation is normally more severe because they affect more than one codon?
Point Mutations
Substitution mutations
mutations where A's are replaced with T's
Frameshift Mutaitons
Which of the following is TRUE about base inversion? (answer can be more than ONE)
Two or more nucleotides are inverted
Small region of DNA breaks off and rotates 180o before rejoin
One type of gene mutation
Changes of amino acid sequence
Mutation is a spontaneous change that occur in ...... (answer can be more than ONE)
Amount of DNA
Arrangement of genes
Structure of DNA
Structure of protein
1. Which ONE is NOT a type of gene mutation?
Base substitution
Base duplication
Base deletion
Base inversion
1. Which of the following are types of chromosomal aberration ? (answer can be more than ONE)
Insertion
Inversion
Deletion
Substitution
1. Induced mutation occur due to
DNA damage caused by mutagen
Natural changes in DNA structure
Errors during DNA replication
Non-disjunction of chromosome
1. Spontaneous mutation occur due to ... (answer can be more than ONE)
DNA damage caused by mutagen
Natural changes due to chemical factor
Errors during DNA replication
Non-disjunction of chromosome
1. Which of the following is NOT physical mutagen?
UV rays
Gamma rays
Colchicine
X-ray
Chromosomal mutation is a change in ____________ (answer can be more than ONE)
Amount of DNA
Arrangement of gene
Number of chromosome
Structure of chromosome
Which ONE is NOT a type of chromosomal aberration?
Deletion
Insertion
Inversion
Translocation
chromosomal alteration occur due to ______________.
crossing over
nondisjunction
anaphase I
prophase II
Which of the following is TRUE about chromosomal aberration? (answer can be more than ONE)
Change in structure of chromosome
Change in number of chromosome
Produce abnormal gamete n-1
One segment of chromosome is removed
Which of the following are effect of aneuploidy? (answer can be more than ONE)
Monosomy 21 (2n-1)
Trisomy 21 (2n-1)
Klinefelter syndrome (44+XXY)
Turner syndrome (44+XO)
The following karyotype illustrates an example of which syndrome?
Turner Syndrome
Klienfelter Syndrome
Edward Syndrome
Down Syndrome
Which of the following is NOT TRUE about non-disjunction?
Turner syndrome is caused by non-disjunction.
Non-disjunction is a point mutation.
Non-disjunction can cause polyploidy.
Non-disjunction occurs during meiosis I or meiosis II.
Hybridisation of plant M (2n=16) with plant N (2n=20) produces a sterile plant P. Non-disjunction occur in plant P produces plant Q.
What is the number of chromosomes in plant P and Q?
P(n+n=18); Q(2n=36)
P(n+n=36); Q(2n=36)
P(n+n-36); Q(2n=72)
P(n+n=72); Q(2n=72)
Which of the following is TRUE about genetic abnormalities and their causes?
Sickle cell anaemia – Base insertion of haemoglobin gene
Down syndrome – Monosomy 21
Klinefelter syndrome – An extra of X chromosome
Turner syndrome – 44 autosomes and XY chromosome
Which of the following could result in polyploidy?
(The answer can be more than ONE)
Random segregation
Non-disjunction
Treatment with colchicine
Hybridization
(CHAPTER 8)
An enzyme that cut DNA molecules at a specific “recognition site” is called
DNA polymerase
Restriction enzyme
RNA polymerase
DNA ligase
Which of the following terms describes the process of the uptake of recombinant plasmids into bacteria cells?
amplification
insertion
transformation
screening
The cell that able to receive recombinant DNA for cloning purpose known as
vector
target DNA
plasmid
host cell
To produce desired protein _____________ should be taken from any source like human, animal and plant.
recombinant DNA
Target DNA
plasmid
ligase
Arrange a correct sequence of process in production of cDNA.
I. Reverse transcriptase used to synthesis single stranded cDNA complementary to mRNA
II. Remove mRNA
III. Extract mRNA from cell
IV. Use DNA polymerase to synthesis second stranded cDNA
I, II, III, IV
III, IV, II, 1
III, I, II, IV
II, III, I, IV
Which of the characteristic is NOT belong to host cell
Able to accept recombinant DNA
Able to accept target DNA
Able to maintain structure of recombinat DNA
Able to express the gene of interest
Which gene is used as selectable marker in a plasmid?
Ampicillin resistance gene
Plasmid resistance gene
Salmonella resistance gene
Penicillin resistance gene
Which of the following is a correct sequence of process in recombinant DNA technology?
I. Isolation of target DNA & vector
II. Introducing recombinant DNA into host cell
III. Fragmentation of target DNA & vector by restriction endonuclease
IV. Ligation of DNA fragment into vector Culturing in host cell to get desired products.
I, II, III, IV
I, III, IV, II
I, II, IV, III
I, IV, II, III
Which technology below would probably be the most important to a person who had diabetes and had to take insulin every day?
using recombinant DNA to produce human hormones from bacteria
testing parents for genetic disorders before they have children
engineering fruits and vegetables that resist insects and other pests
developing ways to identify criminals through DNA fingerprinting
When DNA is manipulated and moved from one source to another it is known as
genetic engineering
electrophoresis
gene therapy
GMO
Which ONE is not used in recombinant DNA technology?
Cloning vector
RNA polymerase
Target DNA
DNA ligase
Which is true about plasmids?
Plasmids can be cut at specific sequences called restriction sites.
Plasmids contain a promoter sequence that defines where transcription begins.
Plasmids can contain an antibiotic resistance gene.
All of the statements are true about plasmids.
Which colonies of bacteria formed after transformation using cold calcium chloride followed by heating?
Bacteria with recombinant plasmid
Bacteria with non-recombinant plasmid
Bacteria without plasmid
Bacteria with two types of plasmid
The process of putting recombinant DNA into an organism is called....
Transformation
Adapter Ligation
Amplification
Screening
What is the purpose of amplification?
To screen for recombinant bacteria
To join the DNA fragment with open plasmid
To produce multiple copies of the desired gene
To prevent the bacteria from reproducing
In gene cloning, antibiotic medium are used to
Selects for plasmids containing particular DNA fragments
Selects for bacteria containing recombinant plasmids
Selects for bacteria lacking plasmids
Selects against plasmids containing human DNA fragments
Which ONE of the following are INCORRECT about screening?
Can identify transformed bacteria that have recombinant plasmid
Blue colonies shows bacteria consist of non-recombinant plasmid
White colonies shows bacteria consist of recombinant plasmid
Bacteria that have non-recombinant plasmid has disrupted lacZ gene
What is the most logical sequence of steps for splicing foreign DNA into a plasmid and inserting the plasmid into bacterium?
I) Transform bacteria with recombinant DNA molecule.
II) Cut the phosphodiester bond of both DNAs using restriction enzymes.
III) Extract plasmid DNA from bacterial cells.
IV) Hydrogen bond formed between the plasmid and DNA fragments.
V) Use ligase to seal plasmid and target gene
I, II, IV, III, V
III, II, IV, V, I
II, III, V, IV, I
III, IV, V, I, II
What is benefits of insulin produced by genetic engineering? (The answer can be more than ONE)
The gene used is from human
Insulin produced similar to human insulin
Non allergic
Cheaper and can reproduce in large amount
Why insulin gene obtained from mRNA rather than DNA?
mRNA doesn’t contain intron
mRNA consist code for specific target gene
Difficult to find one gene among all the genes in the nucleus
mRNA is easy to isolate
Arrange the correct sequence 1-5 of the steps below.
1 – Adapter ligation by adding restriction sequence at both ends of cDNA
2 – mRNA isolated from human pancreatic cell
3 – Reverse transcriptase form first cDNA
4 – Ribonuclease degraded mRNA
5 – DNA polymerase form another strand of cDNA forming double stranded DNA
1, 2, 3,4, 5
2, 3, 4, 5, 1
2, 4, 3, 1, 5
1, 3, 4, 5, 2
What happen to white colonies bacteria after screening?
Human insulin is extracted & purified from E.coli
The bacteria placed in fermentation tank to multiply
Insulin gene will be expressed via transcription and translation to produce insulin
Bacteria is extracted from fermentation tank
An organism in which foreign genes have been incorporated is called a __________.
Recombinant organism
Transgene
Recombinant
Transgenic organism
Which folowing process is catalyzed by a reverse transcriptase?
DNA to mRNA
DNA to cDNA
mRNA to DNA
cDNA to mRNA
Why insulin produced by genetically engineered E.coli is more advantage to insulin obtained from animal sources?
Animal insulin is are difficult to purify
Insulin from recombinant DNA technology has much higher activity level
It contains human gene, reducing the chances of an allergic response
Animal insulin has shorter life span than insulin produced using recombinant DNA technology
In order to obtain the gene of interest from the DNA of organism
The DNA must be cut only at one site near the gene of interest.
The DNA must be cut at two sites to get DNA fragment that contains gene of interest.
The DNA must be cut at any two sites to get any DNA fragments from the organism.
The DNA must not be cut at all, so that the gene of interest is retained.
Name the organism that is commonly used as a host cell.
Escherichia coli
Human pancreatic cell
Agrobacterium tumefaciens
Thermus aquaticus
Which of the following are characteristics of the host cell?
(The answer can be more than ONE)
Receives DNA recombinant through the transformation process.
Maintains the structure of the recombinant DNA from one generation to another.
Amplifies the gene product from recombinant DNA.
Consist of ori that unable them to multiplied by binary fission.
What happen to the transformed bacteria if ampR gene in the plasmid is disrupted?
Died
Unable to hydrolysed X-gal sugar
Reproduce to form blue colonies
Reproduce to form white colonies
Which one of following is TRUE about screening process?
(The answer can be more than ONE)
Transformed bacteria is plated on a medium with ampicillin and X-gal sugar
Bacteria is plated on a medium with cold calcium chloride followed by heating
To identify which bacteria has recombinant plasmid
To identify which bacteria do not have any plasmid
Identify the bacteria colonies appeared white after screening process.
(The answer can be more than ONE)
Bacteria with non-recombinant plasmid have functional lacZ gene
Bacteria with recombinant plasmid able to produce β-galactosidase
Bacteria with recombinant plasmid unable to hydrolyse X-gal sugar
Bacteria with recombinant plasmid have non-functional lacZ gene
Identify the bacteria colonies appeared blue after screening process.
(The answer can be more than ONE)
Bacteria with non-recombinant plasmid have functional lacZ gene
Bacteria with recombinant plasmid able to produce β-galactosidase
Bacteria with non-recombinant plasmid able to hydrolyse X-gal sugar
Bacteria with non-recombinant plasmid have non-functional lacZ gene
Which of the following is TRUE regarding screening by using ampicillin and X-gal?
(The answer can be more than ONE)
Antibiotics will prevent the growth of bacteria without plasmid.
Bacteria cells with recombinant plasmids are resistant to antibiotics and unable to hydrolyse X-gal.
Bacteria cells which contain non-recombinant plasmids are resistant to antibiotics and unable to hydrolyse X-gal.
Bacteria with recombinant plasmids will appears blue colonies.
Why the same restriction enzyme must be used to cut the plasmid and the human insulin producing cell.
(The answer can be more than ONE)
To produce compatible sticky ends
To allow the formation of hydrogen bonds between the complementary sticky ends of both DNA molecules
To allow the formation of phosphodiester bonds between the complementary sticky ends of both DNA molecules
To allow the formation of hydrogen bonds by DNA ligase
An insulin gene is inserted in between ampicillin gene. The recombinant DNA is then transformed into host cell. Will the host cell survive in a medium containing ampicillin?
Yes
No
The statement below shows steps in manipulating bacterial cells to produce insulin. What is enzyme X, process Y and Z?
Gene for human insulin and open plasmid are joined by enzyme X.
Recombinant DNA is inserted into host cell by process Y.
Recombinant bacteria will be identified by process Z.
X: Restriction enzyme, Y: Transformation, Z: Screening
X: DNA ligase, Y: Insertion, Z: Amplification
X: DNA ligase, Y: Transformation, Z: Screening
X: Restriction enzyme, Y: Insertion, Z: Screening
The unpaired nucleotides produced by the action of restriction enzymes are referred as
sticky end
base sequence
single strands
restriction fragments
