VOORBEELD VRAGEN COO COMPUTER TOETSEN GENETICA!
What is the most common result of the insertion of a transposable element into a bacterial genome?
a. Change the level of gene expression
b. Induction of homologous recombination
c. Inactivation of a gene
d. Cause drug resistance
e. Movement of a gene sandwiched between two transposons
In bacteria genes are close together and have no introns; therefore it is highly likely that when a transposon hops into the geneome, it will insert into a gene and inactivate it. In order for a transposon to change the level of gene expression (a), it must insert close to a gene without disrupring. Many transposons do not cary drug-resistance genes and therefor do not cause drug resistance when they are insterted into the genome (d). Induction of homologous recombination (b) or movement of a gene sandwiched between two transposons (e) requires that two independent transposons events involving the same or similar transposons occur.
Correct Answer: Inactivation of a gene
During mutagenic treatment with ethylmethanesulphonate acid (EMS), an guanine become ethylated and will basepair with thymine. What will happen?
a. AT to CG
b. GC to AT
c. TA to AT
d. GT to CA
The ethylated G pairs with T. Effectively, it therefore causes a G to A transition or a change of an GC to an AT basepair.
Correct Answer: GC to AT
What is a provirus?
a. An early evolutionary form of a virus
b. A non-enveloped single-stranded RNA virus.
c. An enveloped single-stranded RNA virus
d. Viral DNA that has integrated into the host chomosome
e. Double stranded DNA virus which produces immediately after infection many virus particles inside the host cell and lyses the host in order to cause new infections.
Correct Answer: Viral DNA that has integrated into the host chomosome.
Transfer of chmososomal genes between E. coli cells occurs only when:
a. The donor cell carries the F+ factor on the bacterial chromosome
b. The receiving cell carries the F+ factor on the bacterial chromosome
c. The donor cell is a male bacterium and the receiving cell a female
d. Both cells have takenup a recombination plasmid
Correct Answer: The donor cell carries the F+ factor on the bacterial chromosome.
Which of the following DNA repair processes can occur only immediately after DNA has been replicated?
a. Crosslink repair
b. Nucleotide excision repair
c. Base excision repair
d. Mismach repair
e. Double strand break repair
Mismach repair. Mismaches occur as e result of replication errors, and thus only on the newly synthesised strand. In mose cells, their repair is thought to require recognition of a newly synthesised DNA strand by nicks left in the sugar-phosphate backbone. These are sealed soon after replication, so mismached repair must occur in the short interval between passage of the replication fork and sealing of the sugar-phosphate backbone. Alle the other repair processes repair damag that may occur at any time during a cell’s life and to either strand of the DNA. Thus their repair processes do not depend on recognizing newly synthesised DNA strands.
Correct Answer: Mismache repair.
Mismache repair of DNA:
a. requires specialized proteins that act before the DNA is replicated
b. requires an undamaged template strand
c. repairs the leading strand to mach the lagging strand
d. makes replication a 100000 times more accurate
e. defects cause xeroderma pigmentosum
A is incorrect, as mismache repair requires specialized repair proteins that act after the DNA is replicated. C is incorrect: it does not repair the leading strand to mach the lagging strand. D is incorrect: mismach repair makes approximatly 100 not 100000 times more acurate. E is incorrect: xerodema pigmentosum is caused by a defect in another type of DNA repair - the nucleotide excision repair system.
Correct Answer: requires an undamaged template strand
a. are found only in eucaryotes
b. can move by either the cut-and-paste mechanism or by a mechnism requiering an RNA intermediate
c. include LINE-1, Alu and Tn10 sequences
d. need to provide their own reverse transcriptase
Retrotransposons move only through an RNA intermediate (b is incorrect). They include LINE-1 and Alu sequences (but not Tn10, wich is a bacterial transposon) and do not necessarily need to provide their own reverse transcriptase so als similar that does encode reverse transciptase is also present in the genome (so C and D are incorrect)
Correct Answer: are found only in eucaryotes
Wich of the following would contribute most to succesfull exon shuffling?
a. Shorter introns
b. a halpoid genome
c. longer exons coding for many domains
d. introns that contain regions of similarity to one another
e. non-modular proteins
Exon shuffling is facilitated by long introns (thus A is incorrect) between short exons that each code for one protein domain (thus C is incorrect). Since exon shuffling can occur via recombination between introns, introns with rewgions of similarity to one another will facilitate shuffling. A haploid genome will probably be less prone to exon shuffling than a diploid genome (thus B is incorrect) because having two copies of each gene allows an organism to keep a backup copy of the gene while it shuffles the other. Exon shuffling is possible only because many proteins are modular, composed of short, folded domains that have discrete functionsl properties (thus E is incorrect)
Correct Answer: introns that contain regions of similarity to one another.
Scientists produce large quantities of RNA by transcription in vitro rather than by expression of cloned genes in vivo primarily because:
a. In vitro transciprtion removes the need to clone the gene for the RNA into a DNA vector
b. In vivo RNA’s are produced at too low levels.
c. In vitro transcription removes the need of cells for your experiment.
d. In vitro transcription removes the need to purify the RNA produced away from the other RNA molcules in a cell
In order to produce an RNA by trnascription in vitro, you must first colne the DNA of the gene you wish to transcribe in order to get a large amount of pure template. Many RNA’s are produced at high levels in cells. Viral RNA polimerses are able to transcribe RNA in vivo to especially high levels since their purpose is to make high levels of viral proteins.
Correct Answer: in vitro transcription removes the need to purify the RNA produced away from the other RNA molecules in a cell
A prartial diploid of genotype I-P+O+Z+ / I+P+O+Z- will show:
a. No production of b-galactosidase
b. Inducible production of b-galactosidase
c. Constitutive production of b-galactosidase
d. Prodcution of misfolded b-galactosidase
Correct Answer: Inducible production of b-galactosidase
For each of the following sentences, fill in the blanks with the correct word selected from the list below. Use each word only once!
A. operator B. operon C. promoter D. represses E. induced F. constitutivly G. allosteric H. negatively I. positively
1. The genes of the bacterial ________ are transcripbed into single mRNA.
2. Many bacterial promoters contain a region known as an ________, to which a specific gene regulatory protein binds.
3. Genes in which transcription is prevented are said to be ________.
4. Bacterial genes are regulated by small molecules such as tryptophan by the interaction of such molecules with ________ DNA-binding proteins such as the trytophan repressor.
Correct Answer: 1B 2A 3D 4G
In the presence of the repressor molecule and absence of tryptophan, the trp operon is:
Correct answer: Induced
In wich of the following instances can the state of chromatin packing differ?
1. Between different cells
2. In different stages of the cell cycle
3. In different parts of the same chromosome
4. In different members of a pair of homologous chromosomes
Wich are correct?
a. 1 and 2
b. 2 and 4
c. 2 and 3
d. 1, 2 and 4
e. all the above
Correct Answer: E
A partial diploid of genotype I+P+O+Z+Y- / I+P-O+Z+Y+ will show:
a. No b-galactosidase or acetylase production at all
b. Inducible production of b-galactosidase and permease
c. Inducible production of b-galactosidase and constitutive production of permease
d. Inducible production of permease and consititutive production of permease
e. Consitutive production of b-galactosidase and permease
Correct answer: b
Some clones from cDNA libaries can have defects because of the way a cDNA library is consturcted. For each of the cases 1-4, indicate which of the problams a-d you might encounter.
1. The mRNA corresponding to the clone you are looking for was degraded at it’s 5’ end by a nuclease
2. The mRNA corresponding to the clone you are looking for was degraged at it’s 3’ end by a nuclease
3. The 5’ end of the reverse transcriptase product of the gene you’re trying to clone hybridizes to sequences in the middle of the gene
4. The gene you are trying to clone has al ong strech of A’s in the middle of the coding sequence.
a. The 3’ part of the gene will be missing
b. The 5’ part of the gene will be missing
c. An internal fragment of the gene will be missing
d. The gene will be missing from the library
Correct answer: 1B 2D 3B 4A
How many telomers are there in a nondividing human liver cell?
Correct answer: E
a. Active genes
b. Inactive genes
c. More inactive genes than active genes
d. More active genes than inactive genes
e. No genes at all
Hetrochromatin represents an inactive state of that protion of the DNA. However, in different celltypes the genes located in it may be activated (euchromatin)
Correct Answer: B
If mottled coloring of Calico cats is due to X-chromosome inactivation, which of the following statements will be true?
a. Only males are mottled
b. Female cats will have the same color as their mother
c. The mottled color is due to X chromosomes repeatedly swiching back and forth between active and inactive states during development.
d. Cats with identical paterns will be rare.
Since the pattern of X-chomosome inactivation is set up randomly over several days of embryonic development, at a stage where the embryo has quite a few cells, it is unlikely that two cats will inactivate the same X chromosome in exactly the same set of cells. This makes C fals also. The other statements are all filase. If mottled coloring is due to X chomosome insctivation, then the coat color gene involved must be on the X chromosome. Since males have only one X chomosome, wich they receiwe from their mother, and do not undergo X-chromosome inactivation, males will not be mottled (A). Female calico cats receive an X chromosome from the father and from the mother, which may have different forms of the coatcolor gene, and so are not necessarily the same color as their mother (B)
Correct answer: D
A super repressor mutation (LacIS) results in:
a. No effect
b. No translation
c. No transcription
d. Nonfunctional b-galactosidase
Correct answer: C
If there are 12 single-chromatid chomosomes in a cell in G1 of the cell cycle, what is the diploid number of chromosomes for the organsism?
Correct answer: B
Which of the following statemenst is true?
a. The mitotic spindle is composed of actin
b. The mitotic spindle is assembled in late mitosis
c. The mitotic spindle is formed also in procaryotes
d. The contractile ring is composed of only actin.
e. The contractile ring is formed only in animal cells.
The contractile ring is foun only in animal cells. The mitotic spindle is composed of microtubules (not actin) (a incorrect), assembles in early mitosis (b incorrect), and is not formed in procatiotes (which have no microubules) (c incorrect). The contractile ring is made up of actin and myosin (d incorrect)
Correct Answer: E
It is possible to measure the DNA content of a single nucleus. Asume that the mesurements ware made on a large number of nuclei in cells of a growing embryo. The measured DNA levels renged between 3.2 picograms and 6.4 picograms per nucleus. One nucleus has 4.8 picograms DNA. What stage of the cell cycle was it in?
Correct answer: B = S-stage.
Suppose that you are a cell biologist determining the length of time a cell takes to pass trough various stages of the cell cycle. From experiments with radioactive thymidine you know it takes 11 hours to get from the biginning of the S to the beginning of the M stage. Observations in the microscope tell you that M takes 4 hours and the entire cycle takes 22 hours; You already know that G1 and G2 take the same amount of time in this cell type. How long does it take for the DNA of the cell to replicate?
The order of the cell cycle is G1-S-G2-M etc. S + G2 takes 11 hours and M takes 4 hours. Since the whole cycle ( = G1 + S + G2 + M) takes 22 hours, G1 + 11 + 4 = 22. G1 is therefore 7 hours. Since it was stated that G1 equals to G2 in length = 7 hours and S + G2 = 11, S must be 4 hours!
Correct answers: 4 hours!
An A/A B/B C/C invidual is paired with an a/a b/b c/c invidual. Assuming no linkage, what will be the expected frequency of A/A b/b C/c inviduals in the F2 generation?
The F1 is A/a B/b C/c. By definition the F2 is selfing of the F2. The frequency of having homozygous allels like A/A or b/b in the F2 is ¼. The frequency of having heterozygous allels like C/c is 2/4 or 1/2 . Because the genes are not linked. They segregate independently. Therefore the frequency of having A/A b/b C/c inviduals in the F2 is ¼ x ¼ x ½ = 2/64
Correct Answer: B
Classical albinisme is a single-gene disorder caused by the lack of an enzyme necessary for the synthesis of melanin pigments. Enzyme production requires the presence of one normal allele. What progeny and in what proportions are exprected from a normally pigmented woman who has an albino husband and an albino father?
a. none albino to all normal
b. ¼ albino to ¾ normal
c. ½ albino to ½ normal
d. ¾ albino to ¼ normal
e. all albino to none normal
The question stating reveals that albinism is a recessive trait. The father and the husband are therefore homozygous for the defective gene (aa). The mother is phenotypicaly not an albino, but since her father is one, she must be heterozygous for the defective allele (Aa).
Any child she’ll have will ge a mutant allele from the father and will have 50% chance of inheritting a seccond defective allele form her. Statistically her offspring will be half albino, half normal. ½ aa and ½ Aa.
Correct answer: C
Among white human beings, when individuals with straight hair mate with those with curly hair, wavy-haired children are produced. If two individuals with wavy hair mate, what phenotypes and ratios would you predict among their offspring?
a. ¼ straight, ¼ curly, ½ wavy
b. ¼ straight, ½ curly, ¼ wavy
c. ½ straight, ¼ curly, ¼ wavy
d. ½ straight, ½ curly, none wavy
Straight hair is for examle HH and culy hh, thus wavy hair is Hh. Mendelian laws say that in this situation the proportion will be: ¼ HH = straight, ¼ hh = curly and ½ Hh wavy.
Correct answer: A
In mammals, there are two sex chromosomes, X and Y, which behave like homologous chromosomes during meiosis. Nroam males have one X chomosome and one Y chromosome, and normal females have two X chromosomes. Males with an extra Y chomosome (XYY) occasionally are found. Which of the following could give rise to such an XYY male?
a. Nondisjunction in the second division of spermatogenisis, normal meiosis in the mother.
b. Nondisjunction in the second meiotic division of oogenisis, normal meiosis in the father.
c. Nondisjunction in the first meiotic division of spermatogenisis, normal meiosis in the mother.
d. Nondisjunction in the first meiotic division of oogenisis, normal meiosis in the father.
Nondisjunction in one of the other meiotic division will give rise to gametes with two sex chomosomes instead of one. Since the only source of Y chomoromes is the father, the gametes that produce an XYY male must be X (a normal egg) and YY. Nondisjunction in the first meiotic division would result in two XY sperm with no sex chromosome. Nondisjunction in the second meiotic division un contrast, could give rise to YY.
Correct answer: A
Made by: EC from p group.