MCQ ppq

Question 1

purine

Answer 1

double ring

Question 2

pyrimidine

Answer 2

single ring

Question 3

who was nucleic acid (DNA) first discovered by?

Answer 3

Friedrich Miescher

Question 4

watson and crick

Answer 4

discovered the structure of DNA,

although its arguable that rosalind franklin discovered it first

Question 5

ribonuclease proteins

Answer 5

ribosome

spliceosome

Question 6

genetic code

Answer 6

crick and brenner

-studies of the rII region of bacteriophage T4, …

Question 7

semi conservative replication

Answer 7

meselson and stahl

-heavy and light nitrogen

Question 8

mbp

Answer 8

megabase pair

Question 9

a megabase par is worth the same as

Answer 9

1 million base pairs

Question 10

upiquitylation

Answer 10

The addition of ubiquitin to a substrate protein

Question 11

Histone N-terminal tails can be modified by

Answer 11

1 acetylation
2 methylation
3 phosphorylation
4 ubiquitylation

Question 12

kilobp

Answer 12

1 thousand kilobase

Question 13

introns can occur in the

Answer 13

5’ UTR, coding region, 3’ UTR

Question 14

where does DNA replication occur

Answer 14

mitochondria - has its own dna separate to the genome

Nucleus

Question 15

If AUG = Met = start codon
UUG = Leu
AAC = Asn
AGC = Ser
GAG = Glu
UAA = Stop
CCC = Pro
Then a mRNA sequence:
[5´cap]-CCC UUG AUG GAG AGC CCC UAA UUG AAC-[poly(A)-tail]- 3´
...in most eukaryotes will code for peptide(s)

Answer 15

Correct answer is Met-Glu-Ser-Pro. The sequences between the start and stop codon will code for this peptide.
The correct answer is: Met-Glu-Ser-Pro

Question 16

On Venus life has an RNA genetic material with six possible bases. Venusian RNA codes for proteins which contain 25 types of amino acid. What is the minimum number of bases in a codon?

Answer 16

Two bases will give 6 x 6 = 36 possible codons which is sufficient for all the possible amino acids.

2

Question 17

A circular bacterial chromosome contains 2 Mb DNA. How long would DNA replication take if replisome speed is 1 kb s-1?

Answer 17

DNA replication is bidirectional from the origin of replication so the total rate of replication is 2 x 1 kb s-1. The whole chromosome will therefore be replicated in 2 x 106 b / 2 x 103 b s-1 = 103 s.

1000 seconds

Question 18

genes in a bacterial operon are

Answer 18

transcribed together, does not contain introns, share a promotor and have a related function

Question 19

In Escherichia coli the lac repressor

Answer 19

binds to the lac operator, preventing transcription

Question 20

dicer

Answer 20

cleaves pre-microRNA to generation miRNA which assembles with a set of proteins to form the RISC complex

Question 21

what proportion of the genome has single nucleotide differences between any two human individuals?

Answer 21

1 in 1000 or 1^-3

any two individual humans have about 3 million SNP differences, which represents approx 0.1% of a 3Gb haploid genome

Question 22

transitions r more likely sub mutations then transversions because

Answer 22

the base substituted in a transvehrsion has a diff shape (e.g. double ring purine to single ring pyrimidine)and is more likely to be discovered by proof reading or repair mechanism (mis match MutS)

Question 23

homologous recombination can lead to

Answer 23

error free repair of double strand breaks and meiotic crossovers

Question 24

recombination occurs during

Answer 24

meiotic prophase 1

Question 25

a single recombination event between a circular bacterial plasmid and a region of homology on a bacterial chromosome leads to

Answer 25

integration of the plasmid in the chromosome

Question 26

non-disjunction during meiosis 2 will result in..

Answer 26

two haploid gametes, one gamete with an extra chromosome copy, and one gamete missing a copy of one of the chromosomes

Question 27

marinas disease is an example of

Answer 27

a disease that has expressivity

Question 28

what indicates that three genes are linked

Answer 28

common, uncommon and rare classes

Question 29

what are the only species not used as model organisms

Answer 29

homo sapiens

Question 30

paralogues

Answer 30

genes which arise after duplication event within a genome

Question 31

A three-point testcross was conducted involving genes F, G, and H. If the most abundant classes in the F2 are fGH and Fgh and the rarest classes are fgh and FGH, which gene is in the middle?

Answer 31

There are three classes of offspring (common, uncommon and rare) for linked genes in three-point testcrosses.
The parental (common) types are:  fGH and Fgh.
The double recombinant (rarest) types are:  fgh and FGH.
Gene F "switches" between parental and double recombinant, therefore this gene must be in the middle.
The correct answer is: F

Question 32

The alleles for eye colour and body colour are on the X chromosome of Drosophila, but not on the Y. Red eye colour (w+) is dominant to white eye colour (w) and tan body colour (y+) is dominant to yellow body colour (y). What offspring would you expect from a true breeding yellow-bodied, red-eyed female and a tan-bodied, white-eyed male?

Answer 32

Female = w+w+ yy Gametes all = w+ y
Male = w y+ Gametes = w y+ or Y
Therefore offspring will be w w+ y y+ females or w+ y males.
Therefore the offspring will be red-eyed tan-bodied females and red-eyed yellow-bodied males.
The correct answer is: The daughters would be tan-bodied, red-eyed; the sons would be yellow-bodied, red-eyed.

Question 33

Lee Hartwell pioneered genetic analysis of the cell cycle. He used the model organism Saccharomyces cerevisiae (budding yeast). To identify cdc (cell division cycle) mutants in this yeast he screened temperature-sensitive mutants for those that exhibited which of the following phenotypes at the non-permissive (restrictive) temperature?

Answer 33

Budding yeast cdc mutants are defective for a cell cycle function, so become blocked at a point in the cell cycle. Importantly the cells continue to grow, showing that the cell cycle block is not a secondary effect of a defect in general metabolism or macromolecular synthesis.
The correct answer is: The cells grew but became blocked at a point in the cell cycle.

Question 34

Agrobacterium tumefaciens is useful for genetic engineering because

Answer 34

Agrobacterium tumefaciens is a natural genetic engineer which is able to incorporate tumour-inducing genes into the genome of plants. This property is exploited by geneticists to introduce DNA into plants (and fungi).
The correct answer is:
It can be used to introduce DNA into plants and some fungi.

Question 35

double crossover gametes are always of the

Answer 35

lowest frequency

Question 36

single crossover gametes are of

Answer 36

medium frequency

Question 37

parental

Answer 37

highest frequency

Question 38

A three-point testcross was conducted involving genes A, B, and C. If the most abundant classes in the F2 are CaB and cAb and the rarest classes are cab and CAB, which gene is in the middle?

A
B
C

Answer 38

The best way to solve these problems is to develop a systematic approach. First, determine which of the the genotypes are the parental gentoypes. The genotypes found most frequently are the parental genotypes. From the table it is clear that the ABC and abc genotypes were the parental genotypes.

Next we need to determine the order of the genes. Once we have determined the parental genotypes, we use that information along with the information obtained from the double-crossover. The double-crossover gametes are always in the lowest frequency. From the table the ABc and abC genotypes are in the lowest frequency. The next important point is that a double-crossover event moves the middle allele from one sister chromatid to the other. This effectively places the non-parental allele of the middle gene onto a chromosome with the parental alleles of the two flanking genes. We can see from the table that the C gene must be in the middle because the recessive c allele is now on the same chromosome as the A and B alleles, and the dominant C allele is on the same chromosome as the recessive a and b alleles.

Question 39

rRNA is synthesised in the

Answer 39

nucleolus

Question 40

what can be spliced

Answer 40

introns and exons

Question 41

deamination of 5methylcytosine forms

Answer 41

thymine

Question 42

what must occur to form thymine from 5methylcytosine

Answer 42

deamination

Question 43

is X inherited from father to son

Answer 43

no- x-linked diseases can not be past from father to son

Question 44

non synonymous substitutions

Answer 44

alter amino acid sequence–> could code for non functional protein and therefore affect cells

Question 45

synonymous substitutions

Answer 45

will not alter amino acid sequence–> most of the time if the last base is wrong in the codon, then the same a.a. will be coded for

Question 46

when a lac operon binds to the promoter, there will be a

Answer 46

lack of expression of the lac operon fine under ALL circumstance e.g. even when glucose is present

Question 47

what other organelles can synthesise proteins

Answer 47

mitochondria

Question 48

next generation sequencing NGS detects

Answer 48

copy number variations

Question 49

RFLP

Answer 49

uses hybridisation to detect specific DNA restriction fragments in genomic DNA

Question 50

if the fro of males affected with an x-linked recessive condition is 0.1 (1/10), what will be the fro of affected females

Answer 50

0.01

Question 51

replisome

Answer 51

carries out replication of DNA e.g. unwinding DNA etc

Question 52

how many chromosomes do each cell have

Answer 52

23 pairs, therefore 46

Question 53

CRISPR is a

Answer 53

histone modifying complex