MCQ ppq Flashcards
purine
double ring
pyrimidine
single ring
who was nucleic acid (DNA) first discovered by?
Friedrich Miescher
watson and crick
discovered the structure of DNA,
although its arguable that rosalind franklin discovered it first
ribonuclease proteins
ribosome
spliceosome
genetic code
crick and brenner
-studies of the rII region of bacteriophage T4, …
semi conservative replication
meselson and stahl
-heavy and light nitrogen
mbp
megabase pair
a megabase par is worth the same as
1 million base pairs
upiquitylation
The addition of ubiquitin to a substrate protein
Histone N-terminal tails can be modified by
1 acetylation
2 methylation
3 phosphorylation
4 ubiquitylation
kilobp
1 thousand kilobase
introns can occur in the
5’ UTR, coding region, 3’ UTR
where does DNA replication occur
mitochondria - has its own dna separate to the genome
Nucleus
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)
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
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?
Two bases will give 6 x 6 = 36 possible codons which is sufficient for all the possible amino acids.
2
A circular bacterial chromosome contains 2 Mb DNA. How long would DNA replication take if replisome speed is 1 kb s-1?
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
genes in a bacterial operon are
transcribed together, does not contain introns, share a promotor and have a related function
In Escherichia coli the lac repressor
binds to the lac operator, preventing transcription
dicer
cleaves pre-microRNA to generation miRNA which assembles with a set of proteins to form the RISC complex
what proportion of the genome has single nucleotide differences between any two human individuals?
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
transitions r more likely sub mutations then transversions because
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)
homologous recombination can lead to
error free repair of double strand breaks and meiotic crossovers
recombination occurs during
meiotic prophase 1
a single recombination event between a circular bacterial plasmid and a region of homology on a bacterial chromosome leads to
integration of the plasmid in the chromosome
non-disjunction during meiosis 2 will result in..
two haploid gametes, one gamete with an extra chromosome copy, and one gamete missing a copy of one of the chromosomes
marinas disease is an example of
a disease that has expressivity
what indicates that three genes are linked
common, uncommon and rare classes
what are the only species not used as model organisms
homo sapiens
paralogues
genes which arise after duplication event within a genome
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?
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
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?
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.
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?
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.
Agrobacterium tumefaciens is useful for genetic engineering because
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.
double crossover gametes are always of the
lowest frequency
single crossover gametes are of
medium frequency
parental
highest frequency
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
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.
rRNA is synthesised in the
nucleolus
what can be spliced
introns and exons
deamination of 5methylcytosine forms
thymine
what must occur to form thymine from 5methylcytosine
deamination
is X inherited from father to son
no- x-linked diseases can not be past from father to son
non synonymous substitutions
alter amino acid sequence–> could code for non functional protein and therefore affect cells
synonymous substitutions
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
when a lac operon binds to the promoter, there will be a
lack of expression of the lac operon fine under ALL circumstance e.g. even when glucose is present
what other organelles can synthesise proteins
mitochondria
next generation sequencing NGS detects
copy number variations
RFLP
uses hybridisation to detect specific DNA restriction fragments in genomic DNA
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
0.01
replisome
carries out replication of DNA e.g. unwinding DNA etc
how many chromosomes do each cell have
23 pairs, therefore 46
CRISPR is a
histone modifying complex
