LQB385 - Mock Exam Flashcards
M stage
(During mitosis (M stage), the chromosomes are most condensed. This is required during this stage of the cell cycle in order to accurately segregate the sister chromatids to opposite poles of the cell - and later to each of the daughter cells.)
b, d
(Euchromatin is open chromatin, from which genes can be readily transcribed. Housekeeping genes are nearly continuously or frequently expressed. Therefore, euchromatin is rich in housekeeping genes.
In contrast heterochromatin is closed chromatin from which genes are not transcribed)
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(In the ribosomes, messenger RNA (mRNA) is translated into protein with the help of transfer RNA (tRNA). The ribosomes themselves contain ribosomal RNA (rRNA). DNA is present in the nucleus, but not in the ribosomes. Lipids are also not present in ribosomes)
4
(Introns are present between every exon. This gene has 5 exons. Therefore, its intron-exon structure is as follows.
Exon 1 - intron 1 - exon 2 - intron 2 - exon 3 - intron 3 - exon 4 - intron 4 - exon 5.
This gene has 4 introns)
C
(Remember that the 5’ UTR is present at the leftside (i.e., 5’ side) of the transcript. At the genomic level, its start can be found at the start of transcription. The 5’ UTR (untranslated region) ends where translation starts, hence at the nucleotide upstream/5’/before the translation start site.)
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(There are two correct answers here: microsatellites and minisatellites, which vary in length)
f.
(All of the mentioned components are important for a PCR to work, except DNA ligase, which is used in cloning procedures in order to ligate (‘connect’) compatable DNA ends together)
c.
(Primers are typically 19-25 nucleotides in length. This length is about optimal when balancing specificity (for binding to a plasmid of genomic DNA), the melting temperature during the PCR and price (when we order primers, we need to pay for them per nucleotide))
e.
Digestion refers to the cutting of DNA (for example by restriction enzymes), while all other terms refer to phases during PCR amplification
b.
(To generate cDNA (complementary DNA), we use mRNA as a template in a reverse-transcription PCR)
b.
(Height is a continuous variable that can be quantified, hence it is a quantitative trait)
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(Julia’s plasmid is 6000 base pairs long. Plasmids are circular DNA molecules. When Julia’s plasmid is cut with a restriction enzyme that cuts the plasmid only once, the plasmid is linearized. Therefore, this reaction produces 1 DNA fragment of 6000 base pairs.
When both enzymes are used, the circular plasmid is cut twice, producing 2 DNA fragments of 500 bp and 5500 bp (together equal to the total length of the plasmid, which is 6000 bp))
e.
(For cloning a DNA insert into a vector (or plasmid), we can digest each with rescrition enzymes, which specifically cut the plasmid and insert. In order to clone the insert into the vector, we require DNA ligase. Note, though, that this will only work if the ends are compatible, in order words, 2 blunt ends (without 3’ or 5’ overhangs) can be ligated together OR 2 sticky ends (with compatable 3’ or 5’ overhangs) can be ligated together)
e.
(Sanger sequencing refers to the traditional sequencing, involving extension and ‘random’ termination at each nucleotide, so that the mixture of end-labelled products can be loaded on a gel and visualised in ordee to the determine the sequence of the template)
b.
(CRISPR is used or has the potential to be used a wide range of areas in society that involve DNA/genetic material that can be edited, including all of the ones mentioned. Home building does not involve this)
b.
(Most cells in living organisms are in G1 or G0 or the cell cycle. At that stage, they carry two copies of each gene (in most circumstances). During S phase, the number of copies increases from 2 to 4. During the next stage, G2, there are still 4 copies. Subsequently, during mitosis, there are still 4 copies too. It is only after cytokinesis (the physical division of the cell into two daughter cells) that there are 2 copies again. At that point, cells are in G1 phase of the cell cycle again)
3 billion basepairs
All are correct
(All are important. The melting tempeture is important to consider during the PCR for chosing the annealing temperature. the GC content (i.e., the number of Cs and Gs in the primers) is important, because it influences the melting tempeture. Primer length also influences the melting temperature.
Primer specificity is important too. In most cases, we want the primer sequence to be 100% complimentary to our target sequence. Primer length also influences specificity)
b.
(The correct answer is (b).
(a) 3C (Chromatin Conformation Capture) identifies DNA-DNA interactions.
(b) ChIP (Chromatin Immunoprecipitation) identifies DNA-protein interactions.
(c) FISH (Fluorecence in situ hybridisation) enable visualisation of loci in the DNA/on chromosomes using a fluorescently-lablled DNA probe.
(d) Next generation sequencing aims to sequence the genome)
Nucleosome
Histone
(Chromatin remodelling involves modification of histone ‘tails’ (e.g., through (de)methylation and (de)acetylation). This affects the nucleosomes and how open or closed the chromatin is.
DNA ligase and exons are unrelated to this.
While DNA replication can only occur in open chromatin (euchromatin), DNA replication errors are independent of chromatin remodelling)
