Private Study Qs

Question 0

Whats the definition of:

pH

Answer 0

pH = conc of H+ ions in solution

Question 1

What is the equation for calculating:
pH
[H+]

Answer 1

pH = -log [H+]

[H+] = 10 (to the power of - pH)

Question 2

Whats the definition of pK

Whats the equation for pK

What does it mean in pK is high vs low

Answer 2

Extent to which acid dissociates (strength of an acid)

pK = -log [Ka] (-log of acid dissociation constant)

High = weaker tendency for acid to dissociate
Low = Stronger tendency for acid to dissociate

Question 3

Whats the Henderson-Hasselbach’s equation

Whats it useful for

Answer 3

pH = pKa + log ([A-] / [HA])
A= conjugate base
HA = acid

estimating pH of a buffer solution
finding the equilibrium pH in acid-base reactions
(used to calculate the isoelectric point of proteins)

Question 4

Why do metabolically active tissues cause a localised fall in blood pH?

Answer 4

Metabolically active tissues make a lot of acidic substances that will lower the pH of the
blood close to them. Examples include lactate, H+, CO2

Question 5

Explain how rapid breathing (hyperventilation) can cause an increase in blood pH.

Answer 5

During hyperventilation the [CO2] (or partial pressure) in the lungs decreases as you are blowing off the gas.
Therefore dissolved CO proportional to the concentration (partial pressure) of the gas in the lungs.
Effectively removing carbon dioxide from the plasma means that the equilibrium position moves so that reaction 1 and 2 reverse to reform carbon dioxide.
This has the effect of removing H+ to rise above the normal 7.4 value.

Question 6

What effect would a pulmonary obstruction have on blood pH?

Answer 6

A pulmonary obstruction causes an increase in [CO2] in the lungs.
Therefore [CO2] in the blood increases, shifting the equilibrium towards production of H+ and HCO3-
Therefore, [H+] increases

Question 7

Aspirin is a weak acid with a pKa of 3.5. It can be absorbed into the blood stream through cells which line the stomach and small intestine. In order to pass through the plasma membrane of these cells the molecule must be in its protonated form.

Why would a deprotonated (ionised) form of aspirin not pass readily through the plasma membrane?

Given that the pH of the stomach is about 1.5 and the pH of the small intestine is about 6, is more aspirin absorbed from the stomach or the small intestine?

Answer 7

The carboxyl group (COOH) is the group that is able to ionize.
The deprotonated form would not pass through a membrane as it is charged.
The protonated form is less polar as it is uncharged and so is more likely to pass through the hydrophobic (non-polar) environment of the cell membrane.

In the stomach, the carboxyl group will be protonated (pH is below the pK of the carboxyl group).
Therefore, the aspirin molecule will be uncharged (non-polar) and will be easily absorbed.
In the small intestine, the carboxyl group will be deprotonated (ionised) and therefore, the aspirin molecule will be charged.
This means that it is less likely to be absorbed.

Question 8

List 3 positively charged amino acids

Answer 8

Lysine
arginine
histidine.

Question 9

List 2 negatively charged amino acids

Answer 9

Glutamate

aspartate

Question 10

In an alpha helix the hydrogen bonds:

A) are roughly parallel to the axis of the helix.
B) are roughly perpendicular to the axis of the helix.
C) occur mainly between electronegative atoms of the R groups.
D) occur only between some of the amino acids of the helix.
E) occur only near the amino and carboxyl termini of the helix.

Answer 10

A

Remember that the R groups are not involved in the determination of protein secondary structure.

Question 11

In an alpha helix, the R groups on the amino acid residues:
A) alternate between the outside and the inside of the helix.
B) are found on the outside of the helix spiral.
C) cause only right-handed helices to form.
D) generate the hydrogen bonds that form the helix.
E) stack within the interior of the helix.

Answer 11

B

In a DNA double helix the bases are stacked within the middle of the helix.

Question 12

List the differences between globular and fibrous proteins.

Answer 12

Fibrous:
Long stands and sheets   
Little or no tertiary structure  
Single type of repeating secondary structure 
Usually insoluble     
Often have a structural role    

Globular:
Compact shape 
Complex tertiary structure 
Several types of secondary structure 
Usually soluble 
Many different roles

Question 13

Studies to find drugs that relieve Alzheimer’s Disease and other amyloidoses have found that small, hydrophobic aromatic compounds can block the formation of amyloid fibrils.
Suggest how these molecules could work.

Answer 13

Amyloid fibrils are formed from proteins that contain beta-sheets as part of their normal structure.
In disease states the fibrils form because beta sheets from one protein molecule associate with the same region from another molecule.
This continues, resulting in the formation of large insoluble aggregates
that form the fibril.
It has been shown that aromatic amino acids are important in stabilising the interaction of the beta-sheets between adjacent proteins.
Therefore, aromatic compounds have been shown to interfere with amyloid fibril formation by blocking the interaction of the aromatic side chains.
One stain for amyloid fibrils, Congo red, is such a molecule and has been shown to block amyloid fibril production in cells.
Of course, one drawback is getting such molecules into the cell in the first place.

Question 14

Use your knowledge of haemoglobin function to explain whether you would advise a free diver (someone who swims underwater on a single breath of air) to hyperventilate prior to a dive

Answer 14

You would advise them not to bother!
Hyperventilation will reduce the concentration of CO2 in the lungs and blood (blown off) but will not significantly affect the concentration of oxygen present in the blood.
Under normal conditions haemoglobin in arterial blood is already saturated with oxygen.
However, hyperventilation can actually decrease the amount of oxygen available.
The removal of CO2 will remove protons because CO2 is in equilibrium with the carbonic acid/bicarbonate buffering system.
Removing protons causes the blood pH to rise.
This would increase the affinity of haemoglobin for oxygen.
The net effect would be that less O2 could be delivered to the tissues.
Elevated levels of CO2 in the blood also results in an urge to breathe. Therefore, reducing CO2 concentrations suppresses this urge and could mean that the diver may lose consciousness and drown.

Question 15

Explain why acidosis can precipitate a sickle cell crisis.

Answer 15

Acidosis causes a shift on Hb conformation to favour the T state (low affinity).
Under these conditions HbS is more likely to form polymerise, thus triggering a sickle cell crisis.

Question 16

Enzyme X exhibits maximum activity at pH = 6.9. X shows a fairly sharp decrease in its activity when the pH goes much lower than 6.4. One likely interpretation of this pH activity isthat:
A) a Glu residue on the enzyme is involved in the reaction.
B) a His residue on the enzyme is involved in the reaction.
C) the enzyme has a metallic cofactor.
D) the enzyme is found in gastric secretions.
E) the reaction relies on specific acid-base catalysis.

Answer 16

B
The pKa of Histidine is about 6.8 so changing the protonation state of this amno acid is likely to have a profound effect on the activity of the enzyme.

Question 17

How can a deficiency of niacin in the diet cause a reduction in the activity of NAD and NADP dependent enzymes?

Answer 17

Niacin (or nicotinic acid) is a vitamin that is synthesised from the amino acid tryptophan.
However, humans cannot synthesise sufficient quantities of this to meet it’s requirements and so most niacin must be derived from dietary sources.
Niacin is needed in the body for the formation of nicotinamide, a component of the coenzymes nicotinamide adenine dinucleotide (NAD) and nicotinamide adenine dinucleotide phosphate
(NADP).
A deficiency in niacin affects all NAD(P) dependent dehydrogenases and can cause the disease pellagra.

Question 18

Define feedback inhibition

Answer 18

The name for the regulatory mechanism in which the final product inhibits activity of the first step.

Question 19

Define isoenzyme

Answer 19

Multiple forms of homologous enzymes found within an organism.

enzymes that catalyse the same reaction but have a different amino acid sequence.
They usually have different kinetic properties.

Question 20

Define zymogen

Answer 20

Enzymes activated by proteolytic cleavage.

Question 21

Which is The less active conformational form of an allosteric enzyme.

Answer 21

T state

Question 22

What is The shape of the kinetic plot of an enzyme that exhibits cooperative binding.

Answer 22

Sigmodal

Question 23

What enzymes that catalyze protein phosphorylation.

Answer 23

Protein kinases

Question 24

What enzyme catalyses Removal of protein phosphates

Answer 24

Protein phosphatases

Question 25

Vitamin required for the clotting pathway.

Answer 25

Vitamin K

Question 26

Modified amino acid found in prothrombin, necessary for

activation.

Answer 26

Carboxyglutamate

Question 27

Hexokinase and glucokinase are isoenzymes that catalyse the phosphorylation of glucose to glucose-6-phosphate. They have Km values for glucose of 0.1mM and 5mM respectively.

Which has a higher affinity for glucose?
Why is this physiologically relevant?

Answer 27

Hexokinase

Under fed and fasted conditions the concentration of glucose in the blood is sufficient that glucose rapidly enters most tissues.
The concentration of glucose in the tissues is therefore usually well
above the Km of hexokinase and so this enzyme, which is present in muscle cells, will be maximally active.
Glucokinase, which is only found in the liver, only becomes active when the concentration of glucose rise significantly above 5mM (the fed state).
The glucose-6-phosphate produced can be used to make glycogen when here is abundant glucose.
When glucose becomes scarce this enzyme is effectively switched off.

Question 28

What effect does phosphorylation have on enzyme activity?

Answer 28

Phosphorylation may either activate or inhibit enzyme activity

For example, phosphorylation is used to regulate glycogen metabolism. When blood glucose levels fall glycogenolysis (glycogen breakdown) is stimulated on the liver. This is achieved by a cascade mechanism that phosphorylates glycogen phsophorylase, ACTIVATING it.

At the same time, glycogenesis (glycogen synthesis) is inhibited. This is achieved by phsophorylation of glycogen synthase, INACTIVATING it.

Question 29

What is meant by feedback inhibition when thinking about regulation of metabolic pathways?

Answer 29

Feedback inhibition is often used in connection of regulation of metabolic pathways.
It refers to the effect where a metabolic intermediate in a pathway can inhibit an enzyme that is present at an earlier stage of that pathway. This provides the cell with a means of controlling the production of
the end products of that pathway.
An example of this is the inhibition of PFK by citrate.
Citrate is part of the Citric acid cycle, part of a groups of pathways involved in the breakdown of glucose to generate energy.
If there is high concentration of citrate in the cell then this is an indication that the cell has plentiful energy.
By inhibiting PFK in glycolysis it ensures that the rate of citrate accumulation is slowed.

Question 30

Explain why genetic defects in factor XI only cause mild thrombosis.

Answer 30

1. Extrinsic pathway can still work.
2. Factor XI does not become significantly active until thrombin catalyse its activation. By then the process of clotting is well under way and so a deficiency of factor XI does not significantly delay.

Remember that the intrinsic pathway is important for sustaining the clotting process.

Question 31

Why is vitamin K given to newborn infants?

Answer 31

1. Milk will not provide all of the vitamin K that is required daily
2. Adults gain vitamin K from intestinal bacteria that synthesise it. Newborns have sterile intestine so cannot obtain it by this source.

Question 32

How many purines can be found in a stretch of 100 bp double stranded DNA?

A) You cannot know as it depends on the DNA sequence
B) You cannot know as it depends on how GC-rich the DNA sequence is
C) You cannot know as it depends on how AT-rich the DNA sequence is
D) 50 purines
E) 100 purines

Answer 32

E

Every base pair consists of one purine and one pyrimidine, so 100 bp are 100 purines and 100 pyrimidines.

Question 33

What is the complementary DNA sequence for 5’GGGATTCC ?

A) 5’CCCTAAGG 
B) 5’CCCUAAGG 
C) 5’GGAATCCC 
D) 5’GGAAUCCC 
E) None of the above

Answer 33

C

5’GGGATTCC 3’
3’CCCTAAGG 5’; which is, when written in the 5’->3’ left->right orientation, 5’GGAATCCC.

Question 34

In the DNA sequence TTGCCCA base G is connected with a covalent bond to…

A) ...base C on the opposite strand 
B) ...base C on the same strand 
C) ...base C and T on the same strand 
D) ...a deoxyribose 
E) ...a phosphate

Answer 34

D

Bonds between the bases within a base pair are hydrogen bonds, NOT covalent bonds, so answer
A is wrong!
Within a DNA strand every base is connected to a sugar-phosphate and those sugar-phosphates
are connected with each other with 5’ to 3’ covalent bonds. Hence the only covalent bond the base
has, is with a deoxyribose.

Question 35

What is chromatin?

Answer 35

the DNA/protein complex that forms a chromosome.

Question 36

Describe in simple terms how eukaryotic DNA is packaged into chromosomes.

Answer 36

DNA wraps around histones to form nucleosomes, which are the beads of ‘beads on a string DNA’.
Nucleosomes can tightly pack together to form 30nm fibres. These fibres and beads on a string are the ‘decondensed forms’ of a chromosome.
Chromosomes can condense much further by several levels of looping of the 30nm fibres.
In highly condensed form chromosomes display their classical ‘chromosome shape’.

Question 37

Why do you think DNA packaging is important?

Answer 37

to keep the DNA molecules organised and protected (remember each
chromosome has one DNA molecule, that means 46 DNA molecules in the nucleus of each human cell!).

Packaging into highly condensed chromosomes is very important in cell division, in order to make sure that this process occurs properly

The level of condensation also helps to regulate gene expression (highly condensed: no gene expression, beads on a string: genes can be expressed).

Question 38

Many nucleoside analogues are frequently used as antiviral and anticancer drugs.
Why do you think nucleoside analogues are used instead of nucleotide analogues, after all nucleotides are the basic units of DNA and RNA.

Answer 38

The negative charge carried by the phosphate groups of nucleotides and nucleotide analogues makes it much more difficult for these molecules to pass membranes
this is easier for nucleosides and nucleoside analogues.
Once in the cell at the right location nucleoside analogues can be
phosphorylated by kinases, which converts them into nucleotide analogues.

Question 39

Name the stages of the eukaryotic cell cycle in the correct order and explain in simple terms what happens at each stage.

Answer 39

Cell cycle stages in order: G1-­‐‑S-­‐‑G2-­‐‑M.

In G1, all cellular contents excluding the chromosomes are duplicated.

In S, DNA replication take place, each chromosome is replicated.
In G2, all DNA is double checked for errors and DNA repair takes place.

In M, mitosis takes place, which is the actual cell division, resulting in two identical daughter cells.

Question 40

What is stage G0, and why does it exist?

Answer 40

G0 (G zero) is a stage where a cell does not prepare to divide or is dividing, i.e. it is a phase where a cell does not receive any signals to ‘enter the cell cycle’.
This could be a temporary phase (the cell starts to prepare for cell division again, once the appropriate signals are received) or for many cells a final, mature stage, e.g. nerve cells and heart muscle.

For the latter, G0 could be viewed as ‘outside the cell cycle’, G0 is the final destination (NOTE, this does not mean these cells are about to die!)
for the former, it is a more that the cell cycle is halted in the early stages G1, and is then often called ‘cell cycle arrest’. In this case, regulation of growth is one of its main purposes.

Question 41

During which phase or which phases of mitosis are the chromosomes heavily condensed?

During  which  phase  or  which  phases  of  mitosis  are  the  chromosomes  replicated?
 
A)  Prophase 
B)  Metaphase 
C)  Anaphase 
D)  Telophase 
E)  Interphase 
F)  None of the above 

Answer 41

A-D

A, B, E:
Chromosomes enter mitosis in replicated form (two DNA molecules each), so they are replicated during prophase and during metaphase.
During anaphase the replicated chromosomes split, the
chomatids separate.
The moment the chromatids are separated, the chromatids are no longer called ‘chromatids’, but they are then called ‘chromosomes’ again (daughter chromosomes that
will go into the daughter cells).
These ‘new chromosomes’ areun-­‐‑replicated form (one
DNA molecules each).
During interphase the chromosomes are in replicated form some of the time; during G1 phase they are not, during S phase replication is taking place, and during G2 phase the chromosomes are replicated.

Question 42

Juvenile cataract shows autosomal dominant inheritance. If an affected woman married a normal man, there is a 50% chance their child will be affected.
what is the probability of the disease occurring in…

All 4 of their children?
Their first great grand child?

Answer 42

0. 5 x0.5 x0.5 x0.5 x = 1/16

0. 5 x0.5 x0.5 = 1/8

Question 43

Explain what is meant by the statement “The genetic loci for NPS and the ABO blood group are linked”.

Answer 43

A genetic locus is a position on a chromosome, each gene has a locus.
If two loci are linked, that means that both loci are on the same chromosome.
This also means that there is no independent segregation of the alleles of the genes during meiosis, as can be observed from their offspring.

Question 44

Describe the key differences between transcription and translation in a human cell.

Answer 44

Transcription:
an RNA molecule is made
takes place in the nucleus
a code is copied
(1 unit=1 base to 1 unit= 1 base)

Translation:
a protein is made  
takes places in the cytoplasm 
a code is translated 
(3 units=triplet -­‐‑> 1 unit=1 amino acid) 

Question 45

Describe in what aspects transcription and translation in a human cell can be considered ‘similar processes’.

Answer 45

In both processes a template (a code) is read to create a macromolecule built up of basic subunits (a polynucleotide or a polypeptide).

Both processes consist of three stages: initiation, elongation and termination (of which especially the first and the last stage are highly regulated).

Both processes require energy and are driven by enzymatic activity.

In both processes the template has ‘surplus code’ that can be used for regulation;
in DNA promoter sequences, terminator sequences and introns;
in mRNAthe 5’UTR and 3’UTR.

Question 46

What is a polysome?

Answer 46

an mRNA template covered with very many actively translating ribosomes.

Question 47

DNA polymerases are capable of editing and error correction, whereas the capacity for error correction in RNA polymerases appears quite limited. Given that a single base error in either replication or transcription can lead to an error in protein synthesis, suggest a possible biological explanation for this striking difference.

Answer 47

A single base error in DNA replication, if not corrected, would cause one of the two daughter cells AND all its progeny, to have a mutated chromosome.

A single base error in transcription would not affect the chromosome, it would lead to a mutated mRNA and to the formation of some defective copies of one protein. But because the turnover of mRNA is relatively rapidly, most
copies of the protein would be normal. Also the progeny of this cell would be normal.

Question 48

During translation the mRNA…

A) is read in a 3’ to 5’ direction.
B) is read in a 5’ to 3’ direction.
C) is made in a 3’ to 5’ direction.
D) is made in a 5’ to 3’ direction.
E) None of the above.

Answer 48

B

mRNA made 5-3 from a 3-5 DNA template strand

Question 49

During transcription DNA…

A) is read in a 3’ to 5’ direction.
B) is read in a 5’ to 3’ direction.
C) is made in a 3’ to 5’ direction.
D) is made in a 5’ to 3’ direction.
E) None of the above.

Answer 49

A

Question 50

During translation DNA…

A) is read in a 3’ to 5’ direction.
B) is read in a 5’ to 3’ direction.
C) is made in a 3’ to 5’ direction.
D) is made in a 5’ to 3’ direction.
E) None of the above.

Answer 50

E

During translation DNA is neither read nor made!

Question 51

During gene expression the process of translocation involves…

A) the movement of the 40S subunit from the 5’cap to the AUG initiation codon.
B) the movement of the ribosome along the mRNA template.
C) the movement of the RNA polymerase along the DNA template.
D) the movement of the growing peptide chain from the tRNA on the P-­‐‑site to the amino acid
connected to tRNA on the A-­‐‑site.
E) None of the above.

Answer 51

B

Question 52

Regulation of gene expression occurs…

A) at the level of chromatin structure.
B) at the transcription level.
C) at the post-­‐‑transcriptional level.
D) at the translation level.
E) at the post-­‐‑translational level.
F) All of the above.

Answer 52

F

In Eukayotes gene expression is highly regulated: through chromosome structure (genes can be
silenced, think of euchromatin and heterochromatin), transcriptionally (think transcription factors
and other promoter binding proteins like activators and repressors), post-­‐‑transcriptionally (think
splicing and RNA stability), translationally (think RNA turn-­‐‑over), and also post-­‐‑translationally

Question 53

True of False:

In Eukaryotes, the translation initiation site is always downstream of the TATA box.

Answer 53

TRUE

Once the direction of transcription of a gene is established, upstream and downstream is set.
The TATA box is a part of the promoter, which is always upstream of the initiation of transcription
and thus also upstream of the initiation of translation.

Question 54

True or False:

Inhibition of transcription would not block protein synthesis.

Answer 54

FALSE

If transcription is inhibited there is no mRNA production and hence no template for translation, so no protein synthesis.

Question 55

True or False:

Eukaryotes contain only one type of RNA polymerase.

Answer 55

FALSE 
Eukaryotes contain three types of RNA polymerase: 
RNA polymerase I: rRNA
RNA polymerase II: mRNA
and RNA polymerase III: tRNA

Question 56

True or false:

The initiation codon is always located in the first exon.

Answer 56

FALSE

Genes can have introns in their 5’UTR so the initiation codon (i.e. initiation of translation) can also be located in the second or third etc exon.

Question 57

True or false:

A deletion of one DNA nucleotide within a gene always changes the reading frame.

Answer 57

FALSE

A deletion of one DNA nucleotide within the ORF will indeed change the reading frame, also one DNA nucleotide deletion in an intron splice-­‐‑site could (but NOT necessarily!) cause a change in reading frame due to the fact that the intron would then not be spliced out.

However:
think of one DNA nucleotide deletions within introns, or in the promoter sequence, or 5’UTR/3’UTR etc.

Question 58

Explain what is fundamentally different between the molecular methods of DNA gel electrophoresis and protein gel electrophoresis (apart from the fact that either DNA or protein is size-separated).

Answer 58

DNA is negatively charged and will always migrate to the positively charged electrode, hence you will separate only (or mainly) on size.

Proteins can be negatively or positively charged, so protein gel electrophoresis can also be used to separate proteins according to their charge (as in IEF)

Question 59

True or false:

Restriction enzymes cut all phosphodiester bonds in the DNA at a restriction site.

Answer 59

FALSE

Restriction enzymes are specific endonucleases that are only able to cut phosphodiester bonds at a specific position within a DNA molecule.
For instance, in the case of the restriction enzyme EcoRI which recognises and cuts the sequence 5’GAATTC3’, only the phosphodiester bond between the G nucleotide and neighbouring (most 5’) A nucleotide is cut in both strands of the double-­‐‑stranded DNA molecule. All other phosphodiester bonds will stay intact.

Question 60

True or false:
Monoclonal antibodies are useful in assays because they are a mixture of antibodies that recognise different epitopes on the same protein.

Answer 60

FALSE

A solution containing a monoclonal antibody will consist of lots of copies of the same antibody. This molecule will recognise a specific epitope on the target protein.

Epitope = the part of an antigen that is recognized by the immune system, specifically by antibodies, B cells, or T cells.

Question 61

For  each  of  the  10 molecular  techniques, state whether  hybridisation  is important for this technique, and if so, briefly explain what is hybridising to what. 
 
(a)  Restriction analysis 
(b)  PCR 
(c)  Southern blotting 
(d)  Northern blotting 
(e)  Western blotting 
(f)  FISH 
(g)  Microarray 
(h)  DNA sequencing (Sanger chain termination) 
(i)  Gene Cloning 
(j)  DNA gel electrophoresis 

Answer 61

a) NO
Restriction enzymes recognise and cut palindromic DNA sequences.

b) YES
In PCR the forward and reverse PCR primers are complementary in sequence to the template sequence you want to amplify. The primers will hybridise to the DNA template each PCR cycle during the annealing phase of the 3-­‐‑temperature cycle, before being extended by Taq DNA polymerase.

c) YES
In Southern blotting (Southern hybridisation) the labelled probe is complementary in sequence to whatever sequence you are interested in. The probe will hybridise to its
complementary DNA sequence that is bound to the membrane (blot).

d) YES
As for southern blotting but hybridises to RNA sequence

e) NO
Western blotting is a technique to detects the presence of proteins using antibodies

f) YES
In FISH (Fluorescent in situ hybridisation) the fluorescently-­‐‑labelled probe is complementary in sequence to whatever sequence you are interested in. The probe will nhybridise to its complementary DNA sequence in situ, i.e. in the cell to the (DNA in the) chromosome(s).

g) YES
In Microarray experiments the fluorescently-­‐‑labelled DNA (either cDNA or genomic DNA) is complementary in sequence to a sequence on the microarray. Each sequence within the
array is known. Somewhat confusingly in microarray the probes are stuck to the array and the samples are added in labelled form. In this case hybridisation will take place between
the labelled (red and green) samples and their complementary probes on the microarray.

h) YES
In Sanger DNA sequencing the primer used is complementary in sequence to the template sequence you want to sequence. The primer will hybridise to the DNA template before being extended by (Taq) DNA polymerase.

i) NO
In gene cloning a (digested) piece of DNA of interest is inserted into a vector (plasmid), which is then used to transform a bacterium.

j) NO 
In DNA gel electrophoresis DNA is separated on size in an electric field. 
 
 
 
 
 

Question 62

A frameshift mutation is caused by…

A) a one-base substitution. 
B) a two-base insertion. 
C) a three-base deletion. 
D) a premature stop codon. 
E) None of the above.

Answer 62

B

A frameshift mutation may cause a premature stop codon, but cannot be caused by a premature stop codon.
A one-base substitution does not change the reading frame, where as a two-base insertion does,
and a three base deletion does not (when 3 bases are deleted one amino acid will be deleted but the reading thereafter will be the same). In fact, any insertion mutation or deletion mutation other than
3 bases or multiples of 3 bases, within the ORF, will cause a frameshift.

Question 63

Deletion and insertion mutations in the same gene at Xp21.2 cause the majority of cases of both Becker’s Muscular Dystrophy (BMD) and Duchenne Muscular Dystrophy (DMD).
BMD is generally less severe than DMD.
Suggest how different mutations in the same exon of the same
gene can have different phenotypic effects.

Answer 63

Deletion and insertion mutations of 3 bases or multiples of 3 bases are in-frame and cause the loss of one or more amino acids, whereas deletion and insertion mutations other than 3 bases or
multiples thereof cause frameshift mutations, which can have serious consequences for the protein produced.
BMD-causing mutations tend to be in-frame, while DMD-causing mutations result in frameshifts.

Question 64

Tay-Sachs disease has an autosomal recessive inheritance pattern. Affected individuals do not often survive to reproductive age (in fact, they often die in early childhood), so you may expect the
Tay Sachs-causing allele to disappear from the population.
However, this has not happened.
Can you think of a reason why Tay Sachs has persisted in humans?

Answer 64

Autosomal recessive alleles can persist within a population by ‘reproductive compensation’ - parents who have lost affected offspring will often opt to have additional siblings; if these siblings
are healthy, they have a probability of 2/3 that they are carriers of an affected allele.
Other factors that might influence the persistence of an autosomal recessive allele within a population include some kind of evolutionary advantage for the carrier status (e.g. survival advantage with respect to malaria for individuals with sickle cell trait)

Question 65

Trinucleotide repeats (triplet repeats) can cause serious neurodegenerative disorders such as Huntington’s disease, Fragile-X syndrome and Myotonic Dystrophy. In Huntington’s Disease the 
triplet CAG is repeated in the HTT gene; the normal range is 9-37 repeats. Additional triplet insertions which bring the number of repeats over 37 will cause late-on set Huntington’s; the disease range is 37-121 repeats. Clearly, establishing the number of repeats at this healthy-disease boundary will be critical in a genetic test. 
What molecular technique(s) would you suggest for such a genetic test? Briefly explain your reasoning.

Answer 65

Establishing whether there are 37 or 38 CAG repeats in the HTT gene can be critical and is probably most accurately done by a PCR reaction spanning that region, followed by DNA sequencing.
When for instance there is a need to establish whether there are 10 or 100 CAG repeats in the HTT gene, the PCR reaction could simply be followed by DNA electrophoresis checking for size of the PCR product.
NOTE: for some triplet repeat diseases the difference in numbers of repeats between healthy and disease state could be far more dramatic, such as Fragile X- syndrome healthy 6-54 and disease 50-
1500! When there are many repeats it is often impossible to use PCR and DNA sequencing, in those cases Southern blotting should be used.

Question 66

True or false:

Duplication of genetic material results in a milder phenotype than deletion of genetic material.

Answer 66

TRUE

Having an excess of genetic material is usually better tolerated than a deletion.
Thinking about aneuploidy syndromes an additional chromosome 21 can be tolerated and causes Down
syndrome, but monosomy 21 is never reported in a live-­‐‑born individual. The same is true for trisomy 13 (Patau syndrome) and trisomy 18 (Edwards syndrome).
The only full monosomy syndrome that is viable involves the X chromosome (45,X Turner syndrome).
Deletion syndromes are more prevalent that duplication syndromes yet it would be expected that the same number of
cases would occur as a duplicated gamete is produced at the same time as deleted gamete.
However, duplications are far less often seen in the laboratory. The likely explanation for this is
that individuals carrying duplications do not come to the attention of clinicians as they have a mild or no abnormal phenotype.

Question 67

True or false:
If a developmentally delayed child is carrying an apparently balanced chromosome inversion, this inversion then cannot account for their clinical phenotype.

Answer 67

FALSE

Although with inversions there may be no loss of DNA (no physical loss of genetic material), there may be loss of genetic information! For instance, when an inversion causes a gene disruption, or causes the creation of a new gene. For this reason inversions can be considered balanced or unbalanced depending on the phenotypic outcome of the rearrangement of genetic material.

Question 68

True or false:
Deletion of the terminal part of a chromosome arm may make a chromosome unstable due to the loss of the telomere.

Answer 68

TRUE

The terminal ends of chromosomes (both on the p-­‐‑arm and q-­‐‑arm) are called telomeres, which are specific DNA sequences that have an important function in protecting the
chromosomes.

Chromosomes without telomeres will be recognised by the cell, and signal DNA-­‐‑damage. The cell will either attempt to repair the DNA damage or it will die. When deletion of terminal part is observed in a now stable truncated chromosome, EITHER the end bit broke off and subsequently a
new telomere has been placed at the broken end (to stabilise it) OR there was in fact a double break, one in the chromosome and one very near to the end; the bit in the middle is lost (deletion) while the remaining two parts of the chromosome rejoin.

Question 69

True or false:

An individual carrying 45 chromosomes can be phenotypically normal.

Answer 69

True

In Robertsonian translocations two acrocentric chromosomes fuse together two form one super-­‐‑chromosome (while both short p-­‐‑arms are lost), so in fact the chromosome number has gone down by one, i.e. an abnormal number of chromosomes, aneuploid. An individual carrying a balanced Robertsonion translocation will have 45 chromosomes, not 46!
For instance 45,XY,der(21;21)(q10;q10) is a male carrying a 21,21 Robertsonian translocation.

Question 70

True or false:

An individual carrying a Y chromosome is always phenotypically male.

Answer 70

Mainly true, but FALSE

Although this statement is true in most cases (normal males XY and individuals carrying chromosomal abnormalities such as XXY and XYY), XY individuals having androgen insensitivity
are phenotypically female. Although the Y chromosome is present in these individuals the androgen insensitivity will cause the development to ‘default to female development’.

In the sporting world there are known cases of female athletes displaying this phenomenon.

Question 71

Consider an XXY and an XYY individual.

What are the phenotypical differences (if any) between these two individuals?

Answer 71

Both individuals will be phenotypically male, however the phenotype of the XXY (Klinefelter’s) will divert from the normal male phenotype from puberty, often increased breast tissue and a less masculine appearance.

XYY individuals are often indistinguishable from normal males, might be a bit taller than average.

XXY are almost always infertile, while XYY are normally fertile.

Question 72

What is (are) the genetic cause(s) for XXY & XYY sex chromosome abnormalities?

Answer 72

Both chromosome abnormalities are caused by meiotic nondisjunction in one of the parents of the individuals concerned.
In the case of XXY nondisjunction could have happened in maternal or paternal gamete formation (i.e. a mutant egg or sperm)
in the case of XYY nondisjunction has happened in paternal gamete formation (i.e. a mutant sperm).

Question 73

The occurrence of uniparental disomy (UPD) – both copies of an individual’s chromosome (or part of a chromosome) come from one parent only – often results phenotypic anomalies. Give two
explanations how UPD may occur.

Answer 73

(1) Trisomy rescue.
An abnormal gamete produced by MEIOTIC nondisjunction has two copies of the same chromosome.
This is fertilised by a normal gamete.
During a MITOTIC cell division one of the three chromosomes is lost.
When the chromosomes that is lost is the only chromosome from one of the parents the conceptus is ‘rescued’ from trisomy but is left with UPD.

(2) Monosomy rescue.
An abnormal gamete produced by MEIOTIC nondisjunction has no copies of one of the chromosomes.
This is fertilised by a normal gamete and the conceptus only has one chromosome from one parent.
This single chromosome is then replicated at a MITOTIC cell
division resulting in two identical copies of a chromosome from one parent.

Question 74

Klinefelter’s syndrome is the most common sex chromosome disorder.
Describe its genetic cause and explain how you could find out whether the cause is maternal or paternal?

Answer 74

XXY syndrome = Klinefelter’s syndrome
The genetic cause of Klinefelter’s syndrome (XXY) is meiotic nondisjunction in the parents of the individuals concerned, which could have happened during maternal gamete formation (oogenesis) or paternal gamete formation (spermatogenesis).

You would have to look in detail at the X chromosomes of the mother, the father, and the XXY offspring.

There may be areas within the X chromosome where variation is observed frequently or genes on X chromosomes where often different alleles are found.

These could then be used as markers to find out whether the extra X chromosome originated from the father or the mother.

Variation on the markers could be studied in various ways, obviously depending on the variation, perhaps a changed restriction site (presence or absence) or ultimately just variation in DNA sequence determined by DNA sequencing after PCR for instance.