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’  are un-­‐‑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  mRNA the  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.