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TEACHERS NOTES
Problem Solving Activities - Suggested Answers
Dolly- set 1
1..
If a fully differentiated mammary gland cell were to divide, what type of cell would
you expect to be produced?
(When cells divide they usually produce cells of the same lineage to the
parental type, so here we would expect to get more cells belonging to the
mammary gland cell lineage).
2.
Donor cells were “starved”, what does this term mean here and why do cell biologists
use this technique ?
(The term “starved” means to maintain cells in serum-poor medium. This makes the
cells exit the cell cycle and enter G0. Cell biologists use this technique to synchronise
the culture, so all the cells in the dish are at the same stage in the cell cycle).
3.
Why did the egg containing the donated nucleus develop into a whole new animal
rather than a ball of mammary gland cells?
( The G0 nucleus was placed in the cytoplasm of an unfertilised egg. Signals from the
egg caused the nucleus to express the genes of a newly fertilised egg ).
4.
Normally penetration of the egg by the sperm cell signals the start of the
developmental process. What can replace this signal when no sperm are involved ?
(Electric pulsing).
5a.
During the set of experiments which resulted in Dolly being born, what percentage
of ‘reconstructed eggs’ resulted in live lambs ?
(1/277 X 100/1 = 0.36%)
b.
Comment on the commercial viability and ethics of this process.
(If the process is to become commercially viable for replicating farm animals the process
must be made more efficient. If nuclear transfer is to be used for producing genetically
modified sheep that secrete human proteins in their milk, then the benefits to thousands
of patients can justify the welfare and financial costs even at todays level of efficiency.
Note: it would be very inefficient to produce new transgenic animals each time. Comment
on ethics).
6.
Dolly is not an exact copy of the Finn Dorset ewe whose mammary cell was used to
create Dolly - explain.
(An exact copy, or clone, is a group of organisms or cells that have arisen from a single
individual by asexual reproduction so all offspring are genetically identical. Dolly is a
product of nuclear transfer. The genes in a “clone” produced by nuclear transfer come
from two sources: the nucleus of the cell used as nuclear donor - which contains
between 50,000 and 140,000 genes, depending on the species - and the mitochondria
of the egg used - which contains between 30 - 40 genes . Dolly is therefore not an
exact genetic copy of the ewe that donated the mammary gland cell used to create
her, but she is 99.94 - 99.97% identical ! ).
Dolly- set 2
1.
Does the DNA fingerprint support or dispute the claim that Dolly is derived from the
nucleus of a Finn Dorset udder cell? Explain.
(Supports the claim. Using 4 different minisatellite probes the fingerprint obtained for
Dolly matches exactly that for the cultured donor cells and the udder from which they
were derived. The bands are the same size and have the same relative intensities).
1
TEACHERS NOTES
2.
The DNA fingerprint analysis includes fingerprints of Finn Dorset sheep from the
same flock as the donor ewe. What do these show and why were they included ?
(The other fingerprints were included as controls. They show that there is variation
within the flock, confirming that Dolly came from a specific and identifiable ewe).
3.
Apart from the 3 experimental lanes, a special relationship exists between 2 other
animals. Which lanes contain this DNA and can you suggest how these animals are
related to each other. Why include these samples ?
( Lanes 2 and 1 are ewe and daughter respectively - they have more bands in common
than do other control samples but are still distinguishable using this technique. Here
4 minisatelliteswere used to ensure accurate results. Including very closely related
sheep as controls demonstrates the accuracy of the test).
4.
Why are two sizes given for each of the microsatellite probes? What does it mean if
both of these are the same ?
(1 from each of the homologous chromosomes. If the two sizes are the same then the
organism is homozygous at that locus).
5.
How do the results from the microsatellite analysis dispute the allegation that Dolly
was derived from a stray fetal cell ?
(If Dolly were derived from a stray fetal cell, she would have derived half of her alleles
from the sire of the fetus and half from the the 6-year-old ewe. Since all markers were
identical to those of the donor tissue and cell culture it is extremely unlikely that this
could have happened - calculated at between 1.1 X 10-6 and 9.2 X 10-6).
6.
Which is the most polymorphic of the markers used in the microsatellite analysis of
Dolly? Why is it better to use markers which are highly polymorphic in this type of
investigation ?
(TGLA122. The more polymorphic a marker, the less likely the same allele will be
present by chance in two individuals).
Dolly- set 3
1.
Why was it important to show that Dolly could breed normally ?
(If the technique of nuclear transfer is to be used commercially then cloned animals
must be able to breed normally. Consider animal welfare issues).
2.
What are “telomeres” and why is it not surprising that when she was one year old
Dolly’s telomeres were about the same as that expected for a 7 year old ewe?
(Telomeres are capping sequences which protect the ends of chromosomes during
cell division but they are gradually eroded away with time. The donor nucleus from
which Dolly was derived came from a 6 year old ewe).
3.
Why are the scientists working on Dolly not overly concerned by the revelations
concerning Dolly’s telomeres ?
(Dolly appears to function normally in all respects including the production of normal
lambs. When a cloned animal is allowed to breed normally, the high telomerase activity
in the cloned germ cells helps ensure any chromosomes passed onto its offspring
have telomeres of normal length).
2
TEACHERS NOTES
4.
Why might a farmer using embryos produced by nuclear transfer from cows in elite
herds see a much bigger jump in performance of the herd compared with a farmer
using artificial insemination ?
(Artificial insemination only supplies half the genes, but nuclear transfer, supplies all
the genes of the donor animal. Farmers who used cloned embryos could lift the
performance of their herds to that of the very best within one generation. Recent
estimates in dairy cattle suggest the performance of the average cow in the UK is
about 10 years behind the best.)
5.
Discuss the moral and ethical issues raised by genetic modification and nuclear transfer
techniques.
(Try to put forward a balanced picture. A useful publication here is “Ethics, Morality
and Animal Biotechnology” a BBSRC publication)
Cystic Fibrosis - Worksheet
This worksheet could also be used to introduce CF while working on Unit 1 in the Advanced
Higher.
1a)
genotype - ff
phenotype - CF sufferer
b)
genotype - Ff
phenotype - normal (carrier)
c)
genotype - FF
phenotype - normal
2.
Ff X Ff
F
f
F
FF
Ff
f
Ff
ff
Expect half of the offspring to be carriers.
3a)
b)
54 000 000
26
=
2.1 million
the chance of two carriers getting married.
(1 in 26 X 1 in 26 = 1 in 676)
4.
CF is an autosomal recessive condition.
5a)
1480 X 3 =4440 nucleotides
b)
4440 - 3 =4437 nucleotides
c)
Faulty gene codes for a polypeptide chain with one less amino acid (phenylalanine)
than the normal CFTR protein.
6.
A couple could be carriers of mutations rarer than the ∆F508 mutation.
3
TEACHERS NOTES
4
Diagnostic Testing for Cystic Fibrosis
1.
Your task is to describe each of the other six individuals shown using the terms
normal, heterozygote, homozygote, compound heterozygote (when two different
mutations are present in the same individual) and the 4 different mutaions. The
first one has been done for you.
Individual 1 is NORMAL.
Individual 2 is NORMAL
Individual 3 is a ∆F508 heterozygote
Individual 4 is a ∆F508 homozygote
Individual 5 is a ∆F508 621+1 compound heterozygote
Individual 6 is a G551D heterozygote
Individual 7 is a ∆F508 G542X compound heterozygote
2.
Which individuals are
a) Normal
Individuals 1 and 2
b) Carriers
Individuals 3 and 6
c) Affected
Individuals 4, 5 and 7
3.
Draw a pedigree of the possible outcome if individuals 3 and 6 were to produce a
family. NB sex of offspring chosen is irrelevant.
AAA
AA
A
AA
AA
AAA
A
AA
AAA
AA
AA
AAAA AA
Symbols used
male
female
Carrier
F508
Carrier
G551D
Affected
individuals
Simon
Compound
Heterozygote
F508/G551D
AFFECTED
Heterozygote
F508/n
CARRIER
AA
AA
AA
Louise
Heterozygote
G551D/n
CARRIER
Homozygote
n/n
NORMAL
TEACHERS NOTES
Cystic Fibrosis part 1: Should Mollie and Carlos have a baby?
•.
Find out what it you can about CF (cystic fibrosis)
(Inherited condition caused by a mutation in a single, large gene. It is caused by a
recessive mutation in a single allele. - the protein coded for by the CF gene - CFTR is a
membrane bound protein involved in the transport of chloride ions from the cell to the
extracellular fluid - symptoms include a build up of sticky mucus in the lungs which is
prone to infection, pancreatic problems which result in a failure to produce enough
digestive enzymes which can lead to nourishment problems and failure to thrive and
there are often problems associated with the male reproductive system. Affected
individuals are require daily physiotherapy and medication. There is no cure and affected
individuals rarely survive beyond their mid-thirties.)
•
How common is CF ?
(It is the most common inherited abnormality among Caucasions affecting 1 in 2,500
births)
•
Carlos is a “carrier” for CF - explain
(A carrier has one allele of the normal CF gene and one mutated copy - heterozygote Carlos is phenotypically normal but can pass the mutated CF gene on to any offspring)
•
Boys and girls have an equal chance of getting CF - what does this tell you about the
affected gene ?
(it is autosomal rather than sex linked)
•.
If Mollie decides to be tested for CF, what are the chances they will have an affected
baby if
a) she is not a carrier ?
(0%)
•
b) she is a carrier ?
(25% 1 in 4 )
•
If they decide to go ahead and try for a baby, several prenatal tests are available, including
alpha-fetoprotein test (AFP), ultrasound, amniocentesis, chorionic villus sampling.
Describe each of these and place them in order from greatest to least risk to the foetus
and usefulness.
(Chorionic villus sampling - adv. very accurate and very rapid results; amniocentesis adv. very accurate but long wait for results; AFP - abnormally high/low levels of this foetal
protein in mothers blood indicate developmental problems - test does not expose foetus
to risk, ultrasound - used to detect developmental problems - test does not expose
foetus to risk).
•
Is it possible for Mollie and Carlos to still have a CF baby even if the normal tests are
negative ?
(Yes - if Mollie is a carrier for an unusual mutation for CF not tested for - this is very
unlikely but the CF gene is very large and several mutations have been described not all
of which are tested for).
•
What are their choices ?
Neither test nor try for their own baby but adopt.
Find out if Mollie is a carrier, if yes do not try for a baby or
do the tests.
Decide to abort a foetus if it is found to be CF, or
prepare themselves for a CF baby.
5
TEACHERS NOTES
•
6
What is the likely outcome if a foetus is found to have 2 mutated CF genes ?
(Don’t know - affected individual could be only mildly affected (asthma/ lung infections)
through to full disease and early death so a baby could be very sick or stay quite healthy.
New treatments are keeping CF sufferers alive much longer - into their early 30’s. In the
next 30 years there may be new treatments available like gene therapy and new drugs
so this could improve further not only extending lifespan but improving health).
•
Other choices include pre implantation diagnosis - what does this entail and what are
the advantages and disadvantages ?
(IVF following mixing of Molly and Carlo’s eggs and sperm in vitro and testing embryo’s
produced for CF prior to implantation advantage - can be sure the implanted embryo is
CF negative (unless new or extremely rare mutation occurs and it is their own genetic
baby. Disadvantages - expensive, time consuming, uncomfortable procedures, high failure
rate).
•
What would you do ?
Cystic Fibrosis part 2:
Mollie, Carlos and their family go for genetic testing
1.
INDIVIDUAL
2.
MARKER SIZE (bp )
Phenotype
Carlos
185/ 182
Carrier
Mollie
185/ 182
Carrier
Hamish
185/ 182
Carrier
Gordon
185
Normal
Morag
182
CF
Why were only white blood cells used for testing ?
RBCs - no nucleus - no DNA.
3.
Describe the process used to produce the information in the tables.
See Student Support Notes for details of the techniques used to do microsatellite
analysis.
4.
Is CF a sex linked disorder? How does the information you have been given support
your answer ?
No it isn’t. Not enough data given to be absolutely sure it isn’t sex linked but Carlos
had an affected brother and daughter. Sex linked disorders are more common in males
but do occur in females.
5.
What is Carlos’ reaction likely to be when he is told about his sons results ?
Results show that like him Hamish is a carrier and will face the same dilemmas he has
faced. Carlos will feel very sad about this but there is a good chance that when Hamish
plans to starts a family his wife will not be a carrier. Carlos will be very happy to hear
that Gordon is completely normal and will not have to face the problems he has had to
face.
TEACHERS NOTES
6.
Describe the mutation detected in the CF test used here.
1 amino acid deletion - 3 base pairs missing in the positive
7.
The insurance company has also seen the test results. Will the results affect the
family’s wish to take out life insurance ?
No. Medical tests had already confirmed Morag’s condition and getting life insurance
for a sufferer of CF will be difficult because CF sufferers do have a much reduced life
expectancy. There is no such effect for carriers, carriers do not show any of the
symptoms of the disorder. The genetic test only differentiates between carriers and
non-carriers.
8.
(a)
1 in 26 of the population are known to be carriers of CF.
What are the chances of two carriers getting married ?
26 X26 = 1 in 676
(b)
What are the chances of Hamish’s future wife being a CF carrier ?
1 in 26
(c)
What proportion of the offspring produced by two carriers will have cystic fibrosis ?
1 in 4
(d)
Why is the proportion given in answer (c) only used as a guideline when counselling
prospective parents ?
Chance involved in which gametes come together during fertilisation.
Gene Therapy
1.
2.
How might the Human Genome Project help sufferers of genetic disorders ?
(The Human Genome Project will provide the complete set of 3 x 109 bases that
make up the human genome. Once a faulty protein has been identified it will be relatively
easy to find the gene responsible and isolate it from the clones generated by the
project.)
What is meant by the term “clone” as used in this passage ?
(A group of genetically identical cells produced from a single cell. In the example
given in the passage the cells are bacteria and they all contain a plasmid carrying the
human CFTR gene).
3(a).
Explain why viruses are being used in the development of gene therapy.
(Viruses are able to get inside living cells and therefore make good vectors of genetic
material. They inject their genetic material into the cell and take over the cells own
machinery for protein synthesis. Using this method of delivery the normal gene should
reach the nucleus of the correct cell and the gene should be expressed appropriately
in the cell.)
(b)
Why is it important that the adenovirus be made safe before use?
(Usually when adenoviruses infect cells they cause disease. The disease causing
portion of the virus needs to be removed before it is safe to use on a patient.)
(c)
Are there any disadvantages of using a viral delivery method?
(Viruses can mutate rapidly and cause disease, there is the possibility that a
virulent particle could be used by accident).
7
TEACHERS NOTES
4(a)
Explain why the structure of a liposome makes it an ideal vector for gene
therapy.
(It has an affinity for the cell membrane which contains a lipid bilayer. Since it is
positively charged it will easily form a complex with negatively charged DNA. Once
inside the cell the liposome will release its DNA passenger and remain near the cell
membrane while the DNA is transported by the cell to the nucleus).
(b)
Give one reason why liposomes might be a more popular than a virus for use as a
vector.
(Liposomes are non-pathogenic).
5.
Why should the type of gene therapy described here for CF be referred to as
a treatment rather than a cure ?
(The patient still carries the faulty gene. The treatment is needed for life because the
gene therapy would have to be readministered at regular intervals to compensate for
cell turnover. Only the lungs are being treated. Other affected organs such as the
pancreas continue to require conventional long term medical intervention).
Biopharmaceuticals Activity
1.
i)
Use the information in figure 1 to answer the following questions:
Which is the most expensive option and what other reason is there for not choosing
this option ,?
(ans. Plasma is the most expensive option. One other problem associated with the
use of blood plasma is that it is difficult to make sure it is free of pathogens).
ii)
The human calcitonin gene has been cloned. Why does the company not even consider
the possibility of using bacteria to produce the calcitonin. It would be much cheaper
to use bacteria ?
(ans. Prokaryotic systems cannot perform the posttranslational modifications required
for proper functioning of a secretory protein).
iii)
Compare the annual cost of producing 40,000 litres of raw material for calcitonin
extraction from mammalian cell culture with that from transgenic livestock.
(ans. cell culture 80 X 40,000 = £3.2 million; transgneics 10 X 40,000 = £400,000 a
difference of £2.8 million).
iv)
What hidden costs are not shown here ?
(ans. The cost of producing the founder animals for the flock of transgenic animals).
v)
What do the genetic engineers have to do to the human calcitonin gene before
microinjection if they want the transgenic animal to produce human calcitonin in its
milk ?
(ans. They must attach it to regulatory sequences from a gene that is normally expresssd
in the animals mammary gland during lactation eg. The regulatory sequences attached
to the sheep alpha-lactalbumin gene if sheep is to be the transgenic animal).
2.
Use the information in figure 2 to answer the following questions:
8
TEACHERS NOTES
a)
9
Why choose a male animal to found G2, the production flock ?
(ans.. A male animal can produce many more offspring than a female animal. Sperm
can be collected, stored and frozen in advance and shipped anywhere).
b)
What is the purpose of producing the females in G1 ?
(ans. To check for stable transmission and expression of the recombinant gene. Also
to decide which of the founders to use as breeding stock for the production flock).
3.
a)
Use figures 2, 3 and 4 to answer the following questions:
How long does it take to go from microinjection of the transgene to production of
raw material in (1) cattle; (2) sheep and (3) pig?
(ans. cattle 78 months; sheep 44 months; pig 37 months)
b)
What are the main (1) advantages, (2) disadvantages of choosing cattle to produce
recombinant protein in their milk?
(ans. Advantages any 2 from: – produce lots of milk, are quite easy to maintain, relatively
easy to produce founders, once the herd is established there is guaranteed high
production for a long time. Need smaller herd sizes than for pigs or sheep.
Disadvantages – Very long time and expensive to establish herd).
c)
A biotechnology company decides to produce flocks of transgenic sheep to produce
human therapeutic proteins, explain why they chose sheep rather than pigs or cattle.
(ans. The time to production is much less than for cattle and the volume of milk produced
is much greater than for pigs. Sheep are easy to maintain).
4.
a)
Use figure 5 to answer the following questions:
Complete the following table for annual production capacity:
Ans.
Rabbits
Sheep
Cows
b)
Milk volume per year
8 litres
300 litres
9,000 litres
Complete the following table for protein yield:
Ans.
Recombinant protein
AAT
Antibiotics
Protein C
Fibrinogen
Antithrombin III
c)
Yield at 5g/l expression
0.04kg
1.5kg
45kg
Yield per year for flock of 50 ewes
450kg
60kg
15kg
75kg
30kg
The annual market for AAT is estimated to be in the order of 1,000kg. The best
mammalian cell culture system working at maximum production levels can only produce
approximately 10-20kg protein per year.
How many (1) mammalian cell culture systems (2) transgenic ewes, are required to
meet the demand for AAT ?
(ans. (1)between 50 and 100 units (2) 112 ewes).
d)
Explain how the technique used to produce Dolly the sheep could be applied to this
technology.
TEACHERS NOTES
10
(ans. Nuclear transfer opens up the possibility of removing and replacing genes as
well as simply adding them as is possible with microinjection. Once an animal with
stable and high expression of the recombinant protein has been identified, unlimited
numbers of genetically identical animals can be “cloned’ from it using nuclear transfer.
This decreases the time from founder to production dramatically).
Comparative Genome Mapping
ANSWER
For option 1
C1
H12
IFNG
MYF6
DCN
From
H12
H22
C18
MIF
NAGA
ADSL
IGL
From
H22
From
H22
CRYBB1
LGALS4
IGF1
TRA
ITPR2
LDHB
GAPD
From
H12
From
H12
ANSWER
For option 2
C1
H12
GAPD
C18
LDHB
From
C1
H22
ITPR2
IFNG
MIF
From
MYF6
C1
IGL
DCN
CRYBB1
NAGA
IGF1
TRA
From
C1
ADSL
LGALS4
1.
From
C18
From
C1
Name the chromosomal mutations involved in the rearrangements you have just made.
(ans. Translocations and 1 inversion).
TEACHERS NOTES
2.
11
Using the data provided is it possible to decide whether the chicken arrangement or
the human arrangement is closer to that of their common ancestor? If not what
other data do you need ?
(ans. No. Need to look at the arrangement in a third species).
3.
Now use the additional information provided on Chromosome sheet 2 to decide whether
the chromosomal arrangement of the chicken or the human is more like the ancestral
arrangement? Explain your answer.
(ans. The chicken arrangement. The conserved region in chromosome 7 from the
Dolphin matches that of chicken chromosome 1).
4.
You should now be able to complete the phylogenetic tree answer sheet.
(ans. See the completed answer sheet).
Phylogenetic tree - completed answer sheet
300 mya
ANCESTOR
65 mya
C1
OR
H12
H22
D7
D7
H12
H22
Present day
Mya - million years ago