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Learning Outcomes
Candidates should be able to:
(a) [PA] describe, with the aid of diagrams, the behaviour of
chromosomes during meiosis, and the associated behaviour of
the nuclear envelope, cell membrane and centrioles (names of
the main stages are expected, but not the sub-divisions of
prophase);(b) explain how meiosis and fertilisation can lead to variation;
(c) explain the terms locus, allele, dominant, recessive,
codominant, homozygous, heterozygous, phenotype and
genotype (see pages 35
37 and 39
41);(d) use genetic diagrams to solve problems involving monohybrid
and dihybrid crosses, including those involving sex linkage,
codominance and multiple alleles (but not involving autosomal
linkage or epistasis);
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Learning Outcomes
Candidates should be able to:
(e) use genetic diagrams to solve problems involving test crosses;
(f) [PA] use the chi-squared test to test the significance of
differences between observed and expected results (the
formula for the chi-squared test will be provided);
(g) explain, with examples, how mutation may affect the
phenotype;
(h) explain, with examples, how the environment may affect the
phenotype;
(i) explain how a change in the nucleotide sequence in DNA mayaffect the amino acid sequence in a protein and hence the
phenotype of the organism;
(j) use the knowledge gained in this section in new situations or to
solve related problems.
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Sexual reproduction
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Meiosis (Middle prophase I)
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Meiosis (late prophase I)
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Meiosis (Metaphase I)
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Meiosis (Anaphase I)
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Meiosis (Telophase I)
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Meiosis (Meiosis II-Prophase II and
Metaphase II)
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Meiosis (Anaphase II)
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Meiosis (Telophase II)
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Meiosis
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Meiosis
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Meiosis
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Meiosis I
During Interphase IChromosomes _________
ie. 2n _____
During Prophase IChromosomes _________
ie. see _______________
____________ chromosomes _________
ie. form _________ / _________
ie. may have __________ at _________* crossovers only occur between NON-SISTER chromatids
Nuclear membrane _______________
double
4n
thicken
sister chromatids
homologous pair up
bivalent synapsis
crossover chiasma
breakdown
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Meiosis I
During Metaphase ISpindle fibers attach
to ___________
(at ____________)
____________chromosomes
______ at _________
During Anaphase I
Spindle fibers _________
____________ chromosomes _________
move to opposite poles
homologous
homologous
centromerekinetochore
align equator
shorten
separate
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Meiosis I
During Telophase IEach separated homologous
chromosome reaches pole
Spindle fibers _________
Nuclear membrane _________
At this time,
~ cells may rest
~ cells may continue immediately into_________
disappear
reform
Meiosis II
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meiosis and fertilisation can lead to
variation
In sexually reproducing organisms, three
processes lead to most genetic variation:
1. independent orientation (independent
assortment )of chromosomes in meiosis
(metaphase I)
2. crossing over of chromosomes in meiosis
(late prophase I)
3. random fertilization
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Independent orientation of chromosomes in meiosis
and random fertilization lead to varied offspring
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Homologous chromosomes can carry
different versions of genes
Separation of homologous chromosomes during
meiosis can lead to genetic differences between
gametes
Homologous chromosomes may have differentversions of a gene at the same locus
One version was inherited from the maternal parent,
and the other came from the paternal parent
Since homologues move to opposite poles during
anaphase I, gametes will receive either the maternal
or paternal version of the gene
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Differing genetic information on
homologous chromosomes
Simplified example in mice:
Coat color and eye color
C (brown) c (white)
E (black) e (pink)
Two possible outcomes for
The genes
C i (d i l h I)
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Crossing over (during late prophase I)
further increases genetic variability
Genetic rearrangement between homologues Sister chromatid exchange
Chiasma Site of crossing over Happens between chromatids within tetrads
Produces new gene combinations Genetic recombination
Can occur several times in various location within onetetrad
Zygote combination possibilities are much greater than 64trillion
Two individuals cannot produce identical offspring fromseparate fertilizations
Everyone is a unique genetic entity never seen before
and never to be seen again
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Figure: How
crossing over
leads to
genetic
recombination
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Random fertilization
The combination of each unique sperm with
each unique
egg increases genetic variability:
Combination 1 + Combination 2 is different from
Combination 3 + Combination 4, etc.
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Paper 4 JO5Figure is a diagram of pair of homologous
chromosomes during meiosis.
(a) State the stage of meiosis shown. [1]
metaphase 1 / (late) prophase 1
(b) Describe what has occurred between the
two homologous chromosomes. [3]- homologous chromosomes) pairing /
synapsis
- to chiasma / crossing over
- exchange of genetic material
- between non-sister chromatids
(c) Explain how this can lead to variation. [2]
- breakage of linkage groups / ref. new
linkage groups ;
- may have different alleles ;
- creates new combinations of alleles ;
- when sister chromatids separate ;
(d) Describe two other sources of variation that
are possible as a result of meiosis. [4]
- idea of random orientation at metaphase
I and II / random alignment of homologous
chromosomes on spindle equator ;
- subsequently leads to independent
assortment ;
- 2n possible combinations when n is
number of chromosome pairs ;
- ref. to chromosome mutation qualified ;
- ref. gametes haploid (so can fuse) ;
- random fusion of gametes ;
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Important terms
Locus- a locus (plural loci) is the specific location of a gene or
DNA sequence on a chromosome.
Allele- a different form of gene (e.g. represented with letters for
the different types of alleles PP, Pp, pp)
Dominant- allele that always expresses itself in the phenotype
when present/ allele which influences the phenotype even in
the presence of a alternative allele.
Recessive- allele which does not have its effect in heterozygote/
allele which only effect in homozygotes/ affects phenotype if
dominant allele is absent.
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Important terms
Codominant- both alleles, influence phenotype/ are
expressed; more than 2 phenotypes possible; phenotype
of heterozygote different from either homozygote
Homozygous- having two identical alleles of gene (PP, pp)
Heterozygous- having two different alleles of gene (Pp)
Phenotype- the observable characteristics of an organism,
Genotype- the alleles possessed by an organism
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G l i i
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Gene loci
Homozygousfor the
dominant allele
Dominantallele
Homozygousfor the
recessive allele
Heterozygous
Recessiveallele
Genotype:
P Ba
P
PP
a
aa
b
Bb
Figure: Matching gene loci on homologous chromosomes.
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Test cross
Test cross always involves crossing an organism
showing the dominant phenotype with one which is
homozygous recessive.
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The inheritance of coat color in horses is complex but all horses have one of
two base colors, red (chestnut) or black. The base color is controlled in a
simple monohybrid way.
Draw a genetic diagram to show how two parents with black coat colorcan produce a chestnut foal and the probability of such an eventoccurring. Choose a letter symbol to represent coat color. [4]
Paper 4 (J09)
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Paper 4 (N04)
Figure shows four generations of a family in which some members of the
family suffer from sickle cell anaemia.
Using the symbols HN for the allele
for normal haemoglobin and HS for the
allele for sickle cell haemoglobin, statethe genotypes of the following
individuals. [1]A:
______________________________C:
______________________________Draw a genetic diagram to show theprobability of the parents A and B
producing another child with sickle cell
anaemia. [4]
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Multiple alleles
Most genes have more than
two alleles, e.g. human
blood groups.
The four groups A, B, AB, O
are all determined by a
single gene.
Three alleles of this exist IA,
IB, and IO.
IA, IB are codominant, while
IO is recessive to both IA, IB .
A diploid cell can carry only
two alleles, the possiblegenotypes and phenotype
are shown in table 17.2.
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S i h it
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Sex inheritance The X chromosome contains many different genes.
One of them is a gene that codes for the production of aprotein needed for blood clotting, called factor VIII.
There are two alleles of this gene, the dominant one Hproducing normal factor VIII, and the recessive one h resultingin lack of it.
People who are homozygous for the recessive allele sufferfrom the disease haemophilia, in which the blood fail to clotproperly.
This gene is on the X chromosome, and not on an autosome,
affects the way that it is inherited. Females, who have two X chromosomes, have two copies of
the gene. Males, have one copy of the gene. Therefore, thepossible genotypes for men and women are different.
Sex linked- is found on a part of the X chromosome not
matched by the Y, and therefore not found on Y chromosome.
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Paper 4 (J08)
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Paper 4 (J08)
Color blindness is a condition
characterized by the inability ofthe brain to perceive certain
colors accurately.
(a) Explain the meaning of the
terms allele and recessive. [2]Allele:
Recessive:
(b) Explain why females are lesslikely than males to have RGC.
[2]
(c) With reference to the figure,
and using the symbols R for
dominant allele and r for the
recessive allele, state the
genotypes of the individuals 1,
4, 6 and 7. [4]
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Dih d id ( ti 1 1 1 1)
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Dihydrid crosses (ratio 1:1:1:1) Dihybrid crosses look at the inheritance of two genes
at once. In tomato plants, there is gene which codes for stem
colour. This gene has two alleles:
A different gene, at a different locus on a differentchromosomw, codes for leaf shape. There are twoalleles:
Dih d id ( ti 1 1 1 1)
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Dihydrid crosses (ratio 1:1:1:1)
What will happen if a plant which is heterozygous for
both of these genes is crossed with a plant with
green stem and potato leaves?
At fetilisation, any of four type of gamete from theheterozygous parent may fuse with the gametes
from the homozygous parent. The genotypes of the
offspring will be:
Dih d id ( ti 1 1 1 1)
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Dihydrid crosses (ratio 1:1:1:1)
From this cross, we would expect approximately equalnumbers of the four possible phenotypes.
This 1: 1: 1: 1 ratio is typical of a dihybrid cross between aheterozygous organism and a homozygous recessiveorganism, where the alleles show complete dominance.
Dih d id ( i 9 3 3 1)
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Dihydrid crosses (ratio 9:3:3:1)
If both parents are
heterozygous. There are sixteen
possibilities.
9 purple, cut:
3 purple, potato:3 green, cut
1 green, potato
9:3:3:1 ratio
This is typical dihydridcross between twoheterozygousorganism.
Dih d id
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Dihydrid crosses
Paper 4 (J06)
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Paper 4 (J06)
The summer squash plant produces fruits that are either
white of yellow in color and are either shaped like a discor a sphere. The dominant phenotypes are white and disc-shaped fruit. Using the symbols A for white and a foryellow and B for disc and b for sphere, draw a geneticdiagram to show what proportion of offspring will have
yellow and sphere-shaped fruit if a white and disc-shapedfruit plant, heterozygous for both genes, is self fertilized.[6]
Paper 4 (J07)
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Paper 4 (J07)
A gene for feather color inchickens is carried on anautosome. This gene has twoalleles, black (CB) and splashed-white (CW). When a malechicken with black feathers ismated with a female chickenwith splashed-white feathers,all the offspring have bluefeathers. This also occurs whena male chicken with splashed-white feathers is crossed with afemale with black feathers.
Another gene may cause stripson feathers (barred feathers).
This gene is carried on the Xchromosome. The allele forbarred feather (XA) is dominantto the allele for non-barredfeathers (Xa). In chickens themale is homogametic and has
two chromosomes while thefemale is heterogametic andhas one X chromosome and oneY chromosome.
Paper 4 (J07)
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Paper 4 (J07) A male chicken with black,
non-barred feathers was
crossed with a female chicken
with splashed-white, barred
feathers. All the offspring had
blue feathers, but the males
were barred and the femaleswere non-barred. Using the
symbols given above, draw a
genetic diagram to show this
cross. [5]
Explain how a farmer coulduse a breeding program to find
out the genotype of a male
chicken with blue, barred
feathers. [3]
P 4 (J07)
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Paper 4 (J07)
The X2 (Chi squared) test
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The X2 (Chi-squared) test
Two plant produced a total of 144 offspring.
If the parents really were both heterozygous, and if the purplestem and cut lead alleles really are dominant, and if the
alleles really do assort independently, then we would expect
the following numbers of each genotype to be present in the
offspring:
But image that, amongst these 144 offspring, the results weactually observed were as follows:
Purple, cut 86 green, cut 24
purple, potato 26 green, potato 8
The X2 (Chi squared) test
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The X2 (Chi-squared) test
We might ask, are these results sufficiently close to ones we
expected that the differences between them have probably justarised by change, or are they so different that something
unexpected must be going on?
We can use a statistical test called the X2 (Chi-squared) test.
This results allows us to compare our observed results with theexpected results, and decide whether or not there is a significant
difference between them.
The first stage in carrying out this test is to work out the expected
results, as were have already done.
These, and the observed results, are then recorded in table like the
one overleaf.
We then calculate the difference between each set of results, and
square each different. (squaring it get rid of any minus signs- it is
irrelevant whether the differences are negative or postive)
Th X2 (Chi d) t t
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The X2 (Chi-squared) test Then we divide each squared
difference by the expected value,and add up all of these answer:
The X2 (Chi squared) test
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The X2 (Chi-squared) test
Now we have X2 value.
We look in a table that relates X2 values to probabilities. The probability given in the table are
E.g. a probability of 0.05 means that we would expect thesedifferences to occur in 5 out of every hundred experiments,or 1 in 20 just by chance.
In biological experiments, we usually take a probability of 0.05as being the critical one.
IfX2 value represents a probabilities of0.05 or larger, thenwe can be fairly certain that the differences between ourobserved and expected results are due to chance- the
differences between them are not significance.
The X2 (Chi squared) test
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The X2 (Chi-squared) test
There is one more aspect of our result to consider, before we
look up our value of X2 in the table- degree of freedom. This takes into account the number of comparisons made.
(Remember that to get our value of X2 , we added up all our
calculated values, so obviously the larger the number of
observed and expected values we have, the larger X2
is likelyto be, we need to compensate for this)
To work out the number of degrees of freedom, simply
calculate the (number of classes of data-1). (4-1)= 3
We can look a the table to determine whether our resultsshow a significant deviation from what we expected.
The X2 (Chi squared) test
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The X2 (Chi-squared) test
Our calculated value is 0.79, the value is a much smaller valuethan the one we have read from the table.
It would be way off the left hand side, representing a
probability of much more than 0.1 (1 in 10) that the differencein our results is just due to chance.
So we can say that the difference between our observed andexpected results is almost certainly due to chance, and thereis no significant difference between what we expect, and
what actually got.
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Paper 4 N09
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Paper 4 N09Complete the missing values in
the table. [3]
Calculate the value for chi-
square. [1]
Table below relates chi-square
values to probability values. As
four classes of data were counted,
the number of degrees of freedom
was 4-1=3. Table gives values of
chi-square where there are three
degrees of freedom.
Explain whether or not the
observed results were significantly
different from the expected results.
[2]
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Mutation
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Mutation
Mutation- an unpredictable change in the genetic material
of an organism. 2 type of mutation
1. Gene mutation- a change in the structure of a DNAmolecule, producing a different allele of a gene.
2. Chromosome mutations- mutation changes in thestructure of or number of whole chromosomes in a cell.
Mutations may occur completely randomly with no obviouscause, but there are several environment factors thatsignificantly increase the chances of a mutation occurring
(mutagen)
1. ionising radiation- (, , )
2. Ultraviolet radiation
Note: Mutagen- a substance that can cause mutation.
Mutation (gene mutation)
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Mutation (gene mutation)
Addition
Mutation (gene mutation)
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Mutation (gene mutation)
Base additions or deletions usually have a very significant
effect on the structure, because they alter every set of threebases that follow them in the DNA molecule. Therefore thefunction, of the polypeptide that the allele codes for.
They are said to cause frame shifts in the code. It mayintroduce a stop triplet part way through a gene, so that a
complete protein is never made at all. Base substitution may have no effect at all.
A mutation that has no apparent effect on an organism is saidto be a silent mutation. E.g. CCT to CCA or CCG makes still
make Gly. However, base substitution can be very large effect. E.g. base
sequence ATG (coding for Tyr) mutated to ATT, this hasproduced a stop triplet, so the synthesis of protein would
sop at this point.
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Sickle cell anaemia
Mutation due to base substitution at the genethat codes for polypeptide chains ofhaemoglobin
CTT which codes for glutamic acid mutates toCAT which codes for valine
Mutant polypeptide chains are less solubleand forms fibres in RBCs Sickle-shapeInefficient in O2 transportation and get stuckin blood capillaries
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Sickle cell anaemia
Recessive allele HS codes for the production of
mutant polypeptide chains that causes sickle
RBCs
HSHS Sickle cell anaemia
HNHS Carrier
HNHN Normal
Common in Africa
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Sickle cell anaemia and malaria
In malarial region, carriers of sickle cell allele(HNHS) can survive better, so HS allele isfrequent
HS
HS
Die from sickle cell anaemia HNHN Die from malaria
HNHS Not suffer from sickle cell anaemia and
less likely to suffer from malaria
Able toreproduce and pass on the HS alleleSelective advantage
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Phenylketonuria (PKU)
Phenylketonuria (PKU) a genetic disease resulting from amutation from in gene that codes for an enzyme involved in
the metabolism of Phenylketonuria.
Mutation happens on gene that codes for production of
enzyme phenylalanine hydroxylase
Complications of PKU:
Less melanin Lighter skin and hair colour
Phenylalanine accumulates in blood and tissue fluid which
cause mentally retarded and brain damage in children.Phenylalanine hydroxylase
Phenylalanine tyrosinemelanin
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Converted
into Malanin
Phenotype affected by both
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1. Diet and nutrients- Height and weight
2. Low temperature
- Dark fur color on Himalayan rabbits
3. High temperature
- Gender of reptiles
4. Ultraviolet rays
- Production of melanin5. Wavelength of light
- Flowers and fruit color
Phenotype affected by bothenvironmental and Gene
d f l
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Production of Insulin
Injection of insulin is needed for diabetes
mellitus patients
Insulin extracted from animal sources is
expensive, has side-effects and ethical /religious issues
Therefore, production of insulin through
genetic engineering is encouraged
Production of Insulin
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Production of Insulin
1. Isolating the insulin gene- mRNA for insulinextracted from cells(human pancreas)
- Incubate mRNA withreverse transcriptaseenzyme
- Complementary DNA(cDNA) is formed
- DNA polymerase is addedto form complete double-helix structure
Production of Insulin
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Production of Insulin
2. Inserting the gene into avector
- Vector Small carrier inwhich fragments of DNAcan be inserted, eg. Plasmid
- Plasmids are centrifugedand isolated from bacteria
- Enzyme restrictionendonuclease is used to cutthe plasmids with stickyends
- DNA ligase to join insulingene into plasmidsRecombinant DNA
Production of Insulin
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3. Inserting the gene into the bacteria
- Plasmids with recombinant DNA mix with bacteria
- Bacteria that take up plasmids are isolated and cultured in
large scale
- Extraction and purification of recombinant insulin
Production of Insulin
Production of Insulin
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Production of Insulin
U f h l
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Uses of gene technology Medical: Synthesis of growth hormones, recombinant Factor
VIII to treat haemophiliac and production of pharmaceuticaldrugs.
The human gene for making factor VIII has been inserted intohamster kidney and ovary cells that are then cultured infermenters. The cells constantly produce factor VIII, which is
extracted and purified before being used to treat people withhemophilia.
Before availability of the recombinant factor VIII, came fromhuman and this carried risks of infection, such as HIV.
Agriculture: Genetically modified crops against pests andinsecticides, genetically modified organisms with desirabletraits and better yields
Industry: Biological washing powder and production ofgenetically modified food