CIE A-level Biology Topic 16: Inherited Change REVISION

CIE International A-level Biology
Topic 16 REVISION
Inherited change

Today’s TOPIC 16
REVISION lesson is based
at your local hospital and
you’re going to begin the
day on a specialist
diseases ward. If two of
the diseases on the ward
today are haemophilia
and Huntington’s, can you
work out which word
should go on the top line
of the sign at the
entrance to the ward?
DISEASES
WARD
___________
GENETIC

Before you’re allowed to meet
some of the cases in the
different rooms of the ward, the
director of the hospital wants to
check that your genetic
terminology is up to scratch and
it just so happens that they love
a quiz competition….

Round 1 is GENETIC
TERMINOLOGY SNAP
and this will enable us
to determine who
knows their allele
from their autosome!

THE RULES
A genetic term and a definition or
description will appear on the board at the
same time. If they match, the first person
to shout out “SNAP” gets 2 TEAM POINTS.
If they don’t match, then shout out
“VARIATION” and in order to get the 2
TEAM POINTS, you have to state the
correct definition of the genetic term and
the term which should have gone with the
definition.

GENOTYPE
The physical
expression
of a trait

GENOTYPE = the genetic makeup of an organism
Did you get the 2 TEAM POINTS?
The physical expression of a trait = PHENOTYPE

ALLELE
An
alternative
form of a
gene

LOCUS
The specific
position of a
gene on a
chromosome

The locus is the position of a single gene on a chromosome and
there may be many alleles of a single gene at that locus
Complete this table by naming each of the alleles
that can be found at a locus from their description
(3 marks)
Two copies of this allele must be found in the genotype
if it is to be expressed in the phenotype
At least one copy of this allele must be found in the
genotype for it to be expressed in the phenotype
Neither alleles are recessive but instead, heterozygotes
will express both phenotypes
RECESSSIVE
DOMINANT
CODOMINANT

LINKAGE
Two or more genes
that are found on
different
chromosomes

LINKAGE = two or more genes that are found on
the same chromosome
Two or more genes that are found on different
chromosomes = GENE INTERACTIONS

The gene which controls the inheritance of the ABO blood
group and the gene which controls the inheritance of a
disease called nail-patella syndrome are both found on
chromosome 9.
State the type of linkage that is demonstrated by
these two genes (1 mark)
autosomal linkage

When studies are carried out on the inheritance of blood groups and
nail-patella syndrome in groups of people, the phenotypes of the
parent are normally exactly the same as the phenotypes of the
offspring and there are very few if any new combinations
Discuss what this suggests about the
loci of the ABO gene and the nail-
patella gene and take a few
moments to see if you can explain
your answer
The loci of the two genes is extremely
close on chromosome 9. As a result, the
chiasma which randomly forms during
crossing over in prophase I of meiosis,
rarely forms between the two loci which
is what would be needed to result in
new combinations of phenotypes called
recombinants

DIPLOID
One complete
set of
chromosomes

DIPLOID = two complete sets of chromosomes
One complete set of chromosomes = HAPLOID

In case you are interested……
Chromosomal mutations can lead
to polyploidy, where complete sets
of chromosomes are duplicated.
Three complete sets of
chromosomes is known as triploid
and four is tetraploid

HETEROZYGOUS
A genotype
that
contains two
different
alleles

HOMOLOGOUS
Pairs of
chromosomes
which are similar
in length, gene
loci and
centromere
location

Terminology check
over, it’s now time
for you to meet the
range of patients
on the ward….

ROOM #1
Alicia Iwenofu
Alicia is suffering from sickle cell
anaemia, a genetic disorder that
results from a substitution
mutation. The inheritance of two
recessive alleles results in an
abnormality in the haemoglobin in
red blood cells and a type of
haemoglobin S. The sickle-shaped
red blood cells are not as efficient
at transporting oxygen around the
body and are prone to getting
stuck in capillaries
WARD
GENETICS

Alicia is planning to start to try for children with
her husband. As Alicia is a sufferer, her husband
was encouraged to undergo a genetic screening
and the results found that he has a heterozygous
genotype for this gene.
Let’s remind ourselves of how to carry out a clear
genetic diagram to determine the % chance of
Alicia and her husband having a child who suffers
from the condition. We will use the letters S and
s to represent the dominant and recessive alleles
respectively

A male who is heterozygous
reproduces with a female sufferer.
What is the % chance of them
having a child who suffers?
#1
Ss ss
x

#2
Ss ss
x
S s s

This might just prove useful……
All of Alicia’s egg cells would have
contained a single recessive allele.
Therefore, there is no need to show
repeats of the same gamete as it will just
lead to more work later on. Importantly,
the % or phenotypic ratio is not affected.
So remember that you only need to show
the different possible gametes that a
parent can make

As this stage of a genetic cross, you will use a diagram called a
Punnett cross. It is named after its inventor, Reginald Punnett,
and this simple but effective diagram allows the different
possible offspring genotypes to be observed for a particular
cross.
The gametes of one parent are
written in here
And the
gametes of the
other parent
are written in
here
And then the gametes are
brought together in here to
show the different possible
genotypes of the offspring
that could result

#3
Ss ss
x
S s
S s
s
s

#4
Ss ss
x
S s
S s
s Ss ss
s

#5
Ss ss
x
S s
S s
s Ss ss
1 out of the 2
offspring genotypes
are sufferers so 50%
chance
s

This might just prove useful……
Alicia’s husband in this example had a
heterozygous genotype. Individuals with
this type of genotype are also known as
(asymptomatic) carriers. They don’t suffer
from any of the symptoms of the disease
but do “carry” one of the recessive alleles
in their genotype. Carriers only exist in
recessive conditions

Sickle cell anaemia is caused by a
gene mutation called a substitution,
where a DNA triplet that read
cytosine - thymine – cytosine
becomes
cytosine – adenine – cytosine.
Use this mRNA codon table to work
out the change in the primary
structure that results (2 marks)

DNA: CTC
RNA: GAG
AA = Glu
DNA: CAC
RNA: GUG
AA = Val

If the substitution had replaced CTC with CTT, the same amino acid
(Glu) would have been coded for and there would have been no change
in the primary structure of the polypeptide. This is known as a silent
mutation
With reference to the genetic code, explain how it
is possible to have a silent mutation (2 marks)
As the genetic code is degenerate, where each amino acid is
coded for by more than one DNA triplet, it is possible for a
mutation to occur that changes a single base on DNA but
still result in the same amino acid in the primary structure.

*
Not all of the 20
amino acids have
more than one
triplet that codes
for it.
Methionine and
Tryptophan only
have 1 triplet
code

Take a break from all of those difficult topic 16 tasks and
questions and just for fun, see if you can work out the
three terms represented below which are somehow related
Two buttons on the top-right of a laptop keyboard
1 2 3 4 5 6 7 8 9 10 11 12 1 2 3 4 5 6 7 8 9 10 11 12
INSERT DELETE
FRAME SHIFT
NON-OVERLAPPING

Challenge your knowledge of DNA, RNA and protein synthesis by carrying out the
following tasks:
#1 For both the non-mutated DNA and the DNA which
has been altered by an insertion mutation, work out the
code on the respective mRNA strands then work out the
primary structure of the resulting polypeptides.
#2 Compare the two primary structures and use your
findings to describe the affect of the frameshift, making
sure to include insertion (deletion) and non-overlapping
too

T A C A C G G T C T T T T C G
T A C A C C G G T C T T T T C G

Due to the non-overlapping nature of the genetic
code, where bases are always read in groups of 3,
an insertion (or deletion) mutation changes the
reading frame and this is called a frame shift. This
results in a completely different translation, as
every triplet and resulting codon after the point of
the mutation is changed meaning different amino
acids will be found in the primary structure
Compare yours against
this

An insertion (or deletion) mutation changes the
reading frame and this is called a frame shift
Do all insertion or deletion
mutations have the same impact on
the primary structure? Discuss
whether the point of the mutation
affects the “size” of the change
NO, not all insertion or deletion mutations have the
same impact and YES, it very much does matter where
the point of the mutation is.
Ultimately, the nearer the beginning of the DNA
sequence that the mutation occurs, the bigger the
effect! For example, we just had an addition mutation
which came after the 5th base which is really near the
start. As a result, of the 210 amino acids that would
be translated, only the 1st amino acid remains
unaffected. That means 209 amino acids would be
changed!

ROOM #2
Julia Robyns
Julia has recently found out that
she suffers from the late-onset
disease called Huntingtons chorea.
This is a genetic disorder that
results in the death of brain cells
and is characterised by a lack of
coordination and dementia. The
gene has its locus on chromosome
4 and a zygote which inherits at
least one dominant allele will
suffer from this currently incurable
disease
WARD
GENETICS

Huntington’s chorea is a genetic disease caused by the
inheritance of at least 1 dominant allele
Julia has been correctly told by another doctor on the ward
that if she reproduces with her husband, who is a non-
sufferer, there is a 50% chance that her offspring will suffer
from Huntingtons.
Use a genetic diagram to work out Julia’s genotype (2 marks)
Use H and h to represent the alleles

Julia must have a heterozygous genotype
(Hh) in order for there to be a 50% chance of
her offspring suffering from the disease.

Another individual who suffers from Huntington’s chorea
has a homozygous dominant genotype. If he reproduces
with a female who doesn’t suffer, what are the % chances
of the child not suffering from the condition? (1 mark)
A 0%
B 25%
C 75%
D 100%
By using key terminology throughout, write a
paragraph that explains why the offspring of this
male will definitely inherit Huntington’s chorea

Compare yours against this
As the male has a homozygous
genotype, all of his sperm cells will
contain a dominant allele. When this
sperm fertilises the egg cell, the
genotype of the zygote will contain at
least one copy of the dominant allele
which is all that is needed for the
disease to develop

The male has a homozygous genotype (HH)
As Huntington’s chorea is a
genetic condition, discuss
what can be deduced about
the genotypes of his parents
The male will have inherited one
dominant allele from his father
and one from his mother, so you
know that both his parents were
at least heterozygotes (Hh) and
were both sufferers too

Most subjects have a
mathematical element
and Biology is definitely
one of those! Round 2 is
called “YOU DO THE
MATH!”

YOU DO THE MATH!
There will be 2 rounds of this competition over the
course of the lesson.
In this round, a series of 4 statements will appear on the
board one at a time and each of them has a number as the
answer.
By considering the numerical signs between each of the
statements, your task is to calculate the final answer.
Buzz in when you are confident about the answer and if
you’re correct, you’ll get 2 TEAM POINTS!

Daughter cells at the end of meiosis
Cell divisions in meiosis
Alleles in a gamete (for each gene)
X
-
Bases that lead to 1 amino acid
X
Discuss why this number is
likely to be important to
someone working in the
genetics ward?

Pair 23
In humans, the 23rd pair of chromosomes are
known as the sex chromosomes and humans can
have either X chromosomes or Y chromosomes.
Females have the genotype XX whereas males have
the genotype XY
As the Y chromosome is much
smaller, it is missing some of
the genes that are carried on
the X. Genes carried on these
sex chromosomes are said to
be sex-linked

ROOM #3
Robert Sheens
Robert suffers from
haemophilia, a condition
which impairs the body’s
ability to clot their blood and
to stop bleeding.
This is a sex-linked condition
where the recessive allele
that causes the condition is
carried on the X
chromosome.
WARD
GENETICS

This is actually fairly important……
When writing the genotype of an
individual in a sex-linked disease, you
must show the sex chromosomes as well
as the alleles
e.g. XHXh

Robert’s girlfriend,Anne, is pregnant and the
couple would like to know what the likelihood is for
the offspring in terms of inheriting haemophilia.
Anne is heterozygous for the haemophilia gene.
Use a genetic diagram to calculate the full
phenotypic ratios of the offspring and use the
alleles R and r to represent the dominant and
recessive alleles respectively
(3 marks)

1 : 1 : 1 : 1
female healthy : female sufferer : male healthy : male sufferer

Am emergency has come in on
the next ward and they are
preparing to give the patient a
blood transfusion. However,
as they are running low on the
reserves of each type of
blood, they really want to
make sure that they give this
patient the correct blood so as
not to waste any of the
precious reserves.
But should they give this
patient A, B, AB or O?

A quick reminder: ABO Human Blood groups
The gene which determines the ABO human blood
groups has its locus on chromosome 9.
The gene locus is represented by I
(isohaemaglutiinogen) and this determines which
antigen is present on the surface of the
erythrocytes.
Three different alleles can be found at this locus,
which are IA, IB and IO.
Genetically, the alleles IA and IB are co-dominant
and IO is recessive to both of them

Three different alleles can be found at the locus on chromosome 9,
and these are IA, IB and IO. Genetically, the alleles IA and IB are co-
dominant and IO is recessive to both of them
Use the information above to copy and complete this table to show
the missing genotypes and phenotypes (4 marks)
Genotype (s) Phenotype
A
IBIB IBIO
IAIB
O
IAIA IAIO
B
AB
IOIO

Fortunately, one member of the patient’s family
has a copy of a genetic tree which shows some of
their blood groups, including the patient who is
labelled member number 5.
Using the example that is shown, work out the
blood group of the patient so that you can inform
the emergency department of the blood type that
should be transfused
If you think you will find these tasks very difficult, ask for a
worksheet labelled STARS which gives a little bit of assistance

One of the offspring, number 7, has the
genotype IOIO, meaning that one of those
alleles was inherited from his father, the
patient (5). In order for offspring (6) to have
the blood group B, they need an IB allele in
their genotype and this must have been
inherited from her father, the patient.
Therefore, the patient has the genotype IBIO
which is blood group B and this is the blood
type which he must have transfused.

The son of the patient, member
number 8, is keen to know his blood
type as everyone else now knows
theirs. As his mother has blood group
A and his father B, use their
genotypes to work out the possible
blood groups of member 8 (2 marks)

As you’ve already encountered,
individuals within a population of a
species may show a wide range of
variation in phenotype. This can
be due to both genetic and
environmental factors
The primary source of genetic
variation is mutation, but that is by
no means the only source…..

Moving quickly onwards with this revision
lesson, have a bit of fun by working out the
process which is represented in the box below
Once you have the name, use your knowledge (and what you
can see) to describe the events of this process
non-sister
chromatids
Crossing over
The exchange of alleles between non-
sister chromatids (during prophase I of
meiosis) which results in genetic
variation due to the production of new
allele combinations called recombinants

Crossing over involves the exchange of alleles between non-sister
chromatids and ensures genetic variation
Explain why the exchange of alleles between sister
chromatids would not result in the genetic variation that is
observed with non-sister chromatids. You should make
reference to a stage of the cell cycle in your answer (3 marks)
During the S phase of interphase, DNA replication occurs where
each chromosome is duplicated. The result is a pair of sister
chromatids which are identical to each other. Therefore, exchange
between sister chromatids would not result in any variation.

Which of the following is the correct order
of the 4 phases in meiosis I? (1 mark)
A metaphase I  anaphase I  prophase I  telophase I
B telophase I  metaphase I  anaphase I  prophase I
C prophase I  anaphase I  metaphase I  telophase I
D prophase I  metaphase I  anaphase I  telophase I

Crossing over involves the exchange of alleles between
non-sister chromatids and ensures genetic variation
In addition to crossing over, state the name of the
other process that occurs during meiosis which ensures
genetic variation and briefly describe what occurs in
this process (2 marks)
Independent assortment of chromosomes is the random
orientation of each chromosome within a homologous pair
at the equator

Using only the stages of meiosis shown at the bottom, state
the stage or stages in which the following events occur:
(i) Independent assortment
(ii) Formation of the spindle apparatus
(iii) Separation of sister chromatids
(iv) Formation of nuclear membranes
(v) Chromosomes pulled to opposite poles
Metaphase I and II
Prophase I
Anaphase II
Telophase II
Anaphase I

I hope you made a note of the key
terminology difference in (iii) and (v).
During anaphase I, chromosomes are
pulled to opposite poles by the spindle
fibres whereas during anaphase II, it is
sister chromatids which are separated

Using the sketch graph for the cell cycle involving
mitosis shown below as a guide, draw one to
represent the changes in the quantity of DNA
during the cell cycle involving meiosis. Make sure
that it is fully annotated

In addition to the contributions
of crossing over and random
assortment to variation, the
random fertilisation of the
haploid gametes (produced by
meiosis) during sexual
reproduction will produce
further genetic variation

An individual’s genotype will
contain 2 alleles for every gene
that is being inherited but
following meiosis, each gamete
will contain 1 allele for every
gene that is being inherited

The next set of inheritance questions will focus on the
inheritance of two genes (dihybrid inheritance).
Which of the following will be correct for the parent’s
genotypes and the gametes they produce? (1 mark)
A The parent’s genotype will contain 4 alleles and each gamete will
also contain 4 alleles (two from each gene)
B The parent’s genotype will contain 2 alleles and each gamete will
contain 4 alleles (two from each gene)
C The parent’s genotype will contain 2 alleles and each gamete will
contain 2 alleles (one from each gene)
D The parent’s genotype will contain 4 alleles and each gamete will
contain 2 alleles (one from each gene)

“The parent’s genotype will contain 4 alleles and each
gamete will contain 2 alleles (one from each gene)”
Imagine that you were looking at the inheritance of
two genes, A and B.
One parent has a genotype that is heterozygous for
both genes.
AaBb
AB
This genotype would produce four different gametes
Ab aB ab

Still looking at the inheritance of the two genes, A and B,
write down all of the different possible gametes that would
result from these genotypes (2 marks)
1. AABb
2. A parent who is homozygous recessive for both
genes
AB Ab
ab
aabb

This way to
our new
FAMILY
GARDEN
The hospital was given a substantial
donation by an anonymous donor and
this money was used to build a large
garden at the back of the hospital for the
use and enjoyment of the patients and
their families.
One of the most recognisable features of
the garden are the large summer squash
that the head gardener grows in the
greenhouse.

The head gardener is keen to learn more about the
genetics behind the inheritance of colour and shape in
these fruits so that he can produce a handout for visitors
who may be interested.
Try question 1 on the dihybrid inheritance worksheet
and calculate the phenotypic ratio.
If you don’t feel confident at this stage to be able to carry out
genetic crosses for the inheritance of two genes, ask for an
assistance sheet labelled STARS

3 x disc-shaped, white
3 x disc-shaped, yellow
1 x sphere-shaped, white
1 x sphere-shaped, yellow
3 : 3 : 1 : 1

YOU DO THE MATH!
Just like last time, a series of 4
statements will appear on the board
and each of them has a numerical
answer. This time the 4 numbers will
form a ratio so the 2 TEAM POINTS will
be awarded to the 1st team to buzz in
with the correct ratio

Types
of
muscle
X
Types
of
muscle
:
Main
types
of
blood
vessel
Amino
acids
coded
for
by 3
DNA
bases
DNA
bases
which
code
for 1
amino
acid
: :
An important
phenotypic ratio
9 : 3 : 3 : 1

Which of the following parent genotypes would
lead to an expected ratio of 9:3:3:1? (1 mark)
A One parent who is heterozygous for the two genes and the other
parent is homozygous dominant for both genes
B One parent who is heterozygous for the two genes and the other
parent is homozygous recessive for both genes
C One parent who is homozygous dominant for both genes and the
other parent is homozygous recessive for both genes
D Both parents are heterozygous for the two genes

Now try question 2 on your
worksheet and prove that the
expected ratio of the cross
between individuals who are
both heterozygous for two
genes is 9:3:3:1

9 : 3 : 3 : 1
purple, long : purple, short : red, long : red, short

9 : 3 : 3 : 1
This is the expected ratio when two heterozygotes for two
genes are crossed
This is a really important ratio to remember. You
may be asked to carry out a statistical test called
the chi-squared test which determines whether
there is a significant difference between the
observed and expected results. If the two parents
involved are heterozygotes for two genes, then
don’t waste time carrying out a genetic cross to find
the expected phenotypic ratio because it is 9:3:3:1

Chi squared test (x2)
x2 = Σ(O-E)2) ÷ E
Σ = sum
O = observed numbers
E = expected numbers

Use the observed results to calculate the
x2 value and use this to accept or reject
the null hypothesis
If you cannot recall how to calculate the expected numbers
for a phenotype, ask for a worksheet labelled STARS which
has the first calculation completed and can be used as a
guide for the work after that

phenotype Observed (O) Expected (E) (O-E)2) ÷ E
purple and smooth 216 (9 ÷ 16) x 381 = 214.3 0.013
purple and shrunken 79 (3 ÷ 16) x 381 = 71.4 0.809
yellow and smooth 65 (3 ÷ 16) x 381 = 71.4 0.573
yellow and shrunken 21 (1 ÷ 16) x 381 = 23.8 0.329
x2 1.724

There were four phenotypes, so the degrees of freedom (df) = 3
As the x2 value is lower than the critical value, the null hypothesis is
accepted as there is no significant difference and any difference is
due to chance

Gene interactions are another
cause of variation where the
presence of a gene at one locus
suppresses or masks the
expression of another gene at a
second locus. This can be known as
epistasis, but you do not need to
be aware of this term for this
specification. Instead, you need to
be able to problem solve…..

Answer the questions on the
worksheet which challenges your
problem solving skills with a gene
interaction that controls the
colour of the coat in Labradors

Q Required response
1a BBEe = black
1b bbEE = chocolate
2 BBee
The presence of two recessive alleles at the second locus
masks the expression at the first locus, no matter which
alleles are present there and results in a yellow coat. If
there is at least one dominant allele at the second locus and
at least one dominant allele at the first locus, then the coat
colour will be black. In order for a Labrador to have a
chocolate coat, they need to have at least one dominant
allele at the second locus and be homozygous recessive at
the first locus

yellow
BbEe
(BE), (Be), (bE), (be) (be)
(BE) (Be) (bE) (be)
(be) (BbEe) (Bbee) (bbEe) (bbee)
(BbEe) (Bbee) (bbEe) (bbee)
black yellow yellow
chocolate

The yellow coat colour is due to the presence of a
pigment called phaeomelanin that is found in the hairs.
The black and chocolate coat colours are due to
different amounts of another pigment, eumelanin, that
is deposited in these hairs. The more eumelanin there
is, the darker the hair.
The diagram shows the action of genes E and B in the
production of the different coat colours.

Explain how
the genotype bbEe
produces a
chocolate coat
colour

Let’s try this together…
Explain how the genotype bbEe produces a chocolate coat
colour
Alleles
Due to the presence
of a dominant allele
(E), the enzyme is
coded for
Pigment
Phaenomelanin is
converted to
eumelanin
Alleles
No dominant allele (B)
means a low amount
of eumelanin is
deposited
COLOUR
The coat is chocolate

Using the example you’ve just seen where the
answer discussed the alleles (and enzymes), then
the pigment and finally the coat colour, answer the
following:
1. Explain how the genotype BbEe produces a
black coat colour (3 marks)
2. Explain how the genotype bbee produces a
yellow coat colour (2 marks)

As there is no dominant allele (E), the enzyme
will not be coded for and so phaeomelanin is
not converted and the coat remains yellow
Due to the presence of a dominant allele (E), the
enzyme is coded for and phaeomelanin is
converted to eumelanin. Due to the presence of
the dominant allele (B), more eumelanin is
deposited in the hairs and the coat colour is
black

State the name of the
disaccharide which is formed of α-
glucose and galactose (1 mark)
lactose

So far this revision lesson has
concentrated on the genetics and
variation that is seen in eukaryotes like
humans, dogs and summer squash.
However, topic 16 expects you to
know about genetics in bacteria and
the lac operon can be used to explain
the genetic control of protein
production in prokaryotes

Round 3 and this is
designed to check
whether your
knowledge is
LACking!

A passage about the Lac Operon will appear on
the board. You have 1 minute to read through
the passage carefully and then prove that
knowledge isn’t LACKING
• If it has been written 100% correctly, simply put a big tick on
your whiteboard to get the 2 TEAM points.
• If there is a mistake, write the mistake on your whiteboard
and the word which actually should have been in the passage
and if both are right, you will get the 2 TEAM points
“Is your knowledge LACKING?”

When lactose is absent,
the regulatory gene is
expressed and the
suppressor protein is
synthesised
When lactose is absent,
the regulatory gene is
expressed and the
repressor protein is
synthesised

The repressor protein binds
to the operator region, but
partially cover the promoter
region too meaning that
RNA polymerase is unable to
bind

As the RNA polymerase cannot
attach, the structural genes Z
and Y are not transcribed and
the enzymes lactose permease
and alpha galactosidase will not
be synthesised
As the RNA polymerase cannot
attach, the structural genes Z
and Y are not transcribed and
the enzymes lactose permease
and beta galactosidase will not
be synthesised

When lactose is present in
the medium, it binds to the
repressor protein, causing
the shape of the other
binding site to change

The repressor protein is now
unable to bind to the
operator region, leaving it
free for the enzyme DNA
polymerase to bind
The repressor protein is now
unable to bind to the
operator region, leaving it
free for the enzyme RNA
polymerase to bind

Transcription and translation of the
structural genes leads to the
synthesis of lactose permease and
beta galactosidase which allow E-coli
to take up the lactose and then break
it down into glucose and sucrose
Transcription and translation of the
structural genes leads to the
synthesis of lactose permease and
beta galactosidase which allow E-coli
to take up the lactose and then break
it down into glucose and galactose

The 4th and final round
will challenge your
Biology, literacy and
numeracy skills as you
have to CONVERT FROM
NUMBERS 2 LETTERS

On the next slide, a series of statements will appear
and each of these statements has a number as
their answer. Those numbers then correspond to a
letter of the alphabet (e.g. 6 = F).
Buzz in when you have converted FROM NUMBERS
2 LETTERS and if you state the correct term, you’ll
get 2 TEAM POINTS.

 Polypeptide chains in haemoglobin
 Carbon atoms in ribose
 Hydrogen atoms in a molecule of glucose
 (Polypeptide chains in collagen) x 4
 Polypeptide chains in a non-quaternary protein

DELLA proteins are repressors of
plant development and inhibit seed
germination, seed growth and
flowering. By binding to a range of
transcription factors, which include
PIFs (phytochrome interacting
factors), they are able to repress the
actions of these proteins that control
the rate of transcription from DNA to
mRNA.

By binding to a range of transcription factors which include
PIFs they are able to repress the action of these proteins.
Suggest how the DELLA proteins can repress
transcription factors like PIFs (2 marks)
They stop the transcription factors from
binding to the DNA or they promote the
degradation of these proteins

Repression of the repressor
State the name of the plant hormone
which promotes the degradation of the
DELLA proteins (1 mark)
gibberellins

Gibberellins activate genes
The binding of gibberellins to a receptor
triggers a sequence of events that catalyses
the addition of a substance called ubiquitin
to the DELLA proteins and promotes their
degradation. As a result, the DELLA
proteins no longer inhibit the transcription
factors, allowing them to bind to bind to
DNA and promote transcription of the DNA
template to mRNA

CIE A-level Biology Topic 16: Inherited Change REVISION

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CIE A-level Biology Topic 16: Inherited Change REVISION