genetic discussion 3 (mutations, DNA damage, diseases)

Question 1

What type of mutation is most likely to cause
KNOCKOUT MUTATION of the gene
(i.e. cause TOTAL LOSS OF GENE FUNCTION)?

What are the causes of this mutation naturally?

Answer 1

A NONSENSE Mutation:
- C. BY CODON ==> STOP CODON

= This CAN OCCUR NATURALLY IN REPLICATION particularly when a
==> BASE IS ADDED or DELETED
==> causing a FRAMESHIFT MUTATION

*STOP CODON is incorporated early in the sequence then this may be called a null or knockout mutation as it COMPLETELY INACTIVATES THE GENE

Question 2

Give an example of how a viral infection can be treated by using a DNA base analogue

Answer 2

= Herpes

==>Herpes are DNA viruses and 
==>use the cell machinery to replicate
==>  ACYCLOVIR looks like a BASE 
used for a growing DNA chain
==> acyclovir picked up by VIRAL POLYMERASE
more than our own DNA polymerase 
more likely to 
==>stop the growing viral DNA.

Question 3

3*. What constitutes a silent mutation and how do these arise?

Answer 3

Silent mutations = mutation that
DOES NOT LEAD TO AN AA CHANGE
in the sequence of amino acids
that make up the genes protein.

This is bec.. THE CODE IS “DEGENERATE”
so a change esp 3RD BASE in a codon
may still lead to the SAME AA

Question 4

4*. What common disease might arise in persons with
— defective NER process?

Explain your answer

*PROCESS nucleotide excision repair (NER)…
==> damaged bases are cut out
within a string of nucleotides
==> replaced with DNA
as directed by the undamaged TEMPLATE STRAND

*NER system used to ..
- remove PYRIMIDINE DIMERS
formed by UV radiation

• remove NUCLEOTIDES
  modified by bulky chemical adducts.

Answer 4

Individuals with a defective NER process would be
at a HIGHER RISK OF SKIN CANCER.

==>bec.. NUCLEOTIDE EXCISION REPAIR (NER) process =
important mechanism that
==>REMOVES DNA DAMAGE = 2 Ts are fused
==>induced by UV LIGHT

==> NER is also damaged
==> SOS repair is used
==> SOS repair does NOT HAVE PROOFREADING
>>> MORE mutations occur
>>> leading ==> CANCER

Question 5

What type of
— repair mechanism
would be used for a
—thymine dimer

and why?

*PROCESS nucleotide excision repair (NER)…
==> damaged bases are cut out
within a string of nucleotides
==> replaced with DNA
as directed by the undamaged TEMPLATE STRAND

*NER system used to ..
- remove PYRIMIDINE DIMERS
formed by UV radiation

• remove NUCLEOTIDES
  modified by bulky chemical adducts.

Answer 5

thymine dimer =
==> 2 ADJACENT THYMINE BASES in the same DNA strand
==>BIND TOG
==>resulting in a thymine DIMER.

==>two bases that need repairing,
==>the (NER) mechanism is used
==>this cleaves the thymine dimer from the DNA strand ==>new DNA is synthesised to replace the dimer.

Question 6

6*.
—How many nucleotides are
replaced with BER DNA repair
(base excision repair)

and which
—enzyme
would probably do this job?

• **difference between
• BER/mismatch repair
• NER nucleotide excision repair

BER/mismatch repair
= one nucleotide is replaced

NER nucleotide excision repair
= several nucleotides are replaced.

Answer 6

= 1 nucleotide is replaced.

==>mutated base is flipped out of helix
==>replaced with a non- mutated nucleotide

multiple enzymes are used but
= DNA POLYMERASE is crucial
*as it has proof reading ability.

Question 7

Explain how

• –smoking causes cancer and
• -mutations in embryos

Answer 7

==>ERRORS IN APOPTOSIS
==>cause UNCONTROLLED CELL GROWTH
==> can cause TUMOURS

p53 gene = ACTS “the GUARDING OF THE CELL”
***MUTATION OF P53 gene
are commonly FOUND IN CANCER

==>BENZOPYRENE is a
POLLUTANT IN CIGARETTE SMOKE
==>acts as an ADDUCT (sticks to a cells DNA)
==> can cause MUTATION

==>Benzopyrene sticks to p53
==>STOPS IT from DOING ITS JOB (protecting cell)

==>benzopyrene ADDUCTS can STICK to 
SPERM & EGG DNA. 
tf person who obtained damage through smoking
==>TRANSFER MUTATIONS 
==> to OFFSPRING

Question 8

From slide 12.

Why can a knockout mutation be both

• a nonsense = CODON to STOP
• or a missense = 1 AA change (but NOT a stop)

but a missense mutation cannot be
- a nonsense mutation?

*** Missense mutation = 1 AA (not result in stop)
= change in one amino acid in a protein
arising from a point mutation in a single nucleotide.

***nonsense mutation = CODON > to STOP
= a codon is changed to a premature stop codon
that results in
“truncation” (Elimination of the N- or C-terminal)
of the resulting protein.

Answer 8

A Knockout mutation
= non functioning gene
- nonsense&raquo_space;> no gene product (stopped early)
- missence&raquo_space;> non-functional protein (same size)

A Missence Mutation
= still a gene product&raquo_space;> non functional

===============
Nonsense 
==> change of nucleotide 
==> leads to an introduction of a stop codon. 
>>> truncation / cut short protein

Missense
==> change of nucleotide
==> leads to a different amino acid, BUT NOT STOP codon.
»> non functional protein

Question 9

From slide 15 - hard.

In the crossover event in meiosis,
parts of mum and dad genes can swap chromosomes,
but the
- gametes still usually have only one copy of the gene.

How do the chromosomes ensure this happens?

Answer 9

In order for crossover to occur
- the genes must be VERY SIMILAR IN THEIR CODE.

crossover only occurs where there are
MATCHING ALLELES

Question 10

From slide 19.

Huntington’s disease is an autosomal dominantly inherited disorder.

What do the words in bold tell us about the disease?

Answer 10

Autosomal - NOT A SEX CHROMOSOME

Dominant - you ONLY NEED ONE COPY
of this gene to have the disease

Question 11

Cystic fibrosis is an
- autosomal recessive disorder.

What does this say about the

• genotype and
• phenotype of a sufferer?

Answer 11

= BOTH PARENTS MUST BE A CARRIER of the
DEFECTIVE GENE
to pass it off to their offspring.

Genotype: both copies of the defective gene must be present

Phenotype: only those individuals with both copies of the gene will have the phenotype of disease

Question 12

From slide 27.

Why is AZT relatively harmless to replication in human cells?

***Zidovudine (ZDV), 
AKA azidothymidine (AZT)
= an antiretroviral medication used to prevent and treat HIV/AIDS.

Answer 12

AZT is relatively harmless to replication in human cells because of the
= WAY IT IS PICKED UP WITHIN THE BODY.

T analog is in the body of a HIV positive individual
= viral HIV enzyme
is 1000 times more likely to pick up this T analog
(than by human cells)

==> AZT has been picked up by the HIV
==> inhibits the replication of the HIV virus
*by putting T analogue in place of thymidine (base)

• some side effects
• relatively harmless to human cells.

Question 13

How does
UV radiation
damage DNA differently to
X-rays?

Answer 13

==> UV radiation 
==> causing a Thymine Dimer 
*result of two T bases 
on the same DNA strand 
being stuck together. 

The thymine dimer doesn’t necessarily damage the DNA
- more the ACTION OF TRYING TO UNBIND the T bases
CAUSES THE DAMAGE

==> X-ray causes
==> breaks to either one or both strands of DNA.
==> tf damage occurs when
- genes are deleted or
- attempted fusion of chromosomes occurs
as it is a complicated process
that could easily go wrong.

Question 14

From slide 35.

How can a defective NER lead to skin cancer?

Answer 14

Defective NER is due to
==> the skin being overexposed to UV rays
==> NER mutates

then
==> the body has to resort to using an extreme form of repair, SOS = alternate polymerases 
==> have no proofreading ability 
==> create many errors 'MUTATIONS'
 in the DNA sequence.

Question 15

Speculate on why a cell may not rely on base excision repair to fix UV damage?

Answer 15

BER Base excision repair
= used when only 1 MUTATED BASE

Thus BER could not be used on UV damage as it results in TWO Ts sticking together. (2 MUTATIONS)

==> NER must be used
==> if this is also damaged
==> then SOS.

Question 16

Slide 36 – hard.

If SOS repair causes the mutations of UV damage, why did it evolve?

***SOS
= repairs severely damaged BASES in DNA
by base excision and replacement,

EVEN IF there is NO TEMPLATE to guide base selection.

This process is a LAST RESORT
for repair and is
OFTEN THE CAUSE OF MUTATIONS

Answer 16

evolved as an EMERGENCY RESPONSE
to repair severely damaged DNA.

==> when too much DNA is damaged by UV
==> too many thymidine dimers need to be replaced
==> SOS repair happens
»> SOS has NO PROOFREADING ability
»>resulting in many errors “MUTATIONS” in the DNA sequence.

This is okay for bacteria
- as they multiply rapidly

not so great in humans as
- it CAN LEAD TO CANCER

Question 17

What is wrong with the enzyme that repairs UV damage?

Answer 17

Nothing wrong with NER

but the risk is that it 
==> MAY ALSO BE DAMAGED 
==> which calls in SOS 
==> which does not have proofreading
==> can cause further mutations 
>>>leading to cancer

Question 18

Why are older people more likely to develop cancer?

Answer 18

==> Random cell mutations accumulate over time
==> immune system becomes less effective as we age
==> tf ability to detect and fight cancer decreases
»> Accumulated mutations may involve
important TUMOUR SUPRESSOR GENES

Question 19

Why don’t we have to test chemicals in animals when they are found to be non-mutagenic in bacteria?

Answer 19

==> test chem on bacteria
==> result no mutations seen 
>>> "non mutagenic chemical"
==> tf will also be a non mutagenic chemical for humans 
(just like in the bacteria)

bec..
None of the non-mutagenic chemicals
shown in bacteria
have been shown to cause cancers in humans

Question 20

20*. Alleles that are not always expressed when present are called?

Answer 20

Alleles that are not expressed (but present)
= “Recessive”

recessive alleles can be expressed if
= 2 COPIES PRESENT
in the organism’s GENOTYPE

Question 21

21*. Draw a punnet square for dwarfism inheritance from 2 heterozygous parents and explain the expected proportions of each genotype in the offspring. What are the expected proportions of the phenotypes?

Answer 21

Dwarfism inheritance:
Genotype: 
= 25% homozygous dominant
= 50% heterozygous dominant
= 25% homozygous recessive

Phenotype:
As one of the infants dies at birth (homozygous dominant), the phenotype of the resulting offspring is
= 2/3 shortened stature
= 1/3 average height

(Danilla edit: I have corrected the way you must express the phenotype as the stillbirth does not give a proportion of children measurable by height. Also avoid the word normal and use average).

     D	d D	DD	Dd d	Dd	dd

Question 22

22*. How many alleles code for hair colour?

Answer 22

= 10
“4 code for black, 1 code for red”

They will be a mix of

• -eumelanin (B and b) alleles = black
• -phomelanin (R and r) alleles = red

Question 23

23*. If a person had the genotype BBbbbbbbRR what hair colour do you think they would have and why ?

Answer 23

LIGHT BROWN/ RED HAIR

==>d. one pair of functional eumelanin (B) allele
==>and the 3 pairs of non-functional eumelanin (b)

means that the person will have a light brown hair colour; functional eumelanin codes for = dark hair (black)
non-functional eumelanin codes for = blonde hair.

They will also have a
==>tinge of red hair
d. final pair of alleles being Phomelanin (R)codes =red hair
leaving only a small amount of red to be seen in the hair colour.

Question 24

Explain how two brown-haired parents could have blonde offspring, given that 4 alleles control brown hair colour. Show the genotypes of the parents and the child.

Answer 24

Parent 1 BBBBbbbbrr

Parent 2 BBBBbbbbrr

Possible offspring gets bbbbr from both paranet s resulting in bbbbbbbbrr

Question 25

What two genotypes lead to an O blood group phenotype?

Answer 25

To be O blood group phenotype you must receive a
= recessive O allele (from each parent)
giving you the genotype = OO.

Additionally if a person is 
Bombay blood group, 
they will be = genotypically A, B or AB, 
==> but because of the mutation 
>>>they will appear O phenotype.

Question 26

Is one gender
more likely to inherit an autosomal recessive disease?

If so, which?

Answer 26

NO
the disease is autosomal and tf NOT ON THE SEX CHROMOSOMES
so one gender is not more likely than the other.

Question 27

Draw the punnet square if a
– person with cystic fibrosis and a
– cystic fibrosis carrier
had children. What are the expected genotype and phenotype proportions?

Answer 27

genotype cystic fibrosis = cc genotype (homozygous rec)
cystic fibrosis carrier = Cc genotype (heterozygous)

Genotype:
= 50% heterozygous carrier Cc,
= 50% homozygous recessive sufferer cc

Phenotype:
= 50% carrier
= 50% affected

    C	c c	Cc	cc c	Cc	cc

Question 28

If Dwarfism is an autosomal dominant trait,

why is not more common?

Answer 28

==>dominant gene if INHERITED FROM BOTH PARENTS

|&raquo_space;> death of the offspring.

Question 29

Why are X-linked dominant disorders so rare?

Give the genotype of a male that could survive such a rare disorder?

Answer 29

Because every child inherits at least 1 X and thus would be affected.

Possible genotype of MALE THAT COULD survivor
= XXY

Question 30

CF is an autosomal recessive disease. Women who have CF are fertile, while men who have CF are usually infertile. What might be a reason for this?

Answer 30

Male CF patients are infertile due to a blockage or absence of the sperm canal. This defect is called congenital bilateral absence of the vas deferens (CBAVD). The vas deferens is a long tube that carries sperm from the testicles through the male reproductive system. The absence of sperm in the semen makes it impossible to fertilize an egg.

Question 31

Now that you have learnt about P53, why does smoking cause skin to look older than it actually does?

Answer 31

P53 required to bring on apoptosis, if it is
==> BLOCKED BY BENZENE then
==> many OLD CELLS REMAIN
==>instead of being apoptosed.

Question 32

Use a punnet square to show how a person with
–blood type A and a person with
–blood type B can have a
child with blood type O.

What proportion of children will have a different blood type and what will be the blood types.

Answer 32

B O
A AB AO
O BO OO
==>2 HETERozygous parents (AO and BO) can
==> produce offspring with ANY of the 4 phenotypes
(AB, A, B, O).

The child had a
= 25% chance of being blood type O.

*25% chance the other offspring would be blood type AB, a 25% chance of being blood type A, and a 25% chance of being blood type B.

Question 33

Use a punnet square to show the proportion of children likely to suffer
—-Huntington’s disease
if one parent is affected.

Answer 33

H = hunington’s disease allele

= 50% chance the child Huntington’s disease
= 50% chance the child will not have the disease.

    h	h H	Hh	Hh h	hh	hh

Question 34

Draw a punnet square of the
—sex linked inheritance of haemophilia

of a carrier female and non-carrier male.

What proportion of children will have haemophilia?

Answer 34

= 25% (50% chance if the child is male).
will have haemophilia
           XH	Xh
XH	XHXH	XHXh
Y	XHY	XhY