6.1.1 Cellular control

Question 1

Describe the 3 types of mutation that can occur in genes.

Answer 1

Substitution:
A nucleotide is replaced with another one. This can change the amino acid coded for. No frameshift occurs.

Insertion:
A new nucleotide is inserted into a codon. This causes a frameshift as all codons change from the point of mutation. Different amino acids coded for.

Deletion:
Nucleotide in codon is removed. This causes a frameshift which changes codons from that point.

Question 2

What effects can genetic mutations have on the proteins coded for? (3)

Answer 2

No effect, silent mutation:
More than one triplet can code for the same amino acid.
A non-coding section of DNA could be altered.

Damaging:
Change to primary, secondary or tertiary structure.
Shape of protein makes function worse.
Deletion/ stop codon can make protein shorter

Beneficial:
Structure of protein can be changed to make function better.

Question 3

What are the causes of mutations? (4)

Answer 3

DNA replication:
DNA might have not been checked properly at checkpoints.
Adding on of base pairs might have not been complementary.

Mutagens:
Agents that cause mutations.

Depurination/ Depyrimidation leads to insertion of nucleotides.

Free radicals (oxidising agents):
Disrupt base pairing and the structure of nucleotides.

Question 4

Examples of mutagens. (4)

Answer 4

Ionising radiation:
Breaks DNA strands, repairing that can cause mutations.

Deaminating agents:
Changes base sequence.

Alkylating agents:
Causes incorrect pairing of bases by alkyl binding to the base.

Viruses:
Inserts its DNA into the host’s genome.

Question 5

What are the types of point mutations? (3)

Answer 5

Silent:
Does not change protein/ its activity as the amino acid could be kept the same or the mutation occurs in the intron.

Nonsense:
Codon codes for a ‘STOP’ codon. This shortens the protein, making it non-functional.

Missense:
When a change in the nucleotide causes the coding of a completely different amino acid or a similar amino acid.

Question 6

Describe two transcriptional factors in Eukaryotes. (4)

Answer 6

Chromatin remodelling:
DNA is either found in ‘euchromatin’- Loosely wound DNA which can be transcribed.
Or ‘heterochromatin’- tightly wound DNA that cannot be transcribed

Histone modification:
Adding phosphate or acetyl groups decreases the +/ve charge of histones which causes DNA to bind less tightly.
Adding methyl groups increase hydrophobic tendency of histone, which increases coiling of DNA.

Question 7

What are the types of genetic regulation? (4)

Answer 7

Transcriptional level:
turning on/off of genes

Post-transcriptional:
modification of mRNA before translation

Translational:
stopping/ starting translation

Post translation:
Modification of proteins after synthesis in translation- occurs in the Golgi apparatus.

Question 8

Describe a transcription factor in Prokaryotes.

Answer 8

Lac Operon:
Group of genes that control the production of enzymes that metabolise lactose.

• Regulatory gene (Lac I): codes for a repressor protein which binds to the control region in the absence of lactose. This prevents RNA polymerase from starting transcription
• Control sites: Consists of the operator and promoter region. The operator region is switched on/off by the presence of the repressor protein
• Structural genes: LacZ, LacY, LacA. Where transcription occurs. Code for proteins involved in the metabolism of lactose.

Question 9

Describe the action of the Lac operon in the absence of lactose.

Answer 9

Regulatory gene expresses the repressor protein.

This binds to the operator region, which prevents RNA polymerase from binding to the promoter region.

Therefore transcription does not occur and no protein is produced.

Question 10

Describe the action of the Lac operon in the presence of lactose. (5)

Answer 10

Lactose binds to the repressor protein and causes it to change shape.

Repressor protein can no longer bind to the operator region.

RNA can now bind to promoter region, therefore translation occurs and structural proteins are transcribed.

LacY codes for Beta galactosidase which breaks down lactose into glucose and galactose.
LacZ codes for Lactose permease which are channel proteins in the membrane that allow the entry of lactose

Question 11

What are the benefits of an operon.

Answer 11

Saves energy:

If a certain product is not needed, then all of the genes expressed in making that product can be turned off.

Question 12

Describe the a regulatory mechanism at post-transcriptional level. (4)

Answer 12

Primary mRNA is produced after transcription and consists of introns(non coding) and exons (coding).

Primary mRNA undergoes splicing which removes introns and leaves exons to form MATURE mRNA.

Alternate splicing also occurs when the order of exons are modified or deleted. This changes the sequence of exons therefore a different sequence of amino acids at translation.

cAMP activates proteins that regulate gene expression by increasing/ decreasing the rate of metabolism.

Question 13

What is a homeobox gene? (3)

Answer 13

Regulatory genes that consists of 180 base pairs.

This codes for a homeodomain which binds to DNA and switches genes ‘on/off’.
This initiates transcription and the development of an organism.

Question 14

Describe how two processes help shape an organism? (5)

Answer 14

Apoptosis and Mitosis:

Mitosis increases cell number via cell division and proliferation. This leads to the overall growth of an organism.

Apoptosis is a programmed cell death which remoevs unwanted material and mitosis adds cells.

Both processes are regulated by hox genes which codes for transcription factors. Cells releases signalling molecules that stimulate mitosis and apoptosis.

Question 15

How is a hox gene different from homeobox gene?

Answer 15

Hox gene is a type of homeobox gene (regulatory gene) that are found in animals, plants and fungi.

The are responsible for the body plans in those organisms.

Question 16

How can different rates of transcription affect the shape of an organism? (4)

Answer 16

The rate determines what enzymes are produced at the right time.

These enzymes are involved in different processes that help with the development of an organism:
Respiration
Further protein synthesis
All vital for the growth of an organism.

Question 17

What is a point mutation?

Answer 17

Occurs when a nucleotide in the DNA sequence is altered by deletion, substitution or insertion.

Question 18

Describe the role of cAMP as post-translational control in the Lac Operon

Answer 18

The rate at which RNA polymerase transcribes genes is very slow.

The binding of cAMP receptor protein (CRP) allows the rate of transcription to speed up.

CRP is only active when cAMP is bound to it.

Question 19

Describe other mechanism for post-translational control. (4)

Answer 19

Addition of non-protein groups to a protein-
E.g carbohydrates (glycoprotein), lipids or phosphates (using protein kinases).
This can be done in the Golgi apparatus

Modification of amino acids by adding or deleting or adding some.

Formation of new bonds: i.e Disulphide bridges.

Folding/ shortening of proteins.

Question 20

Why do homeobox genes have very little mutations across different species of the same kingdom? (3)

Answer 20

The gene is very important in the body plan of all organisms.

A mutation can have a negative effect and alter the body plan.

The mutation of this gene can also alter many other genes.