nucleic acids and genetic structure part 1 (Q&A)

Question 1

describe how mRNA is formed by transcription in eukaryotes ( 5 MARKS)

Answer 1

1) DNA helicase breaks down hydrogen bonds between complementary base pairs splitting the double helix strands

2) only one of those strands acts as a template.

3) complementary RNA joins with the bases on the template DNA strands and forms phosphodiester bonds between nucleotides with the help of RNA polymerase.

4) RNA polymerase works its way down the whole length of the template DNA strand until it reaches the end of the gene ( the stop codon )

5) mRNA strand has been formed/synthesised and RNA polymerase detaches from the DNA so the DNA can go back to its normal DNA helix structure.

6) mRNA then leaves the nucleus to take part in translation however before becomes spliced in RNA splicing to remove the introns in the strand ( the non coding DNA) and normal mRNA is formed

Question 2

describe how a polypeptide is formed by translation of mRNA ( 6 marks )

Answer 2

1. (mRNA attaches) to ribosomes
2. (tRNA) anticodons (bind to) complementary (mRNA) codons;
3. tRNA brings a specific amino acid;
4. Amino acids join by peptide bonds; - this is catalysed by peptide-transferase.
5. (Amino acids join together) with the use of ATP;
6. tRNA released (after amino acid joined to polypeptide);
7. The ribosome moves along the mRNA to form the polypeptide; - once this is correctly folded, it can carry out any function in the cell.

Question 3

describe how a phosphodiester bond is formed between two nucleotides within a DNA molecule

Answer 3

1.Condensation (reaction)/loss of water;
2. (Between) phosphate and deoxyribose;
3. (Catalysed by) DNA polymerase

Question 4

Name the protein associated with DNA in a chromosome.

Answer 4

histone

Question 5

describe how the separation of DNA strands occurs

Answer 5

1.DNA helicase;

2. Breaks hydrogen bonds between base pairs/ AT and GC/complementary bases

Question 6

Describe the role of DNA polymerase in the semi-conservative replication of DNA.

Answer 6

1.Joins (adjacent DNA) nucleotides;
2. (Catalyses) condensation reactions
3. (Catalyses formation of) phosphodiester bonds (between adjacent nucleotides)

Question 7

Describe how an enzyme can be phosphorylated.

Answer 7

1.Attachment/association of (inorganic) phosphate (to the enzyme)
2. (Released from) hydrolysis of ATP

Question 8

Some tumour cells contain higher than normal concentrations of cyclin D.
Use the graph above to suggest why higher than normal concentrations of cyclin D could result in a tumour.

Answer 8

1. Shortens interphase
2. faster cell division (mitosis)
3. results in mass of abnormal cells

Question 9

Define ‘gene mutation’ and explain how a gene mutation can have:
* no effect on an individual
* a positive effect on an individual.

Answer 9

1. change in bases/nucleotides
2. Results in the formation of new allele - recessive so does not influence phenotype

3.genetic code is degenerate

4.does change amino acid but does not change tertiary structure

5. Results in change in polypeptide that positively changes the properties

Question 10

Give three ways in which the DNA in a chloroplast is different from DNA in the nucleus.

Answer 10

DNA in chloroplast is :

1.shorter/smaller
2.circular (linear in nucleic DNA)
3.not associated with a histone

Question 11

Give one other difference between the structure of a DNA nucleotide and the structure of an RNA nucleotide.

Answer 11

DNA contains deoxyribose where as RNA contains only ribose

Question 12

Not all mutations in the nucleotide sequence of a gene cause a change in the structure of a polypeptide.
Give two reasons why.

Answer 12

Triplets code for same amino acid
Occurs in introns /non-coding sequence

Question 13

Name the two scientists who proposed models of the chemical structure of DNA
and of DNA replication.

Answer 13

Watson and crick

Question 14

Use your knowledge of semi-conservative replication of DNA to suggest:
the role of the single-stranded DNA fragments
the role of the DNA nucleotides

Answer 14

Role of single-stranded DNA fragments:
1. Template;
2. Determines order of nucleotides/bases

Role of DNA nucleotides
3. Forms complementary pairs / A – T, G - C

Question 15

Describe the roles of iron ions, sodium ions, and phosphate ions in cells.

Answer 15

Iron ions
1. Haemoglobin binds/associates with oxygen
Sodium ions
2. Co-transport of glucose/amino acids (into cells);
3. (Because) sodium moved out by active transport/Na – K pump;
4. Creates a sodium concentration/diffusion gradient;
Phosphate ions
6. Affects osmosis/water potential
7. Joins nucleotides/in phosphodiester bond/in backbone of DNA/RNA/in
nucleotides

Question 16

The movement of substances across cell membranes is affected by membrane structure. Describe how.

Answer 16

1. Phospholipid (bilayer) allows movement/diffusion of non-polar/lipid-soluble substances
2. Phospholipid (bilayer) prevents movement/diffusion of polar/ charged/lipid-insoluble substances

3.Carrier proteins allow active transport

4. Channel/carrier proteins allow facilitated diffusion/co-transport;
5. Shape/charge of channel / carrier determines which substances move

Question 17

The scientists determined the percentage of heart cells undergoing DNA replication by using a chemical called BrdU. Cells use BrdU instead of nucleotides containing thymine during DNA replication.

(b) Describe how BrdU would be incorporated into new DNA during semi-conservative replication. (5m)

Answer 17

1. DNA helicase
2. Breaks hydrogen bonds (between 2 DNA strands)
3. BrdU complementary to adenine
4. DNA polymerase joins (adjacent) nucleotides (to incorporate BrdU into the new DNA strand)
   5.Phosphodiester bonds form between nucleotides

Question 18

(c) Cells with BrdU in their DNA are detected using an anti-BrdU antibody with an enzyme attached.

Use your knowledge of the ELISA test to suggest and explain how the scientists identified the cells that have BrdU in their D(3m)

Answer 18

1. Add antibody (anti-BrdU with enzyme attached)/DNA
2. Wash (cells/DNA) to remove excess/unattached antibody
3. Add substrate to cause colour change

Question 19

Give two features of DNA and explain how each one is important in the semi-conservative replication of DNA

Answer 19

1. Weak / easily broken hydrogen bonds between bases allow two strands to separate
2. Two strands, so both can act as templates
3. Complementary base pairing allows accurate replication

Question 20

a) The Crohn’s disease symptom of swelling of the lining of the intestines could be triggered by pathogenic bacteria in the intestines (lines 6–10).
Suggest how.

Answer 20

1. (Presence of) antigen of the (pathogenic) bacteria
2. (Causes) more T cells produced / faster T cell production;
3. Against (the pathogen and) normal bacteria;
4. (Long lasting as) cells do not die / live for longer;
5. (More) cytokines / chemicals causing swelling are produced

Question 21

b) Suggest the meaning of ‘a genetic tendency to Crohn’s disease’ (line 10).

Answer 21

1. (Some people) have a mutation / allele / gene
2. (That) increases the chances / risk / makes it more likely for / causes them to have an unusually large T cell response

Question 22

c) Suggest why 5-ASA is only effective in controlling the swelling of the lining of the intestines in some people with Crohn’s disease (lines 11–12)

Answer 22

1. (Some people might) produce (very) large amounts of cytokine / have large amounts of swelling;
2. (That) 5-ASA drugs cannot control / reduce

Question 23

(d) Suggest why 6-MP can be used to control the symptoms of Crohn’s disease (lines 13–15).

Do not include details of enzyme inhibition or protein synthesis in your answer

Answer 23

1. (Lack of adenine and guanine) will slow / stop DNA synthesis / replication (in T cells)
2. Affects T cells more as they cannot recycle nucleotides
3. (6-MP therefore) suppresses / slows the (unusually large) T cell / immune response
4. (So) less cytokine is produced (and therefore less swelling).

Question 24

Give two pieces of evidence from Figure 1 that this cell was undergoing mitosis. Explain your answers.

Answer 24

1. The (individual) chromosomes are visible because they have condensed
2. (Each) chromosome is made up of two chromatids because DNA has replicated
3. The chromosomes are not arranged in homologous pairs, which they would be if it was meiosis

Question 25

When preparing the cells for observation the scientist placed them in a solution that had a slightly higher (less negative) water potential than the cytoplasm. This did not cause the cells to burst but moved the chromosomes further apart in order to reduce the overlapping of the chromosomes when observed with an optical microscope.
Suggest how this procedure moved the chromosomes apart.

Answer 25

1.Water moves into the cells/cytoplasm by osmosis
2. Cell/cytoplasm gets bigger;

Question 26

The dark stain used on the chromosomes binds more to some areas of the chromosomes than others, giving the chromosomes a striped appearance.

Suggest one way the structure of the chromosome could differ along its length to result in the stain binding more in some areas

Answer 26

Differences in base sequences
OR
Differences in histones/interaction with histones

Question 27

In Figure 2 the chromosomes are arranged in homologous pairs. What is a homologous pair of chromosomes?

Answer 27

two chromosomes that carry the same genes.

Question 28

Describe the role of two named enzymes in the process of semi-conservative replication of DNA

Answer 28

DNA helicase - breaks down the hydrogen bonds between complementary bases whilst also unwinding the DNA helix shape.

DNA helicase joins DNA nucleotides together forming phosphodiester bonds between them.

Question 29

Give two ways in which the arrangement of prokaryotic DNA is different from the arrangement of the human DNA in

Answer 29

1. prokaryotic DNA is circular whilst human DNA is linear.
2. prokaryotic DNA is not associated with a protein (histone) whilst human DNA is.

Question 30

Adenosine triphosphate (ATP) is a nucleotide derivative.
Contrast the structures of ATP and a nucleotide found in DNA to give two differences

Answer 30

1. ATP has ribose and DNA nucleotide has deoxyribose;
2. ATP has 3 phosphate (groups) and DNA nucleotide
3. ATP – base always adenine and in DNA nucleotide

Question 31

Give the two types of molecule from which a ribosome is made.

Answer 31

one of RNA and one protein

Question 32

Describe the role of a ribosome in the production of a polypeptide. Do not include transcription in your answer

Answer 32

1. mRNA binds to ribosome;
2. Idea of two codons / binding sites;
3. (Allows) tRNA with anticodons to bind / associate;
4. (Catalyses) formation of peptide bond between amino acids (held by tRNA molecules);
5. Moves along (mRNA to the next codon) / translocation described

Question 33

In a eukaryotic cell, the base sequence of the mRNA might be different from the sequence of the pre-mRNA.

Explain why

Answer 33

introns in Pre- mRNA are removed in RNA splicing

Question 34

ATR works as follows.
* ATR phosphorylates other enzymes involved in repairing DNA.
* ATR also phosphorylates substrates required to repair DNA.
When ATR phosphorylates other enzymes, these enzymes become able to bind to
their substrates.

Use your knowledge of enzyme structure to suggest why.

Answer 34

1. Changes tertiary structure of the enzyme;
   complementary;
2. (Enzyme) active site formed / able to be formed / active site becomes

Question 35

The enzyme-catalysed reactions activated by ATR only occur if the substrates have been phosphorylated.

Use your knowledge of energy changes in enzyme-catalysed reactions to suggest why.

Answer 35

(Phosphorylation) makes substrates more reactive / raises their energy level(s) / lowers activation energy for the reaction

Question 36

Sometimes, a mutagenic agent causes DNA to break. A different enzyme called ATM binds to the broken DNA. This leads to the activation of a protein coded for by a tumour suppressor gene. The effect of ATM binding is to stop cell division until DNA is repaired.

A mutation could result in a person having non-functional forms of the gene that produces ATM.

What can you predict about the possible effects of having a non-functional form of ATM?

Answer 36

1. ATM will not bind to (broken) DNA
2. DNA not repaired / cell still has broken DNA;
3. Cell division continues / tumour forms;
4. Tumour suppressor (gene) not effective / not activated;
5. May have no effect in diploid / heterozygous (organism);
   6.
   (Which) still has a functional ATM / ATM gene

Question 37

The genetic code is degenerate and non-overlapping.
Explain the meaning of both

Answer 37

Degenerate -more than one (base) triplet for each amino acid

Non-overlapping - Non-overlapping: each base is part of only one triplet.

Question 38

DNA is a polymer of nucleotides. Each nucleotide contains an organic base.

Explain how the organic bases help to stabilise the structure of DNA.

Answer 38

1. hydrogen bonds between the complementary base pairs, holding the two strands together.
2. Many hydrogen bonds provides strength

Question 39

Suggest one advantage of showing the genetic code as base sequences on mRNA, rather than triplets on DNA.

Answer 39

(Because) ribosomes assemble polypeptides using mRNA code
OR
DNA has two strands each with a different (complementary) base sequence

Question 40

(d) Use Figure 1, Figure 2 and your knowledge of enzyme action to explain why the arrows point in opposite directions.

Answer 40

1. (Figure 1 shows) DNA has antiparallel strands / described;
2. (Figure 1 shows) shape of the nucleotides is different / nucleotides aligned differently;
3. Enzymes have active sites with specific shape;
4. Only substrates with complementary shape / only the 3’ end can bind with active site of enzyme / active site of DNA polymerase.

Question 41

(a)
HSV infects nerve cells in the face (line 1). Explain why it infects only nerve cells.

Answer 41

1. Outside of virus has antigens / proteins;
2. With complementary shape to receptor / protein in membrane of cells;
3. (Receptor / protein) found only on membrane of nerve cells.

Question 42

(b)
HSV can remain inactive inside the body for years (lines 2–3). Explain why this virus can be described as inactive.

Answer 42

1. No more (nerve) cells infected / no more cold sores form;
2. (Because) virus is not replicating.

Question 43

Suggest one advantage of programmed cell death (line 4).

Answer 43

prevents replication of virus.

Question 44

The scientists concluded that production of this microRNA allows HSV to remain in the body for years (lines 10–12).

Explain how this microRNA allows HSV to remain in the body for year

Answer 44

mRNA
1. (Binds) by specific base pairing;
2. (So) prevents mRNA being read by ribosomes;
3. (So) prevents translation / production of proteins;
4. (Proteins) that cause cell death

Question 45

During replication, the two DNA strands separate and each acts as a template for the production of a new strand. As new DNA strands are produced, nucleotides can only be added in the 5’ to 3’ direction.

Use the figure in part (a) and your knowledge of enzyme action and DNA replication to explain why new nucleotides can only be added in a 5’ to 3’ direction.

Answer 45

1. Reference to DNA polymerase;
2. (Which is) specific;
3. Only complementary with / binds to 5’ end (of strand);
4. Shapes of 5’ end and 3’ end are different / description of how different.

Question 46

The events that take place during interphase and mitosis lead to the production of two genetically identical cells.
Explain how.

Answer 46

1.DNA replicated;

2. (Involving) specific / accurate / complementary base-pairing;
3. (Ref to) two identical / sister chromatids;

Question 47

The student cut thin sections of tissue to view with an optical microscope. Explain why it was important that the sections were thin.

Answer 47

1. to allow more light through (transparent
2. few layers of cells can be viewed

Question 48

What does the graph show about the growth of roots? Use the data to explain your answer

Answer 48

• faster mitosis at 0.2mm
• no mitosis after 1.6mm therefore roots grow by mitosis.