Biological Molecules exam qs Flashcards
(98 cards)
1
Q
The general structure of a fatty acid is RCOOH. Name the group represented by COOH
A
Carboxyl
accept carboxylic acid
2
Q
Describe how you would test for the presence of a lipid and how you would recognise a positive result (2 marks)
A
- Add ethanol/alcohol then add water and shake/mix
OR
Add ethanol/alcohol and shake/mix then pour into/add water; Reject heating the emulsion test
Accept ‘Add Sudan III and mix’
Ignore a second shake - White/milky (emulsion)
OR
(emulsion) test turns white/milky;
Ignore cloudy
Reject precipitate
Accept (for Sudan III) top (layer) red
3
Q
Describe how a triglyceride molecule is formed (3 marks)
A
- One glycerol and three fatty acids;
- Condensation (reactions) and removal of three molecules of water;
- Ester bond(s) (formed);
Accept all marks in suitably labelled diagram OR in a balanced equation
4
Q
Describe how an ester bond is formed in a phospholipid molecule (2 marks)
A
- Condensation (reaction)
OR
Loss of water; - Between glycerol and fatty acid;
Accept labelled diagram
5
Q
Scientists investigated the percentage of different types of lipid in plasma membranes from different types of cell. The scientists expressed their results as Percentage of lipid in plasma membrane by mass. Explain how they would find these values (2 marks)
A
- Divide mass of each lipid by total mass of all lipids (in that type of cell);
- Multiply answer by 100.
6
Q
Cholesterol increases the stability of plasma membranes. Cholesterol does this by making membranes less flexible. Suggest one advance of the different percentage of cholesterol in red blood cells compared with cells lining the ileum (1 mark)
(membrane of red blood cell of mammal had 23% cholesterol, whereas membrane of cell lining ileum of mammal had 17% cholesterol)
A
Red blood cells free in blood / not supported by other cells so cholesterol helps to maintain shape;
Allow converse for cell from ileum – cell supported by others in endothelium so cholesterol has less effect on maintaining shape
7
Q
E. Coli has no cholesterol in its cell surface membrane. Despite this, the cell maintains a constant shape. Explain why. (2 marks)
A
- Cell unable to change shape;
- (Because) cell has a cell wall;
- (Wall is) rigid / made of peptidoglycan / murein.
8
Q
Newborn babies can be fed with breast milk or with formula milk. Both types of milk contain carbohydrates, lipids and proteins.
• Human breast milk also contains a bile-activated lipase. This enzyme is thought to be inactive in milk but activated by bile in the small intestine of the newborn baby.
• Formula milk does not contain a bile-activated lipase.
Scientists investigated the benefits of breast milk compared with formula milk.
(a)
The scientists used kittens (newborn cats) as model organisms in their laboratory investigation.
Other than ethical reasons, suggest two reasons why they chose to use cats as model organisms. (2 marks)
A
Two suitable suggestions; E.g.
1. (Are mammals so) likely to have same physiology / reactions as humans;
2. Small enough to keep in laboratory / produce enough milk to extract;
3. (Can use a) large number.
Ignore references to ethical issues
9
Q
Before starting their experiments, the scientists confirmed that, like human breast milk, cat’s milk also contained bile-activated lipase.
To do this, they added bile to cat’s milk and monitored the pH of the mixture.
Explain why monitoring the pH of the mixture could show whether the cat’s milk contained lipase (2 marks)
A
- Hydrolysis of lipids produces fatty acids;
- Which lower pH of mixture.
10
Q
From how many molecules is a triglyceride formed
A
4
11
Q
The structure of a phospholipid molecule is different from that of a triglyceride. Describe how a phospholipid is different (2 marks)
A
- Phosphate / PO4;
“It” refers to phospholipid - Instead of one of the fatty acids / and two fatty acids;
Accept minor errors in formula. Do not accept phosphorus / phosphorus group.
12
Q
Omega-3 fatty acids are unsaturated. What is an unsaturated fatty acid?
A
Double bond(s);
(Bonds) between carbon;
C=C bond(s) = 2 marks
‘No’ C=C bond(s) disqualifies 1 mark only
Accept: does not contain maximum number of H for 1 mark Neutral: contains C=O bonds
13
Q
Starch, glycogen, deoxyribose, DNA helicase
Which of these substances contains only carbon, hydrogen and oxygen?
A
Starch, glycogen and deoxyribose
all 3 are needed for the mark
14
Q
Starch, glycogen, deoxyribose, DNA helicase
Which of these substances is made from amino acid monomers?
A
DNA helicase
15
Q
Starch, glycogen, deoxyribose, DNA helicase
Which of these substances is found in both animal cells and plant cells?
A
Deoxyribose and DNA helicase
16
Q
Hydrogen bonds are important in cellulose molecules. Explain why. (2 marks)
A
- Holds chains / cellulose molecules together / forms cross links between chains / cellulose molecules / forms microfibrils, providing strength / rigidity (to cellulose / cell wall);
2 . Hydrogen bonds strong in large numbers;
Principles here are first mark for where hydrogen bonds are formed and second for a consequence of this.
Accept microfibres
17
Q
A starch molecule has a spiral shape. Explain why this shape is important to its function in cells. (1 mark)
A
Compact / occupies small space / tightly packed;
Answer indicates depth required. Answers such as “good for storage”, “easily stored” or “small” are insufficient.
18
Q
A precipitate is produced in a positive result for reducing sugar in a Benedict’s test. A precipitate is solid matter suspended in solution. A student carried out the Benedict’s test. Suggest a method, other than using a colorimeter, that this student could use to measure the quantity of reducing sugar in a solution. (2 marks)
A
- Filter and dry (the precipitate);
Accept: correct reference to evaporation after filtration - Find mass/weight;
19
Q
In an investigation, a student wanted to identify the solutions in two beakers, A and B. She knew one beaker contained maltose solution and the other beaker contained glucose solution. Both solutions had the same concentration.
She did two separate biochemical tests on a sample from each beaker.
Test 1 – used Benedict’s solution to test for reducing sugar.
Test 2 – added the enzyme maltase, heated the mixture at 30 °C for 5 minutes, and then used Benedict’s solution to test for reducing sugar. Maltose is hydrolysed by maltase.
The student’s results are shown in the table below.
(Beaker A - red in Test 1 and red in Test 2)
(Beaker B - red in Test 1 and dark red in Test 2)
Explain the results for beaker A and B in the table (2 marks)
A
- A = glucose and B = maltose;
- Because more sugar/precipitate after hydrolysis/maltase action; Accept ‘higher concentration of sugar’ for ‘more sugar’ Accept ‘break down’ for hydrolysis
20
Q
Use of a colorimeter in this investigation would improve the repeatability of the student’s results.
Give one reason why. (1 mark)
A
Quantitative
OR
(Colour change is) subjective;
Accept: accurate/precise
Standardises (the) method;
21
Q
In Test 1, the student used a measuring cylinder to measure 15 cm3 of solution from a beaker. The measuring cylinder gives a volume with an uncertainty of ±1 cm3. She used a graduated syringe to measure 5.0 cm3 of Benedict’s solution. The graduated syringe gives a volume with an uncertainty of ± 0.5 cm3. She mixed these volumes of liquid to do the biochemical test.
Calculate the percentage error for the measurements used to obtain a 20 cm3 mixture of the solution from the beaker and Benedict’s solution. Show your working. (2 marks)
A
16.67 − 17 = 2 marks;
(cumulative percentage error of both measuring vessels)
If incorrect final answer, accept for 1 mark: 0.167 − 0.17 (not a percentage)
(1/15 + 0.5/5) x 100
OR
evidence of
1/15 + 0.5/5
(correct understanding, but not calculated)
Ignore: ± (plus or minus) in answer
22
Q
What is a monomer? (1 mark)
A
(a monomer is a smaller / repeating) unit / molecule from which larger molecules / polymers are made;
Reject atoms / elements / ’building blocks’ for units / molecules
Ignore examples
23
Q
Lactulose is a disaccharide formed from one molecule of galactose and one molecule of fructose.
Other than being disaccharides, give one similarity and one difference between the structures of lactulose and lactose (2 marks)
A
Similarity
1. Both contain galactose / a glycosidic bond;
Ignore references to hydrolysis and / or condensation
Difference
2. Lactulose contains fructose, whereas lactose contains glucose;
ignore alpha/beta prefix for glucose
Difference must be stated, not implied
24
Q
Glycogen and cellulose are both carbohydrates. Describe two differences between the structure of a cellulose molecule and a glycogen molecule. (2 marks)
A
- Cellulose is made up of β-glucose (monomers) and glycogen is made up of α-glucose (monomers);
- Cellulose molecule has straight chain and glycogen is branched;
- Cellulose molecule has straight chain and glycogen is coiled;
- glycogen has 1,4- and 1,6- glycosidic bonds and cellulose has only 1,4- glycosidic bonds;
Ignore ref. to H bonds / microfibrils
25
Starch is a carbohydrate often stored in plant cells. Describe and explain **two** features of starch that make it a good storage molecule (2 marks)
Any **two** from:
1. Insoluble (in water), so doesn’t affect water potential;
2. Branched / coiled / (α-)helix, so makes molecule compact;
**OR**
Branched / coiled / (α-)helix so can fit many (molecules) in
small area;
3. Polymer of (α-)glucose so provides glucose for respiration;
4. Branched / more ends for fast breakdown / enzyme action;
5. Large (molecule), so can’t cross the cell membrane
*Require feature and explanation for 1 mark
1. Accept Ψ or WP
1. Accept Insoluble so doesn’t affect osmosis
1. Do not allow ref to ‘doesn’t affect water leaving cells
4. Ignore ‘surface area’
4. Accept ‘branched so glucose readily released’*
26
In mammals, in the early stages of pregnancy, a developing embryo exchanges substances with its mother via cells in the lining of the uterus. At this stage, there is a high concentration of glycogen in cells lining the uterus.
Describe the structure of glycogen (2 marks)
1.
Polysaccharide of α-glucose; **OR**
polymer of α-glucose;
2. (Joined by) glycosidic bonds
**OR**
Branched structure;
27
In mammals, in the early stages of pregnancy, a developing embryo exchanges substances with its mother via cells in the lining of the uterus. At this stage, there is a high concentration of glycogen in cells lining the uterus.
During early pregnancy, the glycogen in the cells lining the uterus is an important energy source for the embryo.
Suggest how glycogen acts as a source of energy. Do **not** include transport across membranes in your answer. (2 marks)
1. Hydrolysed (to glucose);
*Ignore ‘Broken down’*
2. Glucose used in respiration;
*’Energy produced’ disqualifies mp2**
28
Name the monomers from which a maltose molecule is made (1 mark)
Glucose (and glucose)
29
Name the type of chemical bond that joins the **two** monomers to form maltose (1 mark)
(α1,4) Glycosidic
30
Explain **one** way in which starch molecules are adapted to their function in plant cells (2 marks)
1. Insoluble;
2. Don’t affect water potential;
**OR**
3. Helical;
*Accept form spirals*
4. Compact;
**OR**
5. Large molecule;
6. Cannot leave cell.
31
Explain how cellulose molecules are adapted to their function in plant cells (3 marks)
1. Long and straight chains;
2. Become linked together by many hydrogen bonds to form fibrils;
3. Provide strength (to cell wall).
32
Glycogen and cellulose are both carbohydrates.
Describe two differences between the structure of a cellulose molecule and a glycogen molecule. (2 marks)
1. Cellulose is made up of β-glucose (monomers) **and** glycogen is made up of α-glucose (monomers)
2. Cellulose molecule has straight chain **and** glycogen is branched
3. Cellulose molecule has straight chain **and** glycogen is coiled
4. Glycogen has 1,4- and 1,6- glycosidic bonds **and** cellulose has only 1,4- glycosidic bonds
*ignore ref. to H bonds/microfibrils
33
Starch is a carbohydrate often stored in plant cells.
Describe and explain two features of starch that make it a good storage molecule.
[2 marks]
Any **two** from:
1. Insoluble (in water), so doesn’t affect water potential
2. Branched/coiled/ (α-)helix, so makes molecule compact
**OR**
2. Branched/coiled/ (α-)helix, so can fit many (molecules) in small area
3. Polymer of (α-)glucose so provides glucose for respiration
4. Branched / more ends for fast breakdown / enzyme action
5. Large (molecule), so can’t cross cell membrane
34
polysaccharides such as glycogen amylopectin and amylose are formed by polymerisation of glucose
With reference to fig 2.1
i) describe how the structure of glycogen differs from the structure of amylose (2 marks)
ii) describe the advantages for organisms in storing polysaccharides, such as glycogen, rather than storing glucose (3 marks)
i) *assume answer is about glycogen*
Branched;
1-6, glycosidic, links/bonds;
Not, coiled / helical;
ii) compact so large quantity can be stored
insoluble so no osmotic effect;
glucose would lower water potential; A decrease, more negative (so) water would enter and cell volume would increase;
(So) plant cells would need thicker cell walls / animal cells might burst; glucose reactive molecule
35
Describe how the structure of glycogen is related to its function (4 marks)
1. Helix/coiled/branched so compact;
2. Polymer of glucose so easily hydrolysed;
3. Branched so more ends for faster hydrolysis;
4. Glucose (polymer) so provides respiratory substrate for energy (release);
5. Insoluble so not (easily) lost (from cell)
OR
Insoluble so does not affect water potential/osmosis
*1. Accept description of ‘compact’, eg many glucoses packed closely/densely/tightly*
36
Compare and contrast the structure of starch and the structure of cellulose (6 marks)
Both polysaccharides
OR
Both are glucose *polymers*
OR
Both are made of glucose *monomers*;
2. Both contain glycosidic bonds (between monomers);
3. Both contain carbon, hydrogen and oxygen/C, H and O;
4. Starch has α-glucose **and** cellulose has β- glucose;
5. Starch (molecule) is helical/coiled **and** cellulose (molecule) is straight;
6. Starch (molecule) is branched **and** cellulose is not/unbranched;
7.Cellulose has (micro/macro) fibrils **and** starch does not;
*must include 1,2 OR 3 to achieve 6 marks
All statements must be clearly comparative or linked by the candidate, not inferred from separate statements
Additional mark point
8. Starch has 1-6 glycosidic bonds and cellulose does not
OR
Starch contains two types of molechle and cellulose contains one type of molecule
OR
Starch is amylose and amylopectin and cellulose is one type of molecule*
37
Name the type of peptidase which will hydrolyse the bond labelled G in figure 5 (1 mark)
*in the diagram, G is pointing to a bond between two amino acids*
Endo(peptidase)
*correct spelling*
38
Figure 1 shows the structure of a fat substitute
This fat substitute cannot be digested in the gut by lipase. Suggest why. (2 marks)
1. (Fat substitute) is a different/wrong shape/not complementary;
OR
Bond between glycerol/fatty acid and propylene glycol different (to that between glycerol and fatty acid)/no ester bond;
2. Unable to fit/bind to (active site of) lipase/no ES complex formed
*if wrong bond name given (e.g. peptide, glycosidic), then penalise once*
39
This fat substitute is a lipid. Despite being a lipid, it cannot cross the cell-surface membranes of cells lining the gut.
Suggest why it cannot cross cell-surface membranes.
[1 mark]
It is hydrophilic/is polar/is too large/is too big;
*ignore ‘is not lipid soluble’
40
name one chemical element found in all amino acids but not in monosaccharides (1 mark)
Nitrogen
41
What type of chemical reaction occurs to form a dipeptide?
Condensation reaction
42
Describe the difference between the structure of a triglyceride molecule and the structure of a phospholipid molecule. (1 mark)
1. In phospholipid, one fatty acid replaced by a phosphate;
*Ignore references to saturated and unsaturated
1. Accept Pi/PO43- / P
1. Reject P/Phosphorus Accept annotated diagrams*
43
Animal fats contain triglycerides with a high proportion of saturated fatty acids. If people have too much fat in their diet, absorption of the products of fat digestion can increase the risk of obesity. To help people lose weight, fat substitutes can be used to replace triglycerides in food.
Describe how a saturated fatty acid is different from an unsaturated fatty acid. (1 mark)
Saturated single/no double bonds (between carbons)
OR
Unsaturated has (at least one) double bond (between carbons)
*accept hydrocarbon chain/R group for ‘between carbons’ for either
Accept Sat = max number of H atoms bound
‘It’ refers to saturated
44
Describe the biochemical texts you would use to confirm the presence of lipid, non-reducing sugar and amylase in a sample (5 marks)
LIPID
1. Add ethanol/alcohol **then** add water **and** shake/mix
2. Milky/white emulsion
NON-REDUCING SUGAR
1. Do Benedict’s test and stays blue/negative
2. Boil with acid then neutralises with alkali
3. Heat with Benedict’s test and becomes red/orange (precipitate)
AMYLASE
1. Add biuret (reagent) and becomes purple/violet/mauve/lilac
2. Add starch, (leave for time), test for reducing sugar/absence of starch
45
explain how the structure of fibrous proteins is related to their functions (5 marks)
1. Long chains of amino acids;
2. Folding of chain into a coil / folds / helix / pleated sheet;
3. Association of several polypeptide chains together;
4. Formation of fibres / sheets explained;
5. H bonds / Disulphide bonding (In context);
6. Fibres provide strength (and flexibility);
7. Sheets provide flexibility;
8. Example e.g. keratin in hair, collagen in bone; (MUST be in context)
9. Insoluble because external R-groups are non-polar
46
Explain what causes the secondary structure to differ in length from the primary structure (1 mark)
Peptide is coiled/folded
47
Explain what is meant by the tertiary structure (1 mark)
Way in which whole molecule is folded/globular shape/folding of secondary structure/further folding/structure held by ionic/disulfide bonds
48
Heating may affect the tertiary structure of a protein. Explain how.
Causes bonds which hold the tertiary structure/named bond;
To break;
Shape no longer maintained/protein denatured
49
Describe the induced-fit model of enzyme action (2 marks)
1. Active site not initially complementary
2. Active site changes shape/is flexible
50
A quantitative Benedict’s test produces a colour whose intensity depends on the concentration of reducing sugar in a solution. A colorimeter can be used to measure the intensity of this colour.
The scientist used quantitative Benedict’s tests to produce a calibration curve of colorimeter reading against concentration of maltose.
Describe how the scientist would have produced the calibration curve and used it to obtain the results in Figure 4.
Do not include details of how to perform a Benedict’s test in your answer. (3 marks)
1. Make/use maltose solutions of known/different concentrations (and carry out quantitative Benedict’s test on each);
2. (Use colorimeter to) measure colour/colorimeter value of each solution and plot calibration curve/graph described;
3. Find concentration of sample from calibration curve;
*2.Axes must be correct if axes mentioned, concentration on x-axis and colorimeter reading on y-axis’
51
7.1 explain how the active site of an enzyme causes a high rate of reaction (3 marks)
52
Name the chemical elements in a non-reducing sugar (1 mark)
Carbon, hydrogen, oxygen
53
A student investigated the effect of pH on the activity of the enzyme amylase.
The tubes were made from Visking tubing. Visking tubing is partially permeable. She added an equal volume of amylase solution and starch to each tube. She added a buffer solution at pH2 to tube A. She added an equal volume of buffer solution at pH8 to tube B. After 30 minutes, she measured the height of the solutions in both tubes. She then tested the solutions in tubes A and B for the presence of reducing sugars.
Describe how the student would show that reducing sugars were present in the solution
1. Add Benedict’s;
2. Heat;
3. Red/orange/yellow/green (shows reducing sugar present);
*Hydrolyse with acid negates mp1
2. Accept warm, but not an unqualified reference to water bath
3. Accept brown*
54
A student investigated the effect of pH on the activity of the enzyme amylase.
The tubes were made from Visking tubing. Visking tubing is partially permeable. She added an equal volume of amylase solution and starch to each tube. She added a buffer solution at pH2 to tube A. She added an equal volume of buffer solution at pH8 to tube B. After 30 minutes, she measured the height of the solutions in both tubes. She then tested the solutions in tubes A and B for the presence of reducing sugars.
After 30 minutes, the solution in Tube B was higher than that of Tube A.
i) explain why the solution in Tube B was higher
ii) the student concluded from her investigation that the optimum pH of amylase was pH8. Is this conclusion valid? Explain your answer
i)
1. Starch hydrolysed / broken down / glucose/maltose produced;
2. Lower water potential;
3. Water enters by osmosis
ii) only 2 pHs studied / more pHs need to be tested
*accept: different amylase may have a different optimum pH*
55
Suggest 1 advantage of using a pH meter rather than a pH indicator in this experiment (1)
Numerical readings / Not subjective / Gives quantitive data / Gives continuous data (1)
Greater accuracy (1)
56
Explain why the pH decreases when the lipase is added to the milk (1)
Fatty acids produced (1)
57
The pH of milk before and after using lipase is recorded.
Suggest why the pH remained constant after 2 minutes (1)
No more fatty acids produced (1)
All triglycerides used up (1)
Enzyme denatured due to pH (1)
58
Name the type of bond hydrolysed when a short chain of amino acids is removed (1)
Peptide (1)
59
Replication of mitochondrial DNA (mtDNA) is different from that of nuclear DNA. box
The replication of the second strand of mtDNA only starts after two-thirds of the first strand of mtDNA has been copied.
A piece of mtDNA is 16 500 base pairs long and is replicated at a rate of 50 nucleotides per second.
Tick ( ) the box that shows how long it would take to copy this mtDNA. [1 mark]
550 seconds
60
give two features of dna and explain how each one is important in the semi-conservative replication of DNA
1. Weak / easily broken hydrogen bonds between bases allow two strands to separate / unzip;
*may appear in the same feature*
2. Two strands, so both can act as templates;
*may appear in the same feature*
3. Complementary base pairing allows accurate replication
*Allow description of complementary base pairing and accurate replication*
61
The enzymes DNA helicase and DNA polymerase are involved in DNA replication. Describe the function of each of these enzymes (2 marks)
1. DNA helicase – (unwinding DNA and) breaking hydrogen bonds / bonds between chains / bases / strands;
2. DNA polymerase – joins (adjacent) nucleotides OR forms phosphodiester bond / sugar-phosphate backbone;
1. Accept H bonds.
1. Accept hydrolyses for breaks
2. Reject forms hydrogen bonds (between nucleotides / bases)
62
Adenosine triphosphate (ATP) is a nucleotide derivative.
Contrast the structures of ATP and a nucleotide found in DNA to give two differences. (2 marks)
ATP has ribose and DNA nucleotide has deoxyribose;
2. ATP has 3 phosphate (groups) and DNA
nucleotide has 1 phosphate (group);
3. ATP – base always adenine and in DNA nucleotide base can be different / varies;
*Both parts of each MP needed
3. Reject Uracil / U
3. Accept C, T or G for different bases
Accept annotated diagram for any of the three marks*
63
give the two types of molecules from which a ribosome is made (2 marks)
1. RNA/rRNA
2. proteins
*reject tRNA and mRNA
Ignore amino acids*
64
complete table 2 to give four structural differences between a DNA molecule and an mRNA molecule
1. DNA has deoxyribose, mRNA has ribose;
2. DNA has thymine, mRNA has uracil
3. DNA long, mRNA short;
4. DNA is double stranded, mRNA is single stranded;
5. DNA has hydrogen bonds, mRNA has no hydrogen bonds
OR
DNA has (complementary) base pairing, mRNA does not;
*must be comparisons
ignore splicing/introns
4. Accept ‘double helix’ for ‘double stranded’ and ‘single helix’ for ‘single stranded’*
65
Name the type of bond between
Complementary base pairs:
Adjacent nucleotides in a DNA strand:
(2 marks)
Hydrogen bonds
Phosphodiester bonds
66
Describe how an ATP molecule is formed from its component molecules (4 marks)
1. and 2. Accept for 2 marks correct names of three components adenine, ribose/pentose, three phosphates;;
3. Condensation (reaction);
4. ATP synthase;
* 1. and 2. Accept for 1 mark, correct name of two components
1. and 2. Accept for 1 mark, ADP and phosphate/Pi
1. and 2. Ignore adenosine
1. and 2. Accept suitably labelled diagram
3. Ignore phosphodiester
4. Reject ATPase*
67
The length of a gene is described as the number of nucleotide base pairs it contains.
Use information in above diagram to calculate the length of a gene containing 4.38 × 10^3 base pairs. (2 marks)
(1.7nm = 5 base pairs)
Correct answer for 2 marks = 1489/1489.2;; Incorrect answer but for 1 mark accept:
876
OR
1861 - 1862
68
Meselson and Stahl investigated whether DNA replicated in a conservative or semiconservative way
a) What is meant by the term semiconservative replication? (2 marks)
Each strand of DNA used as a template to make a new DNA strand;
New DNA (mols) made of an old/original strand linked to a new strand; *(not: if ref. to new DNA strand)*
69
The bacterium Escherichia coli (E.coli) was cultured in a nutrient broth, containing the heavy isotope as a source of nitrogen 15N instead of the normal 14N. After several generations all of the DNA in all of the bacteria contained the heavy isotope 15N. They were then washed and transferred to a 14N medium and allowed to replicate. After each generation, bacteria were removed and ruptured to release the DNA. The DNA was then placed in a medium and spun in a centrifuge. The position of the DNA in the medium was then determined.
i) Name the part of the DNA molecule which contained the 15N. (1 mark)
ii) if they wanted to show the relative position of DNA from different samples of bacteria, suggest **two** variables which would need to be controlled in the centrifugation process (2 marks)
i) nitrogenous bases/organic bases/purines and pyrimidine
bases/ all four named 1 (not: bases/letters only/nucleotide)
ii) spin (at same) speed; (Same) time;
(Same) density/concentration of gel; (Same) temperature
(not: pH/ref. volume or mass)
70
Explain **five** properties that make water important for organisms (5 marks)
1. A **metabolite** in condensation/hydrolysis/ photosynthesis/respiration;
2. A solvent **so** (metabolic) reactions can occur
OR
A solvent **so** allowing transport of substances;
3. High (specific) heat capacity **so** buffers changes in temperature;
4. Large latent heat of vaporisation **so** provides a cooling effect (through evaporation);
5. Cohesion (between water molecules) **so** supports columns of water (in plants);
6. Cohesion (between water molecules) **so** produces surface tension supporting (small) organisms;
*3. For ‘buffer’ accept ‘resist’.
4. Reject latent heat of evaporation
5. For ‘columns of water’ accept ‘transpiration stream’. Do not credit ‘transpiration’ alone but accept description of ‘stream’.
5. For ‘columns of water’ accept ‘cohesion-tension (theory)’.
5. and 6. For cohesion accept hydrogen bonding
Ignore reference to pH. Allow other suitable properties but must have a valid explanation.
For example
• ice floating so maintaining aquatic habitat beneath
• water transparent so allowing light penetration for photosynthesis*
71
Describe the process of semi-conservative replication of DNA (5 marks)
1. DNA helicase unwinds DNA/double helix
**OR**
DNA helicase breaks hydrogen bonds;
2. Both strands act as templates;
3. (Free DNA) nucleotides line up in complementary pairs/A-T and G-C;
4. DNA polymerase joins nucleotides (of new strand);
5. Forming phosphodiester bonds;
6. Each new DNA molecule consists of one old/original/template strand and one new strand
*2. Accept description of ‘template, e.g. exposes bases on single (polynucleotide) strands
4. Reject forms hydrogen bonds/joins bases*
72
Describe how an enzyme is phosphorylated (2 marks)
-phosphate group binds to the hydroxyl group of an amino acid
-ATP gives up the phosphate (ATP —> ADP + Pi)
+Enzyme-substrate complex is formed more easily
73
State and explain the property of water that helps to prevent temperature increase in a cell (2 marks)
1. High (specific) heat capacity
2. Buffers changes in temperature
*accept ideas such as a lot of energy needed/gained to change temperature*
74
Describe how mRNA is produced from an exposed template strand of DNA
Do NOT include DNA helicase or splicing in your answer. (3 marks)
1. (Free RNA) nucleotides form complementary base pairs;
2. Phosphodiester bonds form;
3. By (action of) RNA polymerase;
75
Two enzymes, P and Q, are proteins with quaternary structure which catalyse the same reaction, but they have different amino acid sequences.
Define the quaternary structure of a protein. (1 mark)
More than 1 *polypeptide*
76
Explain how two enzymes with different amino acid sequences can catalyse the same reaction. (2 marks)
1. (Both) active sites have similar/identical tertiary structures
OR
(Both) active sites have isentical amino acid sequences
2. (So) form enzyme-substrate complexes (with the same substrate)
*1. Ignore shape for tertiary structure
1. Accept (both) have active sites that are complementary to different parts of the substrate;
1. Accept attach/bind for complementary
2 Accept E-S for enzyme-substrate*
77
Scientists investigated ribosomal RNA in liver cells.
Figure 10 shows the method they used to isolate the ribosomes from the liver cells. The detergent dissolves lipids.
The scientists broke open the cells to produce a suspension of cell contents. Describe how the scientists would remove large organelles from this suspension of
cell contents (2 marks)
1. Use centrifuge/centrifugation at slow/low/increasing (sequence of) speed(s);
2. Large/dense organelles (removed) in (first/early) pellet
OR
Less dense organelles (removed) in supernatant
OR
Small organelles (removed) in supernatant
78
Complete Table 3 to give three differences between DNA molecules and tRNA molecules. (3 marks)
DNA v tRNA
1. Deoxyribose v ribose;
2. Double-stranded v single-stranded;
3. Many nucleotides v few ;
4. Thymine v uracil;
5. Linear v clover leaf (structure)
OR
Double helix v clover (leaf structure);
6. Does not bind to amino acid v does bind to
amino acid;
7. No exposed bases v anticodon;
79
Contrast the structures of DNA and mRNA molecules to give three differences. (3 marks)
1. DNA double stranded/double helix and mRNA single-stranded;
2. DNA (very) long and RNA short;
3. Thymine/T in DNA and uracil/U in RNA;
4. Deoxyribose in DNA and ribose in RNA;
5. DNA has base pairing and mRNA doesn’t/ DNA has hydrogen bonding and mRNA doesn’t;
6. DNA has introns/non-coding sequences and mRNA doesn’t;
80
Cells constantly hydrolyse ATP to provide energy. Describe how ATP is resynthesised in cells (2 marks)
1. From ADP and phosphate
2. By ATP synthase
3. During respiration/photosynthesis
81
Give two ways in which the hydrolysis of ATP is used in cells. (2 marks)
1. To provide energy for other reactions/named process;
2. To add phosphate to other substances and make them more reactive/change their shape
82
Scientists investigated the hydrolysis of sucrose in growing plant cells by an enzyme called SPS.
Name the products of the hydrolysis of sucrose (2 marks)
1. Glucose
2. Fructose
83
Read the following passage.
Alzheimer’s disease leads to dementia. This involves small β-amyloid proteins binding together to form structures called plaques in the brain.
Nerve cells in the brain produce a large protein called amyloid-precursor protein that has a complex shape. This protein is the substrate of two
different enzymes, α-secretase and β-secretase. These enzymes are 5 normally produced in the brain. One product of the reaction catalysed by β-secretase is a smaller protein that can lead to β-amyloid protein formation. Many people with Alzheimer’s disease have mutations that decrease α-secretase production, or increase β-secretase production.
One possible type of drug for treating Alzheimer’s disease is a competitive 10 inhibitor of β-secretase. When some of these types of drugs were trialled on patients, the trials had to be stopped because some patients developed serious side effects.
Use information from the passage and your own knowledge to answer the following questions.
a) Suggest how amyloid-precursor protein can be the substrate of two different enzymes, α-secretase and β-secretase (lines 3–5).
[2 marks]
b) One product of the reaction catalysed by β-secretase is a smaller protein (lines 6–7). Describe what happens in the hydrolysis reaction that produces the smaller protein from amyloid-precursor protein. (2 marks)
c) Many people with Alzheimer’s disease have mutations that decrease α-secretase production, or increase β-secretase production (lines 8–9).
Use the information provided to explain how these mutations can lead to
Alzheimer’s disease.
d) One possible type of drug for treating Alzheimer’s disease is a competitive inhibitor of β-secretase (lines 10–11).
Explain how this type of drug could prevent Alzheimer’s disease becoming worse (2 marks)
e) When some of these types of drugs were trialled on patients, the trials were stopped because some patients developed serious side effects (lines 11–13).
Using the information provided, suggest why some patients developed serious side effects (1 mark)
a) 1. Different parts/areas/amino acid sequences (of amyloid-precursor) protein
2. Each enzyme is specific/fits=binds/complementary to a different part of the APP
b) 1. Peptide bond broken
2. Using water
c) 1. Mutations prevent production of enzyme(s)/functional enzyme
2. Increase in β-secretase leads to faster/more β-amyloid production
OR
Decrease in alpha-secretase leads to more substrate for β-secretase
3. (Leads to) more/greater plaque formation
d) 1. (Inhibitor) binds to / blocks active site of β-secretase/enzyme
2. Stops/reduces production of β-amyloid/plaque
e)
1. Some β-amyloid required/needed (to prevent side effects)
OR
(Some) β-secretase needed;
2. Leads to build-up of amyloid-precursor protein (that causes harm)
OR
Too much product of α-secretase (causes harm)
84
Some DNA nucleotides have the organic base thymine, but RNA nucleotides do not have thymine. RNA nucleotides have uracil instead of thymine.
Give *one* other difference between the structure of a DNA molecule and the structure of an RNA nucleotide
Deoxyribose in DNA and ribose in RNA;
85
Not all mutations in the nucleotide sequence of a gene cause a change in the structure of a polypeptide. Give two reasons why. (2 marks)
1. Triplets code for same amino acid
2. Occurs in introns /non-coding sequence;
*1. Accept: DNA/code/triplets are degenerate
1+ 2. Reject codons (as question states within genes)
2. Ignore junk DNA
2. Reject : multiple repeats*
86
C is a protein with a carbohydrate attached to it. This carbohydrate is formed by joining
monosaccharides together.
Name the type of reaction that joins monosaccharides together. (1 mark)
Condensation
87
Explain DNA replication in terms of unwinding the double helix and separation of the strands by helicase, followed by formation of the new complementary strands by DNA polymerase.
A. helix is unwound;
B. two strands are separated;
C. helicase (is the enzyme that unwinds the helix separating the two strands);
D. by breaking hydrogen bonds between bases;
E. new strands formed on each of the two single strands;
F. nucleotides added to form new strands;
G. complementary base pairing;
H. A to T and G to C;
I. DNA polymerase forms the new complementary strands;
J. replication is semi-conservative;
K. each of the DNA molecules formed has one old and one new strand;
88
Describe the role of DNA polymerase in DNA replication (1 mark)
Joins nucleotides (to form new strand).
Accept: joins sugar and phosphate / forms sugar-phosphate backbone
Reject: (DNA polymerase) forms base pairs / hydrogen bonds
89
Other than being smaller, give two ways in which prokaryotic DNA is different from eukaryotic DNA. (2 marks)
1. Circular / non-linear (DNA);
Accept converse for eukaryotic DNA
Ignore: references to nucleus, binary fission, strands and plasmids
2. Not (associated) with proteins / histones;
Accept does not form chromosomes / chromatin
3. No introns / no non-coding DNA.
Accept only exons
Q Neutral: no ‘junk’ DNA
90
The table shows the percentage of each base in the DNA from three different organisms.
Humans and grasshoppers have very similar percentages of each base in their DNA but they are very different organisms.
Use your knowledge of DNA structure and function to explain how this is possible (2 marks)
1. Have different genes;
*Reject: different alleles*
2. (Sobases / triplets) are in a different sequence / order;
*Accept: base sequence that matters, not percentage*
3. (So) different amino acid (sequence / coded for) / different protein / different polypeptide / different enzyme.
*Unqualified ‘different amino acids’ does not gain a mark
Reject: references to different amino acids formed
Ignore: references to mutations / exons / non-coding / introns*
91
The DNA of the virus is different from that of other organisms. Use the table above and your knowledge of DNA to suggest what this difference is. Explain your answer. (2 marks)
1. A does not equal T / G does not equal C;
Accept: similar for equal
Accept: virus has more C than G / has more A than T
2. (So) no base pairing;
3. (So) DNA is not double stranded / is single stranded.
92
Scientists determined that a sample of DNA contained 18% adenine.
What were the percentages of thymine and guanine in this sample of DNA? (2 marks)
Thymine: 18%
Guanine: 32%
93
The arrows in Figure 2 show the directions in which each new DNA strand is being produced.
Use Figure 1, Figure 2 and your knowledge of enzyme action to explain why the arrows point in opposite directions. (4 marks)
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
94
The table below shows the percentage of bases in each of the strands of a DNA molecule. Complete the table.
Strand 1: 16% A
Strand 2: 21% C and 34% G
Strand 1: 34% C, 21% G and 29% T
Strand 2: 29% A, 16% T
95
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. (4 marks)
Reference to DNA polymerase;
(Which is) specific;
Only complementary with / binds to 5’ end (of strand);
Reject hydrogen bonds / base pairing
Shapes of 5’ end and 3’ end are different / description of how different
96
A polypeptide has 51 amino acids in its primary structure.
(i) What is the minimum number of DNA bases required to code for the amino acids in this polypeptide? (1 mark)
(ii) The gene for this polypeptide contains more than this number of bases. Explain why. (1 mark)
(i) 153;
(ii) Some regions of the gene are non-coding / introns / start / stop code / triplet / there are two DNA strands;
*Allow addition mutation
Ignore unqualified reference to mutation
Accept reference to introns and exons if given together Ignore ‘junk’ DNA / multiple repeats*
97
Describe and explain the link between oxygen concentration, rate of respiration and rate of uptake of potassium ions. (4 marks)
*refer to table in question*
1. greater rate of oxygen consumption / leads to greater rate of respiration and greater rate of uptake;
*(allow this mark even if spread through account but cause and effect must be within the correct context)*
2. oxygen required for respiration;
3. respiration produces ATP / releases energy;
*(ignore ref to producing or making energy)*
4. potassium ions taken up by active transport / against concentration gradient;
98
Describe the structure of DNA (5 marks)
1. Polymer of nucleotides;
*Accept ‘Polynucleotide’
Accept for ‘phosphate’. phosphoric acid*
2. Each nucleotide formed from deoxyribose, a phosphate (group) and an organic/nitrogenous base;
3. Phosphodiester bonds (between nucleotides);
4. Double helix/2 strands held by hydrogen bonds;
5. (Hydrogen bonds/pairing) between adenine, thymine and cytosine, guanine;
