Danho
ZIMSEC A Level · 6030/2 · N2023

Biology Paper 2 November 2023

Questions
34
Total marks
60
Syllabus code
6030/2

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Questions
34
Pass mark
21
Sit this paper

Answer every question in the printed order, get marked at the end, then see the answers.

The questions

Question 101

[1 marks]lipids
Fats and oils are both triglycerides. They differ in that
  1. Aa fat is liquid at room temperature while an oil is solid at the same temperature
  2. Ba fat contains glycerol while an oil is built from fatty acid molecules alone
  3. Ca fat is made only by animals while an oil is made only by bacteria and fungi
  4. Da fat is solid at room temperature while an oil is liquid at the same temperature

Question 102

[1 marks]lipids
One property of a phospholipid is that
  1. Athe whole molecule mixes freely with water in every part of the cell
  2. Bthe whole molecule repels water, which is why it is stored as a droplet
  3. Cit dissolves in water only after it has been heated above forty degrees
  4. Done end of the molecule mixes with water while the other end does not

Question 103

[1 marks]lipids
Fig.1.1 shows a glycerol molecule and a fatty acid molecule. Name the bond formed when the two combine chemically.

Answer this when you sit the paper.

Question 104

[2 marks]lipids
A triglyceride is formed from glycerol and fatty acids when
  1. Athree fatty acids join to one glycerol molecule and three molecules of water are taken in
  2. Bthree fatty acids join to one another in a ring and one molecule of glycerol is released
  3. Cthree fatty acids join to one glycerol molecule and three molecules of water are released
  4. Done fatty acid joins to three glycerol molecules and three molecules of water are released

Question 105

[1 marks]lipids
Name the type of reaction in which a bond forms between glycerol and a fatty acid and a molecule of water is released.

Answer this when you sit the paper.

Question 201

[2 marks]microscopy and slide preparation
In preparing a permanent slide of a tissue, one stage is carried out to remove water and prevent bacterial decay, and a later stage is carried out to make the material transparent. These two stages are
  1. Astaining and mounting respectively
  2. Bclearing and dehydration respectively
  3. Cdehydration and clearing respectively
  4. Dfixation and sectioning respectively

Question 202

[1 marks]microscopy and slide preparation
Fixation is carried out when a temporary slide is prepared because
  1. Ait colours the different organelles so that they can be told apart under the objective
  2. Bit kills the tissue quickly and preserves the structures close to their living state
  3. Cit removes the water from the tissue so that the specimen will not decay in storage
  4. Dit dissolves the cell walls so that the individual cells can be spread out on the slide

Question 203

[3 marks]microscopy and slide preparation
The electron microscope differs from the light microscope in that
  1. Ait uses a beam of electrons and has a much higher resolution, but living material cannot be examined with it
  2. Bit uses a beam of electrons and has a much higher resolution, and living material is examined easily with it
  3. Cit uses a beam of light focused by glass lenses, so it resolves detail down to a fraction of a nanometre
  4. Dit uses a beam of electrons focused by glass lenses, so specimens can be viewed in their natural colours

Question 204

[1 marks]microscopy and slide preparation
Regulation of the fluidity of a cell surface membrane is important because
  1. Athe membrane must stay flexible enough for vesicles to form and transport proteins to work
  2. Bthe membrane must stay completely rigid so that the cell keeps a fixed shape at all times
  3. Cthe membrane must dissolve in the cytoplasm whenever the cell is about to divide in two
  4. Dthe membrane must conduct heat away from the cytoplasm as fast as respiration produces it

Question 301

[1 marks]DNA structure and replication
Name the type of bond that holds a base on one strand of a DNA molecule to the base facing it on the other strand.

Answer this when you sit the paper.

Question 302

[1 marks]DNA structure and replication
The purine bases of a DNA molecule are
  1. Aguanine and thymine
  2. Badenine and cytosine
  3. Cadenine and guanine
  4. Dcytosine and thymine

Question 303

[2 marks]DNA structure and replication
Base pairing within the double helix is important because
  1. Aa large purine pairs with a small pyrimidine, so the two strands stay an even distance apart
  2. Btwo large purines pair together, so the helix is wide enough for the sugars to lie inside it
  3. Cthe paired bases are joined by covalent bonds, so the strands cannot be parted for copying
  4. Dthe paired bases carry opposite electrical charges, which holds the strands in a tight coil

Question 304

[1 marks]DNA structure and replication
DNA replication is described as semi conservative because
  1. Ahalf of the bases in each new strand are copied and the other half are guessed
  2. Beach new molecule keeps one strand of the original and one strand newly built
  3. Ceach new molecule is built entirely from nucleotides that were not there before
  4. Done of the two new molecules is the original and the other is entirely new

Question 305

[2 marks]DNA structure and replication
During genetic counselling a couple can be given
  1. Aa promise about which of their children will be affected and which will not be affected
  2. Ba test that changes the sex of the embryo where the disorder is carried on the X chromosome
  3. Cthe chance that a child of theirs will inherit a disorder, worked out from the family history
  4. Da treatment that replaces the faulty allele in every cell of a child before it is born

Question 401

[2 marks]dihybrid inheritance and the chi-squared test
A cross produced offspring in four phenotype classes, counted as 33, 23, 28 and 16, giving 100 offspring in total. If the expected ratio of the four classes is 1:1:1:1, what number is expected in each class?

Answer this when you sit the paper.

Question 402

[2 marks]dihybrid inheritance and the chi-squared test
In a cross the observed numbers in four phenotype classes are 33, 23, 28 and 16, and 25 offspring are expected in each class. Calculate the chi-squared value, taking chi-squared as the sum of the difference squared divided by the expected number.

Answer this when you sit the paper.

Question 403

[2 marks]dihybrid inheritance and the chi-squared test
A chi-squared value of 6.32 is obtained from a cross with three degrees of freedom, for which the critical value is 7.815. It follows that
  1. Athe difference between the observed and expected numbers is significant and needs another explanation
  2. Bthe observed numbers match the expected ratio exactly, so no difference between them arose at all
  3. Cthe cross must be repeated, because a chi-squared test gives no answer below the critical value
  4. Dthe difference between the observed and expected numbers is not significant and is due to chance

Question 501

[2 marks]photosynthesis and limiting factors
Besides temperature and light intensity, the rate of photosynthesis in a green plant is affected by
  1. Athe oxygen concentration of the air and the amount of nitrogen dissolved in the soil water
  2. Bthe number of stomata that are closed and the thickness of the waxy cuticle on the leaf
  3. Cthe mass of starch already stored in the leaf and the age of the roots holding the plant
  4. Dthe carbon dioxide concentration of the air and the amount of chlorophyll in the leaf

Question 502

[2 marks]photosynthesis and limiting factors
In Fig. 5.1, curve A was obtained at high light intensity and is still climbing steeply at 30 degrees Celsius, while curve B was obtained at low light intensity and is almost flat across the whole range. At 30 degrees Celsius the limiting factors are
  1. Alight intensity for curve A and carbon dioxide concentration for curve B
  2. Blight intensity for curve A and temperature for curve B
  3. Ctemperature for curve A and light intensity for curve B
  4. Dcarbon dioxide concentration for curve A and temperature for curve B

Question 503

[2 marks]photosynthesis and limiting factors
At high light intensity the rate of photosynthesis falls away above about 35 degrees Celsius. This decline happens because
  1. Athe chlorophyll molecules are bleached, so light of every wavelength passes straight through the leaf
  2. Bthe carbon dioxide in the air becomes too concentrated for the leaf to take any more of it in
  3. Cthe light reaching the leaf is reflected once the temperature rises past the optimum for the plant
  4. Dthe enzymes of photosynthesis are denatured, so their active sites no longer fit the substrate

Question 601

[2 marks]haemoglobin and oxygen dissociation
Fig. 6.1 shows two oxygen dissociation curves at sea level, curve A lying to the left of curve B, so that at any given oxygen tension curve A is the more saturated of the two. The curve for sickle cell trait haemoglobin is
  1. AB, because it lies to the right and so is less saturated than normal haemoglobin at a given tension
  2. BB, because a curve to the right shows that the haemoglobin takes up oxygen at a lower tension
  3. CA, because it lies to the left and so holds on to oxygen more tightly than normal haemoglobin
  4. DA, because a curve to the left shows that the haemoglobin becomes saturated only at high tension

Question 602

[2 marks]haemoglobin and oxygen dissociation
An oxygen dissociation curve for haemoglobin is S shaped rather than a straight line because
  1. Athe first oxygen molecule to bind blocks the other three binding sites until the tension rises
  2. Bhaemoglobin can carry oxygen only when carbon dioxide is present in the plasma around it
  3. Cthe four polypeptide chains of a haemoglobin molecule take up oxygen one after the other slowly
  4. Dthe binding of the first oxygen molecule changes the shape of the haemoglobin and eases the next

Question 603

[2 marks]haemoglobin and oxygen dissociation
A person with sickle cell trait travels from sea level to a high altitude. The likely implication is that
  1. Athe low oxygen tension leaves the abnormal haemoglobin poorly saturated, so red cells may sickle
  2. Bthe low oxygen tension makes the abnormal haemoglobin fully saturated, so the red cells return to normal
  3. Cthe high air pressure at altitude forces extra oxygen into the red cells, which then rupture at once
  4. Dthe cold air at altitude thickens the plasma, so the sickle shaped cells pass through capillaries easily

Question 701

[2 marks]cytokinesis in plant cells
Fig.7.1 shows a dividing plant cell in which X labels the outer boundary of the cell and Y labels one of the two dark round bodies lying above and below the row of Golgi vesicles. X and Y are
  1. Athe cell surface membrane and the vacuole respectively
  2. Bthe cell wall and the nucleus respectively
  3. Cthe cellulose cell wall and the chloroplast respectively
  4. Dthe middle lamella and the nucleolus respectively

Question 702

[2 marks]cytokinesis in plant cells
During cytokinesis in a plant cell, Golgi vesicles gather along the equator because
  1. Athey fuse there and their contents form the cell plate that becomes the new wall between the cells
  2. Bthey fuse there and release enzymes that digest the old cell wall so the two cells can separate
  3. Cthey carry the chromosomes to the two poles of the cell before the new wall is laid down
  4. Dthey store the water that is needed to fill the two vacuoles of the newly formed daughter cells

Question 703

[2 marks]cytokinesis in plant cells
A cleavage furrow does not form during cytokinesis in a plant cell because
  1. Athe large permanent vacuole fills the cell completely and leaves no room for a furrow
  2. Bplant cells divide by meiosis alone, and a cleavage furrow forms only after mitosis
  3. Cthe rigid cellulose wall cannot be drawn inwards by a ring of contracting microfilaments
  4. Dthe plant cell has no cell surface membrane that a ring of microfilaments could pull on

Question 801

[1 marks]menopause and hormone replacement therapy
Hormone replacement therapy is given during the menopause in order to
  1. Asupply testosterone so that the woman's muscles and bones are strengthened during her later years
  2. Bsupply oestrogen, and often progesterone, to relieve the symptoms of the woman's own falling levels
  3. Csupply follicle stimulating hormone so that the ovaries begin to release eggs once again each month
  4. Dsuppress the remaining oestrogen so that menstruation is brought to an end more quickly than usual

Question 802

[1 marks]menopause and hormone replacement therapy
The menopause occurs in a woman because
  1. Aher uterus lining becomes too thick for an egg to implant, so the cycle comes to a stop
  2. Bher body begins to produce antibodies against her own eggs, which are then destroyed
  3. Cher ovaries run out of follicles able to respond to FSH, so the oestrogen level falls
  4. Dher pituitary gland stops producing FSH and LH, so the follicles she still has cannot ripen

Question 803

[3 marks]menopause and hormone replacement therapy
Advantages of the menopause to a woman include that
  1. Amenstruation ends, bone density rises steadily and the risk of heart disease falls year by year
  2. Boestrogen levels rise, the skin becomes more elastic and the ovaries release two eggs a month
  3. Cthe ovaries enlarge, hot flushes cease within a week and the uterus lining thickens permanently
  4. Dmenstruation ends, contraception is no longer needed and the risks of a late pregnancy are avoided

Question 901

[1 marks]alternation of generations and plant classification
Fig.9.1 shows a plant in which a stalked structure grows up out of a low, lobed body that is anchored to the ground by fine rhizoids rather than by true roots. The phylum to which this organism belongs is
  1. ABryophyta
  2. BConiferophyta
  3. CFilicinophyta
  4. DAngiospermophyta

Question 902

[2 marks]alternation of generations and plant classification
Fig.9.1 shows alternation of generations, with A the stalked structure that grows up from the low, rooted body B and produces spores. A and B are
  1. Athe gametophyte and the sporophyte respectively
  2. Bthe sporophyte and the gametophyte respectively
  3. Cthe sporophyte and the protonema respectively
  4. Dthe zygote and the gametophyte respectively

Question 903

[3 marks]alternation of generations and plant classification
Features that allow mosses and liverworts to survive on land include
  1. Aa waxy cuticle that limits water loss, rhizoids that anchor the plant, and spores with resistant walls
  2. Ba waxy cuticle that limits water loss, xylem vessels that carry water, and seeds with a food store
  3. Ctrue roots that reach deep water, lignified stems that stand upright, and flowers that attract insects
  4. Da thick woody bark that resists drying, a taproot that stores starch, and pollen carried on the wind

Question 1001

[2 marks]infectious disease and epidemiology
Ebola is transmitted from one person to another by
  1. Aeating food grown in soil that has been contaminated, and walking barefoot over infected ground
  2. Bcontact with the blood or other body fluids of an infected person, and handling infected bushmeat
  3. Cthe bite of a mosquito that has fed on an infected person, and drinking untreated river water
  4. Dbreathing air in a room where an infected person has been, and touching a surface days afterwards

Question 1002

[3 marks]infectious disease and epidemiology
The distribution of Ebola is highest in the Congo region because
  1. Athe region has more mosquitoes than anywhere else, and it is the mosquito that carries the virus about
  2. Bthe forest holds the fruit bat reservoir, bushmeat is widely hunted, and health services are stretched thin
  3. Cthe climate there is dry enough for the virus to survive on open ground for months between outbreaks
  4. Dthe population there has no immune system response to any virus, so every person exposed is infected

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