Cape Biology 2018 u1 p2 Ms

02107020/CAPE/KMS

C A R I B B E A N

E X A M I N A T I O N S

C O U N C I L

CARIBBEAN ADVANCED PROFICIENCY EXAMINATION®

BIOLOGY

UNIT 1 - Paper 02 KEYS AND MARK SCHEME MAY/JUNE 2018

2018

The world’s largest digital library

Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.







- 2 02107020/CAPE/KMS 2018 BIOLOGY

UNIT 1 – PAPER KEY AND MARK

02

SCHEME

Question 1 Specific Objectives: Objectives: 1.1, 1.2, 1.4 (a)

Correct structure clearly labelled to show: 1. Phosphate 2. Glycerol 3. Fatty acid tails

Each label - 1 mark 4. Hydrophilic and hydrophobic labelled correctly - 1 mark

[4 marks] Accept labelled diagram with polar head and hydrophobic hydrophobic tails -2 marks [Kink in tails not necessary, R-group not necessary and can be replaced with an O-, CHs can replace the zig-zagged lines in the fatty acid tails]. Correct chemical structure without labels – 2 marks Correct symbolic diagram, without/incorrect labels - 1 mark









02

SCHEME

Examples of use of marking scheme for 1a: A.

  

correct chemical structure with three parts labelled= 3 marks identification identification of hydrophobic and hydrophilic parts= 1 mark NOTE: if there were no labels — the structure = 2 marks the

B.

 

correct symbolic diagram without labels-1 identification identification of hydrophobic and hydrophilic parts =1









02

SCHEME

C.



Correct symbolic diagram with all three parts labelled =2 marks



Plus identification of hydrophilic and hydrophobic areas =1

D.

1 mark for identification of areas (accept if hydrophilic/hydrophobic hydrophilic/hydrophobic areas are correct but only 1 tail is depicted) 

(b) Orientation 1.

Bilayer -

composed of two

layers

of

phospholipids









02

SCHEME

Question 1 cont’d (c)

Cholesterol - 1 mark

[1 mark] (d)

Intrinsic proteins 1. Intrinsic proteins span the entire length of the lipid bilayer 2. Facilitate movement of substances across the membrane via channels (or

carriers) 3. Mechanism of facilitated movement via carriers (flip-flop mechanism) and/or channels (water-filled) 4. Types of molecules transported: Substances that cannot pass through the hydrophobic lipid bilayer / such as ions, hydrophilic and lipid insoluble substances (give at least one example) 5. Passive or facilitated diffusion requires no energy (ATP) inputs 6. Active transport requires the use of ATP to move substances across the membrane.

Any 4 points - 1 mark each (Accept well-annotated diagram for points 1-3) (e)

(i)

[4 marks] marks]

Triglyceride Triglyceride / triacylglycerid triacylglyceride e / triacylglycerol - 1

[1 mark]

(ii) Function 1. Energy:

Because

triglycerides

contain

three

fatty

acids

they

are

highly concentrated sources of energy. / Because the fatty acid tails are hydrophobic, they are stored anhydrously. As a result, more triglycerides can be packed in adipocytes. 2. Long-term storage: anhydrous state – no interference with metabolism. 3. Insulation / Protection / Buoyancy: due to lipid properties. 4. Nutrition: Nutrition: Triglycerides needed in body in order to absorb lipid l ipid soluble Vitamins A, D, E and K. 5. Metabolic water: produced when fat is broken down

Feature must be linked to function to get the mark Any 2 points points - 1 mark each [2 marks] Total 15 marks









02

SCHEME

Question 2 (a) (i) Position of alleles

1. 2. 3.

A: RR – homozygous round-seeded variety B: rr – homozygous homozygous wrinkle-seeded variety C: Rr (or rR) – heterozygous variety

Each correct pair including location – 1 mark All 3 allelic allelic pairs correct but but incorrect loci loci – 2 marks 2 of 3 correct pairs (locus is incorrect) – 1 mark (Do not award marks if alleles are outside chromosomes) [3 marks]

(ii)

Punnett square Parents

Rr

x

rr

r

r

R

Rr

Rr

r

rr

rr

Rr – round-seeded, rr – wrinkle-seeded;











02

SCHEME

Question 2 cont’d (iii) Gene and allele 1. Gene – a sequence/portion/region of DNA nucleotides that codes for a polypeptide/RNA polypeptide/RNA molecule (that assigns a p articular aspect of character)/ character)/ a short piece of DNA, which is responsible for the inheritance of a particular characteristic. characteristic. 2. Allele – one of two or more alternative forms (variants) of a gene (wild-type is the most common form)

Each point – 1 mark 1 mark for partial answer

(b) (i)

Scaled drawing

[2 marks]









02

SCHEME

Question 2 cont’d (ii)

Cell cycle stages

1. Cell D – (Mitosis) Prophase 2. Cell E – (Mitosis) Anaphase 3. Cell F – Interphase

1 mark each (Do not accept any meiosis stages e.g. Anaphase 1 or II)

[3 marks] Total 15 marks









02

SCHEME

Question 3 (a) (i) Bar graph Number of maternal amalgam fillings and concentration of mercury in maternal and foetal blood.

1. Correct labels on X and Y-axis 2. All bars plotted correctly (Maternal and foetal bars drawn next to each other for comparison) 3. Key for identification of maternal and foetal mercury concentration bars (or bars labelled correctly)

1 mark each









02

SCHEME

(iii) Movement of mercury from mother to foetus 1. 2. 3.

4.

Active process The foetal blood always has a higher concentration of mercury than that of the mother This implies that mercury has has to be pumped across across the placenta from the mother against a concentration gradient to the foetal blood supply. If it were passive, the concentration of mercury would have been equal in both mother and foetus (or concentration in foetus will be lower than in the mother).

Any 3 points points – 1 mark each [3 marks] (b)

Labels A – intine (Accept: cytoplasm - since the line seems to be touching the cytoplasm just beneath the intine) B – exine C – tube nucleus D – generative nucleus / generative cell

3-4 points correct – 2 marks (1-2 correct – 1 mark)

[2 marks]

(c) (i) Double fertilization 1. Pollen grain germinates to form pollen tube / On descent, the generative nucleus divides by mitosis producing two male haploid gametes. 2. Once in the embryo sac of the ovule, one male haploid gamete fuses with ovum nucleus (to form the diploid zygote). 3. The other haploid male nucleus fuses with the diploid/secondary/endosperm nucleus in the embryo sac (to form the triploid endosperm).









02

SCHEME

Question 4 (a)

Competitive and non-competitive inhibition of enzyme activity 1. 2. 3. 4. 5. 6.

7.

8.

Enzymes are globular proteins / 3-D shape / tertiary structure Enzymes have active site on the 3-D structure to which a specific substrate attaches for reaction to occur Competitive inhibitors have a similar shape to the substrate molecule Competitive inhibitors fit temporarily onto the active site and compete with substrate for entry to the active site Effect of competitive inhibitors decrease as the substrate concentration is increased / effects are reversible Non-competitive inhibitors bind to another another part of the enzyme molecule (allosteric site) / or binds permanently to active site Distortion Distortion of the (3-D) structure of enzyme can occur if the inhibitor binds to other part (non- acti ve site) of enzyme molecule substrate Distortion of the enzyme structure can prevents the substrate

from binding Effects of non-competitive inhibitors are not reversed if substrate concentration is increased / irreversible 10. Non-competitive Non-competitive inhibition can be reversible if inhibitor binds briefly to the enzyme/ nonreversible if i t binds permanently 11. Competitive inhibition inhibition has no effect on Vmax but increases Km 12. Non-competitive Non-competitive inhibition has no effect on Km but lowers Vmax.

9.

Any 7 points points that include include or suggest suggest highlighted highlighted phrase phrase – 1 mark each [7 marks]

(b)

Role of tissues in supporting the function of the root 1. Epidermis – outermost layer of root can offer some protection as a physical barrier / Some cells have root hair extensions.











02

SCHEME

5. Xylem tissue – vascular tissue - contains dead, empty cells with lignified side walls and no end walls (vessel element). Lignin gives strength to the organ for anchorage. Vessel elements arranged end to end to form continuous continuous tube for water transport throughout the plant 6. Phloem tissue – vascular tissue – contains phloem sieve elements. Living cells with perforated end walls arranged end to end. Allows sucrose solution to flow to root and other parts of plants to supply energy and building materials. 7. Exodermis - found under the e pidermis, has suberin. protection against microorganisms, prevents water loss from plant. 8. Meristematic tissue - found in root tips, for growth/mitosis

Any four tissues tissues well discussed (linking structure/lo structure/location cation to role) – 2 marks each Partial discussion – 1 mark each [8 marks] Total 15 marks









02

SCHEME

Question 5 (a)

Synthesis of RNA 1. The information information stored in the gene is in the form of a sequence of bases which is used to sequentially arrange amino acids to synthesize a protein. 2. RNA /mRNA is the molecule which is used as the template to organize the primary structure of proteins. 3. Process by which DNA is used to synthesize RNA is called transcription transcription / producing mRNA with a complimentary base sequence to one strand of DNA 4. DNA strand unzips so that both strands separate (catalyzed by DNA helicase / RNA polymerase) 5. One strand is used as the template strand. 6. Complementary Complementary bases (ribonucleotides) are added to strand being copied (in 3’to 5’ direction, catalyzed by RNA polymerase).

7. Base pairing: A with U; C with G 8. Condensation reactions occur (phosphodiester bonds form) between bases (in a 5’to 3’ direction)to form mRNA

9. mRNA strand is identical to the coding/sense strand 10.

Three bases (triplet codon)on the strand code for 1 amino acid

Any 7 points points - 7 marks








UNIT 1 – PAPER

(b)

02

KEY AND MARK SCHEME How proteins are responsible for phenotype

1.

Phenotype is the external observable characteristics of an organism.

2.

Phenotype is determined by the kind of proteins produced in the body

3.

Proteins are expressions of genes / a particular gene codes for a particular protein.

4.

Cells of a similar phenotype form tissues (different tissues form organs, organs form organ systems), which compose the organism.

5.

For example, the protein haemoglobin is packed into an erythrocyte giving it its characteristic red colour. Accept ( any

appropriate example e.g. sickle cell anemia).

[5 marks] Any 5 points points – 1 mark each Total 15 marks









02

SCHEME

Question 6 (a) Steps in plant tissue culture and key scientific principle 1. 2. 3. 4.

Removal of small group of cells from plant (called explant) Disinfection of the explant immediately after removal from plant Immerse explant in sterile aerated solution/culture medium containing hormones and nutrients Undifferentiated cells of explant divide repeatedly to form callus – can be maintained indefinitely indefinitely in culture

5.

Callus cells sub-cultured on sterile medium and induced to form

6.

shoots and roots by varying plant growth substances Plantlets transplanted to sterile soil once they are large enough

Any 5 points points – 1 mark each 7.

Key scientific principle underlying this technique: Plant tissues are totipotent – Each cell contains all the information required to produce the entire plant (totipotency)

Principle well explained – 2 marks Partial explanation – 1 mark

(b)

[7 marks]

Structural features of the sperm cell and secondary oocyte in humans related to function









02

SCHEME

There are 4 components of each listed feature: Sperm cell Secondary oocyte 1. Structure 2. Function

3. Structure 4. Function

Any 3-4 components components correct – 2 marks Any 2 components components correct – 1 mark [8 marks] Total 15 marks

Cape Biology 2018 u1 p2 Ms

Recommend Documents