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

- 2 02107020/CAPE/KMS 2018 BIOLOGY

UNIT 1 – PAPER KEY AND MARK

02

SCHEME

Question 1 Specific 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 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

- 3 02107020/CAPE/KMS 2018 BIOLOGY

UNIT 1 – PAPER KEY AND MARK

02

SCHEME

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

  

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

B.

 

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

- 4 02107020/CAPE/KMS 2018 BIOLOGY

UNIT 1 – PAPER KEY AND MARK

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 areas are correct but only 1 tail is depicted) 

(b) Orientation 1. 2.

3.

Bilayer - composed of two layers of phospholipids (Accept labeled diagram) Polar/hydrophilic heads face outward    (to the internal and external aqueous media)

Tails from each layer face inward   to each other (and exclude water) Each point - 1 mark [3 marks]

- 5 02107020/CAPE/KMS 2018 BIOLOGY

UNIT 1 – PAPER KEY AND MARK

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]

Triglyceride / triacylglyceride / 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: Triglycerides needed in body in order to absorb lipid 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 - 1 mark each [2 marks] Total 15 marks

- 6 02107020/CAPE/KMS 2018 BIOLOGY

UNIT 1 – PAPER KEY AND MARK

02

SCHEME

Question 2 (a) (i) Position of alleles

1. 2. 3.

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

Each correct pair including location – 1 mark  All 3 allelic pairs correct but incorrect 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; Expected offspring phenotype ratio is 1:1 (phenotypes must be stated) 1. 2. 3.

Correct genotype of parents Correct use of Punnett square (even with incorrect parents) Correct expected ratio

Each point - 1 mark [3 marks]

- 5 02107020/CAPE/KMS 2018 BIOLOGY

UNIT 1 – PAPER KEY AND MARK

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 molecule (that assigns a p articular aspect of character)/ a short piece of DNA, which is responsible for the inheritance of a particular 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

[2 marks]

1 mark for partial answer

(b) (i)

Scaled drawing

1. 2.

3. 4.

Cells drawn to acceptable relative proportions Cell contents representative of cell cycle stages shown in micrograph (prophase, anaphase, interphase) e.g. For prophase thick short lines or crosses representing chromosomes and for interphase thin and diffused (or circle) Neat and tidy drawing (clear, single and continuous lines) Feasible magnification given (if cells are similar in size to those in micrograph, the magnification is x600 to x800; i f cells drawn to twice the size as in the micrograph can be x1200 to x1600) – Accept feasible value.

Calculation: Not necessary to show

Each point – 1 mark [4 marks]

- 6 02107020/CAPE/KMS 2018 BIOLOGY

UNIT 1 – PAPER KEY AND MARK

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

- 7 02107020/CAPE/KMS 2018 BIOLOGY

UNIT 1 – PAPER KEY AND MARK

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 [3 marks] (ii)

Conclusions based on Comparison 1. 2.

The concentration of mercury in foetal blood is higher than in the maternal blood (or vice versa) As the number of maternal amalgam fillings increase, the concentration of mercury increases in both maternal and foetal blood.

Each point – 1 mark [2 marks]

- 8 02107020/CAPE/KMS 2018 BIOLOGY

UNIT 1 – PAPER KEY AND MARK

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 to be pumped 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 – 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).

Each point – 1 mark [3 marks] (ii)

Significance of double fertilization In some crop seeds such as rice, wheat and maize, the endosperm  remains after the embryo has matured and provides the main source of nutrition for human consumption / In some crops the nutrients in endosperm is transferred to the maturing embryo (cotyledons) and this is used for human consumption.

Complete explanation – 2 marks Comment: endosperm should be identified to obtain the two marks.

1 mark for partial outline

[2 marks] Total 15 marks

- 9 02107020/CAPE/KMS 2018 BIOLOGY

UNIT 1 – PAPER KEY AND MARK

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 part of the enzyme molecule (allosteric site) / or binds permanently to active site Distortion of the (3-D) structure of enzyme can occur if the inhibitor binds to other part (non- acti ve site) of enzyme molecule 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 inhibition can be reversible if inhibitor binds briefly to the enzyme/ nonreversible if i t binds permanently 11. Competitive inhibition has no effect on Vmax but increases Km 12. Non-competitive inhibition has no effect on Km but lowers Vmax. 9.

 Any 7 points that include or suggest highlighted 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. Provide larger surface area for the uptake of water and nutrients (minerals) 2. Parenchyma –  beneath the epidermis (Cortex). These cells are relatively unspecialised. They have thin cell walls and facilitate the movement of water. They provide support when cells are turgid. They may also serve for storage of starch (energy reserve) 3. Endodermis – single layer of cells surrounding the stele / vascular tissue. Cell walls are waterproofed to form the Casparian strip. Controls the movement of water and mineral salts into the vascular tissue. 4. Pericycle –  layer of cells just beneath the endodermis. Cells remain capable of dividing during plant growth. Gives rise to branches in root – increasing surface area and providing better anchorage

- 10 02107020/CAPE/KMS 2018 BIOLOGY

UNIT 1 – PAPER KEY AND MARK

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

- 11 02107020/CAPE/KMS 2018 BIOLOGY

UNIT 1 – PAPER KEY AND MARK

02

SCHEME

Question 5 (a)

Synthesis of RNA 1. The 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 / 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 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 - 7 marks Alternative diagram

Diagram: 3 marks

[10 marks]

- 12 02107020/CAPE/KMS 2018 BIOLOGY

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 – 1 mark each Total 15 marks

- 13 02107020/CAPE/KMS 2018 BIOLOGY

UNIT 1 – PAPER KEY AND MARK

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 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 – 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

Feature Overall structure and size

Structure and function of Structure and function of human sperm cell human secondary oocyte Head, mid piece, flagellum (Smaller, Length = 60 µm, Size of head 4 µm) -  provides motility

Spherical cell surrounded by zona pellucida / follicle cells OR  Larger (Width =120-140 µm) - Limited motility / store

nutrients / sperm access Nucleus

Haploid (23 chromosomes)/ has completed meiosis II (contains highly condensed DNA/histones -reduce mass)

Haploid (23 chromosomes) / has not completed meiosis II – provide set of

 maternal/female chromosomes/genetic material chromosomes/genetic  material – delivers paternal/male

Cell membrane

Glycoproteins complementary to proteins on oocyte (zona pellucida) membrane – facilitate union with

oocyte Mitochondria

Many (arranged spirally around axial filament) - to provide energy for

swimming/movement

Microvilli - to absorb nutrients /

has proteins that bind to  proteins on sperm cells (Many) mitochondria - to provide energy (for development, metabolism etc.)

- 14 02107020/CAPE/KMS 2018 BIOLOGY

UNIT 1 – PAPER KEY AND MARK

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