Biology Revision Unit 1

Front 1

Suggest why the initial rate of reaction was measured in this investigation.

Back 1

1. it allows a comparison to be made (with and without copper ions) ;
2. substrate concentration being the same (with and without copper ions) ;
3. the rate of reaction changes with time ;
4. because substrate is being used up ;

Front 2

Describe a test that could be used to show that starch has been broken down.

Back 2

1. use of iodine solution ;
2. observation / colour change;
3. credit any valid experimental details ;

Front 3

Suggest one reason why the double membrane is not clearly visible all around the
mitochondrion labelled A.

Back 3

1, resolution not high enough
2, damage
3, angle of section
4, poor printing of photograph

Front 4

draw and label a diagram to show the structure of a chloroplast

Back 4

1. {envelope / double membrane} clearly shown ;
2. granum clearly shown ;
3. {granum / thylakoid(s)} labelled ;
4. {stroma / ribosomes / starch grain / DNA / lipid droplet / {double
/ inner / outer} membrane / envelope / intergranal lamellae}
correctly labelled ;

Front 5

Give an account of the events that take place during prophase and metaphase of mitosis.

Back 5

1. during prophase chromosomes becoming visible ;
2. centrioles move to opposite poles ;
3. formation of spindle ;
4. disappearance of nucleolus;
5. breaking down of nuclear membrane in prophase or
nuclear envelope is broken down by metaphase ;
6. at metaphase chromosomes attached to spindle fibres ;
7. chromosomes lined up at equator ;

Front 6

Explain what is meant by the term tissue.

Back 6

same cells of similar function

Front 7

Some polar molecules cross the membrane by combining
with transport proteins which can change their shape. When this movement occurs
down a concentration gradient, the process is called ...............................................
and when it occurs against a concentration gradient the process is called
................................................

Back 7

facilitated diffusion, active transport

Front 8

which is structural carbohydrate? Glycogen or Cellulose

Back 8

cellulose

Front 9

suggest why the clotting time of the human blood with snake venom added was compared with the clotting time in the presence of Factor Xa

Back 9

1, if the snake venom has similar effects as a known
clotting factor, an idea of its mode of action can be
worked out
2, we can see how deadly the snake is
3, compare to normal clotting process
4, possible use as medication
5, for research into antidotes

Front 10

describe the structure of enzyme

Back 10

1. an enzyme is a protein ;
2. they have 3D / tertiary / globular structure ;
3. intramolecular bonds holding structure in place ;
4. between the R groups ;
5. active site ;
6. specificity of active site ;

Front 11

how the enzyme in the snake venom could be involved in the blood clotting process

Back 11

1. it is one of the enzymes in the clotting process has active site complementary to one of the substrates ;
2. it activating other enzymes ;
3. effect on platelets ;
4. it triggers the clotting process ;

Front 12

mammalian gametes are formed by meiosis and have the ability to fuse to form a zygote. explain why it is important that gametes are produced by meiosis.

Back 12

1, halves the chromosome number to make it haploid cells
2, randomly assorts the chromosomes to provide genetic variation
3, the diploid number is maintained in the zygote

Front 13

explain how the genetic material from a mammalian sperm cell is able to enter an ovum to bring about the process of fertilisation

Back 13

1, acrosome
2, digestive enzymes
3, break down jelly surrounding ovum and allow head to reach the ovum membrane
4, genetic material passing through ovum membrane

Front 14

outline the events that occur in mitosis from the start of prophase up to the end of metaphase

Back 14

1, chromosomes shorten, coil and condense, thus each chromosome become visible as two chromatids
2, nuclear membrane breaks down
3, centrioles position themselves at the opposite end of the cell
4, spindles formed from microbubules between centrioles
5, chromosomes line up on the equator and attached by centromeres

Front 15

explain why pectinase does not hydrolyse the cellulose found in the orange peel cell walls

Back 15

shape of enzyme active site fits pectin / does not fit cellulose /

reference to specificity of enzymes ;

Front 16

suggest why chopping up the oranges could speed up hydrolysis

Back 16

1. increases the surface area ;
2. more substrate available / increases the number
of enzyme-substrate complexes / collisions between enzyme and substrate

Front 17

the pectinase shows a small increase in concentration over the 12 hours of the hydrolysis. suggest why the concentration of the pectinase increases slightly

Back 17

1. hydrolysis uses up water ;
2. evaporation of water ;
3, osmosis into orange ; so same number of the enzyme molecules but in less solvent, thus concentration of the pectinase increases;
4. pectinase released from orange tissues;

Front 18

how the cell differ when it is in anaphase

Back 18

1, centromeres have separated
2, spindle fibres visible between chromosomes
3, chromosomes are shorter/ longer in late anaphase
4, groups of chromosomes pulled toward each pole of cell

Front 19

explain how mitochondria 's structures helps to bring about fertilisation

Back 19

provide energy to enable sperm to reach ovum

Front 20

where do unsaturated lipids have their double bonds?

Back 20

have double bonds between carbon atoms and between carbon and oxygen atoms

Front 21

primary structure of a protein?

Back 21

linear sequence of amino acids joined by peptide bonds

Front 22

the investigation was repeated using DNA polymerase and starch solution but no reaction took place. explain why no reaction took place.

Back 22

1, DNA polymerase has a different shaped active site
2, so starch cannot bind
3, enzyme-substrate complexes cannot form
4, enzyme specificity

Front 23

glycoproteins and phospholipids are molecules found in cell surface membranes. give one function of a glycoprotein.

Back 23

act as receptors

Front 24

describe the structure of a phospholipid molecule

Back 24

1. two fatty acid ‘tails’ ;
2. glycerol ;
3. phosphate ;

Front 25

three steps to destroy a tumour

Back 25

1, stop oxygenated blood reaching tumour
2, anaerobic respiration to tumour tissue
3, build-up of lactic acid

Front 26

disaccharides can be split by what bonds?

Back 26

hydrolysis of glycosidic bonds

Front 27

amylose is an example of a ?saccharide?

Back 27

polysaccharide

Front 28

the role of starch is to?

Back 28

store energy in plants

Front 29

proteins are polymers of amino acids joined by peptide bonds formed between the?

Back 29

carboxyl group and amino group

Front 30

the 3D structure of a protein is held together by?

Back 30

disulphide bridges, hydrogen and ionic bonds

Front 31

describe the structure of phospholipid

Back 31

1. phosphate head ;
2. two fatty acid tails ;
4. location of glycerol ;
5. ester bonds ;

Front 32

explain how the properties of phospholipid contribute to the structure of the cell membrane?

Back 32

1. hydrophilic part, polar, charged ;
2. hydrophobic part, non polar, uncharged ;
3. orientation of molecule in
relation to water;
4. aqueous environment is on two sides

Front 33

describe the role of proteins in active transport

Back 33

1. molecule binds carrier proteins;
2. carrier protein changes shape ;
3. molecules move against a concentration gradient;
4. use of ATP / energy ;

Front 34

activation energy

Back 34

energy needed for a reaction to occur by increasing the number of collisions;

Front 35

explain why it is necessary to measure the initial rate of reaction when investigating the effect of enzyme concentration on the rate of reaction

Back 35

1. there should be enough substrate molecules to saturate the enzyme ;
2. to ensure that substrate is not a limiting factor;
3. highest rate / decreases after initial rate ;
4. as reaction proceeds substrate concentration
decreases ;
5. as substrate gets used up by enzyme
6. substrate concentration should be constant in each test;

Front 36

in this investigation, the substrate concentration was a factor that was kept constant, suggest other factors thatshould be kept constant. for each factor, state how it can be kept constant

Back 36

pH ;
buffer ;

temperature ;
water bath ; not room temperature

time of reaction ;
stopwatch ;

volume of {enzyme / substrate} ; not amount
measuring cylinder / pipette ;

type of enzyme ;
same batch of enzyme ;