EXAM QUESTIONS &
ANSWERS(RATED A+)
Explain how the active site of an enzyme causes a high rate of reaction - ANSWER1.
Lowers activation energy;
2. Induced fit causes active site (of enzyme) to change shape;
3. (So) enzyme-substrate complex causes bonds to form/break;
Accept: description, of induced fit
Accept: enzyme-substrate complex causes stress/strain on bonds.
A student investigated mitosis in the tissue from an onion root tip.
(a) The student prepared a temporary mount of the onion tissue on a glass slide.
She covered the tissue with a cover slip. She was then given the following
instruction.
"Push down hard on the cover slip, but do not push the cover slip sideways."
Explain why she was given this instruction. - ANSWER1. Push hard - spread /
squash tissue;
2. Not push sideways - avoid rolling cells together / breaking chromosomes.
Neutral - to see cells clearly
A student investigated the effect of hydrogen peroxide concentration on the rate of
this reaction. He used catalase from potato tissue.
The student:
• put five potato chips in a flask
• added 20 cm3 of 0.5 mol dm-3 hydrogen peroxide solution to the flask
• measured the time in seconds for production of 10 cm3 of oxygen gas
• repeated this procedure with four different concentrations of hydrogen peroxide
solution.
(b) Other than those stated, give one factor the student would have controlled in his
investigation. - ANSWERSize/dimensions /mass/variety of potato
OR
Temperature (of solution/flask)
OR
pH (of solution);
Accept : weight of potato
Ignore : amount of potato
Ignore concentration/ volume of catalase
Galactose has a similar structure to part of the lactose molecule.
Explain how galactose inhibits lactase. - ANSWER1. Galactose is a competitive
inhibitor / attaches to the active site (of lactase);
2. Fewer enzyme substrate complexes formed.
A technician investigated the effect of temperature on the rate of an enzyme-
controlled reaction.
, At each temperature, he started the reaction using the same volume of substrate
solution and the same volume of enzyme solution.
Give one other factor the technician would have controlled. - ANSWERConcentration
of substrate solution / of enzyme solution / pH.
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. - ANSWER1. (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
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. - ANSWER1. Hydrolysis of lipids produces fatty acids;
2. Which lower pH of mixture.
Lipase is an enzyme that hydrolyses triglycerides.
A student investigated the hydrolysis of triglycerides in milk by human lipase at 20
°C.
He recorded the pH of a sample of milk before and after adding lipase. He used a pH
meter to record pH.
Suggest one advantage of using a pH meter rather than a pH indicator in this
experiment. - ANSWER(a) Any one from:
1. Numerical readings / not subjective / colour change subjective / gives quantitative
data / not qualitative / gives continuous data;
2. Greater accuracy;
Accept greater precision
Explain why the pH decreases when the lipase is added to the milk - ANSWERFatty
acids produced;
A student prepared a stained squash of cells from the tip of an onion root and
observed it using an optical microscope.
During the preparation of the slide, he:
• cut the first 5 mm from the tip of an onion root and placed it on a glass slide
• covered this tip with a drop of stain solution and a cover slip
• warmed the glass slide
• pressed down firmly on the cover slip.
He identified and counted nuclei in different stages of the cell cycle.