ANSWERS TO IN-CHAPTER QUESTIONS

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Q1 Faeces are not waste products
of biochemical reactions and so are not excreted from the body. Faeces are made
of undigested food, bacteria and dead cells from the lining of the gut – the
elimination of faeces is a process known as egestion.
Q2 A
particular amylase enzyme works at the particular place in a starch molecule
which has the complimentary shape to that of the active site of the amylase
enzyme. Hence the specific 3D shape due to tertiary level structure of the
protein determines the activity of the enzyme.
Q3. Lactose
intolerant people lack lactase but can eat yoghurt because the lactose has. been broken down during the
fermentation process to form glucose and galactose therefore no lactose will be
present needing lactase to break it down
Q4. Lactose reduced milk is
sweeter than fresh milk because fresh milk contains the disaccharide lactose
while lactose reduced milk has the
monosaccharides glucose and galactose. In lactose reduced milk there
are twice as many molecules that cause sweetening than there are in fresh milk
and so lactose reduced milk is sweeter.
Q5. Endopeptidases are first released and then exopeptidases as this is more efficient
since the action of
endopeptidases is to
hydrolyse peptide bonds within the protein molecule thereby creating more
terminal bonds for the exopeptidases to hydrolyse. If exopeptidases
were the only enzymes secreted the protein digesting process would be much
slower as the amino acids would be removed two at a time, one from each end of
the protein molecule.
Q6. a. When a
molecule of triglyceride is broken down by lipase 3 molecules of fatty acids
and one molecule of glycerol. Therefore a total of 4 molecules.
b. When a molecule of triglyceride is broken down by lipase to form
monoglyceride and fatty acids one monoglyceride molecule is formed and two
fatty acid molecules giving a total of three molecules.
Q7. The small intestine is long
and circular in cross section so its internal surface area for absorption is
large. The surface of the epithelial cells adjacent to the lumen of the gut is
greatly expanded by the presence of folds called villi. Each villus has its surface
area still more increased by the presence of microvilli.
Q'8. By the time food
is halfway along the small intestine it will have received digestive
juices from:
·
Salivary glands
·
Gastric glands in the stomach
·
Pancreas
·
Liver (bile)
·
Intestinal glands of the duodenum and ileum
Q9. Adding more
glucose to the ORT mixture would not make the treatment more effective as there
would need to be more sodium to help take the glucose through the carrier
molecules. Only then would the concentration of glucose increase in the cells
and cause osmosis to take place so that water would be absorbed into the cells
lining the gut. The extra glucose would
lower the water potential in the gut and so make the absorption of water
harder.
Q10. If the stomach of a newborn
mammal does not secrete hydrochloric acid it will not provide the correct pH
for the activation of pepsinogen and so pepsin will not be formed and
proteins will not be digested, allowing for antibodies in the mother’s milk to
be absorbed by the new-born child and so protect it. If trypsin is not present in the small intestine then again
proteins will not be digested into amino acids and so the new born will not be
able to absorb amino-acids. However,
proteins can be absorbed directly through the epithelial cells by pinocytosis.
The large surface area of the small intestine mean that there can be many
pinocytotic channels present.
Q11. If digestive juices are
secreted when no food is present in the gut then
would all be wasted.
Q12. a. I would expect the
diameter of the pupil to change as a result of the nervous system because the
change needs to be quick as high light levels could permanently damage
the retina and light levels can change quite suddenly.
b. Hormones are more likely to control
the concentration of a chemical within the blood since they travel in the blood
to the target organs and the change would be gradual – a form of homeostasis.
Q13. The units of energy are shown as kilojoules per 100ml so that the foods may be directly compared.
Q14. a. As 4.2 joules
of energy raises the temperature of one ml of water one degree Celsius it is
necessary to multiply the increase in temperature by the volume of water. The answer is 20 x 30 x 4.2 = 2520 Joules. This needs to be divided by 1000 to get the
answer in kiloJoules = 2.52kJ
b. The energy content calculated in a.
above should be divided by the mass to
get kJ per g
Q15. Because it is essential in the
diet – cannot be made through transamination.
Q16. Women in the age range 15 - 50 years menstruate and lose blood - and hence
red blood cells - on a regular basis and need to have a higher input of iron so
they can form haemoglobin, a component of red blood cells.
Q17. Pregnancy will
affect BMR increasing it because a pregnant woman is heavier, has a larger
surface area and growth of the foetus is taking place. She will also be more active a.k.a. ‘nesting
syndrome’.
Q18. If she is playing tennis,
6.9 x 260kJ = 1794kJ.
Q19 a. She will need 40 x 0.76 = 30.4
grams of animal protein.
b. If
she gets her protein from cereals and vegetables, she will need 30.4
x 1.2 = 36.48 grams.
Q20. A normal distribution would
show the mean, median and mode all at the same value: the histogram shows the
median and modal group to be 20 - 40 ml, whilst the
mean value is the 40 to 60 group.
Q21. The mean blood
loss when IUDs are fitted is 90 ml and so more iron must be included in the
diet if the women are not to become
anaemic.
Q22.
27% of 222 mg watercress is absorbed = 59.94g
46% of 122 mg of
skimmed milk is absorbed = 56.12g
Therefore the 100g portion of
watercress contains more absorbable calcium than the skimmed milk.