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
· Salivary glands
· Gastric glands in the stomach
· 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.
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.