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LIVER + KIDNEY MOCK EXAM
NAME:______________________________________________________
GROUP:__________________________________
DATE:_________________________________________
MARKS__________________OUT OF_______________________________
________________________%
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1.
The graph shows the rate of glucose absorption in and excretion from a mammalian kidney in
relation to the glucose concentration in the plasma.
Glucose
excreted
800
600
Rate of glucose
reabsorbtion or 400
excretion/
mg minute –1
Glucose
reabsorbed
200
0
0
200
400
600
800
Plasma glucose concentration/mg 100 cm–3
(a)
Draw a line on the graph to show the rate of filtration of glucose in the renal capsule.
(1)
(b)
In which part of the nephron is glucose reabsorbed?
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(1)
(c)
Explain the shape of the glucose reabsorption curve.
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(3)
(Total 5 marks)
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2.
(a)
Small desert mammals can survive without drinking. They meet their water requirements
mainly from metabolic water.
(i)
What is metabolic water?
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(1)
(ii)
Describe one other way in which these animals are able to gain water.
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(1)
(b)
The fat-tailed mouse has a characteristic store of fat within its tail. It has been suggested
that stores of fat are ways of providing desert-living animals, like the fat-tailed mouse,
with water.
The table gives some figures relating to the metabolism of fat over a period of time.
Mass of fat respired (g)
Mass of water formed on oxidation (g)
Volume of oxygen consumed (cm3)
Mass of water evaporated from lungs (g)
1.06
1.13
2.13
1.80
Use the figures in this table to explain why:
(i)
one gram of fat can produce more than one gram of water on oxidation;
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(1)
(ii)
the suggestion that stores of fat are ways of providing desert-living animals with
water cannot be correct.
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(2)
(Total 5 marks)
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3.
Answers should be written in continuous prose. Credit will be given for biological accuracy, the
organisation and presentation of the information and the way in which the answer is expressed.
Read the following passage.
American scientists have now moved a step closer to developing artificial blood by producing
clumps of haemoglobin molecules in the form of tiny bubbles. These ‘microbubbles’ are
chemically stable and have excellent oxygen-carrying ability.
Real blood has a number of disadvantages when used for the treatment of human patients. It
must be screened for disease, its blood group must be checked and carefully matched to the
recipient and it has a very short shelf-life, even when stored in a refrigerator. Pure haemoglobin
has none of these disadvantages. It is disease free and can be used to treat anyone, regardless of
blood group. It does have one major disadvantage, however. It is broken down into a form which
is readily removed from the blood when it flows through the kidneys.
(a)
Describe how haemoglobin normally loads oxygen in the lungs and unloads it in a tissue
cell.
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(7)
(b)
Haemoglobin in solution in the blood is broken down by enzymes into smaller molecules.
Explain how these smaller molecules are removed from the blood in the kidneys.
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(5)
(Total 12 marks)
4.
(a)
Ethanol in alcoholic drinks suppresses the secretion of the hormone ADH. Explain the
effect that consumption of alcoholic drinks is likely to have on the concentration of urine.
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(2)
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(b)
Diuretics are drugs which are used to decrease the volume of tissue fluid by increasing the
rate of urine production. Frusemide is a diuretic which prevents the uptake of sodium ions
from the loop of Henle. Explain how Frusemide increases the rate of urine production.
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(2)
(c)
Urinary retention is the inability to empty the bladder. It usually results from the failure of
the bladder muscle to contract sufficiently. One drug used to treat urinary retention has
molecules that are very similar in shape to those of acetylcholine. Suggest how treatment
with this drug may prevent urinary retention.
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(2)
(Total 6 marks)
5.
Read the following passage.
Amino acids contain from 0.07 to 0.32 grams of nitrogen per gram of amino acid. When
they are broken down in the body, a problem arises. The nitrogen is attached to hydrogen
so the easiest way to deal with it is to convert it to ammonia. No ATP is needed to do this
and ammonia is very soluble in water but it is extremely poisonous. Although one part in
20000 in our blood will kill us, organisms that develop and live in water, such as
freshwater fish, can cope.
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Land animals need to conserve water. Ammonia in the blood would be toxic so urea is
made instead. Urea is about 400 times less poisonous than ammonia. It is fairly soluble in
water but the synthesis of each molecule of urea uses the energy from four molecules of
ATP.
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Birds and reptiles develop inside eggs which contain just enough water to see them
through to the hatching period. During this time the developing embryos produce
nitrogenous waste products but cannot excrete them. These animals make urate. Synthesis
of urate uses the energy from eight ATP molecules but it is almost insoluble in water and
therefore not poisonous.
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From what has been said so far you might expect humans not to excrete urate. But human
urine does contain some urate. This comes from the diet and also from the turnover of
cells in the body. Urate is a purine, as are the nucleotide bases, adenine and guanine. We
do not have the enzymes to break down purines. The best we can do is to convert them to
urate. Urate is excreted in the urine but problems can arise because of its low solubility in
water. At 10°C, for example, the solubility of its sodium salt is 12 mg dm–3, at 30°C it is
45 mg dm–3. If too much urate gets into the urine, it precipitates as kidney stones. Many
animals avoid this problem by converting urate into more soluble compounds but humans
and apes lack the enzymes to do this.
Some people have so much urate in their blood that it precipitates within the body. In the
extremities of the body such as the joint of the big toe it can cause excruciating pain, a
condition known as gout. The pain is caused by the body’s reaction to insoluble urate
crystals. Treatment aims to lower the concentration of urate in the blood. This is done by
inhibiting the action of the enzyme that forms it. The patient is given a drug called
allopurinol which has a very similar molecular shape to hypoxanthine, the normal
substrate of this enzyme.
Traditionally gout has been associated with rich foods and high living. Excessive
consumption of alcohol probably does encourage attacks of gout in genetically susceptible
individuals. One reason for this is that alcohol inhibits the production of ADH and thus
leads to dehydration.
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Source: adapted from Biological Sciences Review, September 1996
Using information in the passage and your own knowledge, answer the following questions.
(a)
Explain what causes different amino acids to contain different proportions of nitrogen.
(2)
(b)
(i)
The larval stage of a toad is an aquatic “tadpole” which excretes 60% to 80% of its
nitrogenous waste as ammonia; the adult toad is terrestrial and excretes its
nitrogenous waste mainly as urea. Suggest the advantage of this change in pattern
of nitrogenous excretion.
(3)
(ii)
Explain why the developing embryos of birds and reptiles “produce nitrogenous
waste products but cannot excrete them” (lines 12- 13)
(2)
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(iii)
Explain why “you might expect humans not to excrete urate” (line 16).
(2)
(c)
Microorganisms can be grown to provide food. If this food is protein, the product is called
single-cell protein (SCP). Unfortunately, the prokaryotic organisms used for SCP
production have a very high nucleic acid content. Use the material in the passage to
explain why this can lead to human health problems.
(2)
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(d)
(i)
Suggest an explanation for the fact that gout usually affects the extremities of the
body such as the big toe (lines 26-27).
(2)
(ii)
Use your knowledge of enzyme structure to explain why allopurinol is effective in
treating gout.
(4)
(e)
Describe the mechanism by which excessive consumption of alcohol encourages attacks
of gout.
(3)
(Total 20 marks)
6.
The diagram shows the structure of glutamic acid. Glutamic acid is an amino acid.
COOH
(CH 2 )
H
N
H
C
H
O
C
OH
Describe what happens to this amino acid in the process of deamination in the liver.
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(Total 2 marks)
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7.
A reaction involving the amino acid alanine occurs in liver cells.
CH 3
NH 2
C
CH 3
COOH + NAD + H 2 O
Reduced NAD + NH 3 + O
H
alanine
(a)
C
COOH
pyruvic acid
What is the name of this type of reaction?
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(1)
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(b)
Choose two of the products of this reaction and suggest what will happen to them.
1 ..................................................................................................................................
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2 ..................................................................................................................................
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(2)
(Total 3 marks)
8.
The diagram shows the structure of a nephron (kidney tubule).
A
B
C
E
F
Direction
of blood flow
(a)
D
Name the major artery of which A is a branch.
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(1)
(b)
Name the process that takes place in the part of the nephron labelled B.
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(1)
(c)
Give the letter or letters which represent the region or regions of glucose reabsorption.
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(1)
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(d)
If humans drank only sea water, they would be unable to remove the excess salt from their
bodies. Marine mammals, such as seals and whales, are able to remove excess salt
because region D is relatively longer in these mammals than in humans. Explain why the
longer length of region D enables marine mammals to remove all of the excess salt from
their bodies.
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(3)
(Total 6 marks)
9.
The African lungfish lives in water but is able to survive for periods of time in the mud of dried
up rivers. In water, the lungfish excretes most of its nitrogenous waste in the form of ammonia.
When the river dries up, the lungfish burrows deep into the mud and curls up into a ball. In this
state, the lungfish excretes only small amounts of ammonia. Instead it produces urea which
accumulates in its tissues.
When rain refills the river, the lungfish excretes the accumulated urea and then returns to
excreting large amounts of ammonia.
S
(a)
(i)
Suggest how curling up into a ball increases the lungfish’s chance of survival.
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(1)
(ii)
Explain the advantage of changing the method of excreting nitrogenous waste when
the river dries up.
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(2)
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(b)
The diagram shows some of the steps in the production of nitrogenous excretory products
by the liver cells of the lungfish.
NAD
reduced NAD
water + amino acids
ADP
pyruvic acid + ammonia
Deamination
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ATP
urea
Ornithine cycle
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(i)
Use the equation to suggest one disadvantage of excreting urea rather than
ammonia.
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(1)
(ii)
The lungfish requires water for the deamination of amino acids. Name one
metabolic process that replaces water used for deamination.
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(1)
(Total 5 marks)
10.
Most nitrogenous waste material comes from surplus protein in the diet.
(a)
Figure 1 shows some of the important steps in the formation of urea in mammals.
Protein
Amino
acids
C
D
Ammonia
Urea
Figure 1
S
(i)
Why are protein molecules considered to be polymers?
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(1)
(ii)
Name process C. ...............................................................................................
(1)
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(iii)
Describe what happens to the part of the amino acid molecule labelled D.
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(2)
(b)
Tadpoles of the common frog live in freshwater ponds. Over a period of weeks, they
undergo metamorphosis as they develop into adult frogs and move onto land. Figure 2
shows the proportions of nitrogenous waste excreted as ammonia and as urea during the
time after the tadpole hatches from the egg.
100
Urea
80
Nitrogenous 60
waste excreted
as ammonia or
urea /%
40
20
0
Ammonia
0
10
20
30
40
50
60
70
80
Days after hatching from egg
90
100
110
Figure 2
(i)
Describe the changes in proportions of the two excretory products over the period
shown.
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(2)
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(ii)
Suggest an explanation for the changes in the proportions of ammonia and urea
excreted.
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(4)
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(c)
In the first convoluted tubule of a human nephron, sodium ions, glucose molecules and
water molecules are reabsorbed into the blood plasma. Figure 3 illustrates how these
substances are reabsorbed.
H 2O
Y
Gl
uco
se
So
d
iu
m
io
n
Wall of first
convoluted
tubule
of nephron
Capillary
X
Carrier protein
Figure 3
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(i)
Explain the large number of the organelles labelled X in these cells.
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(2)
(ii)
Give two differences between the process by which water enters the capillary from
the epithelial cell and that by which glucose and sodium leave the epithelial cell.
1 ........................................................................................................................
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(2)
(iii)
Explain the importance of the structures labelled Y on the epithelial cells.
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(1)
(Total 15 marks)
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11.
(a)
The flowchart summarises some of the events involved in the control of the water
potential of blood plasma.
Detected by
receptors in the
hypothalamus
Secretion of ADH
by pituitary gland
Permeability of
collecting duct
Reabsorption of
water
Increases
Plasma water
potential normal
Plasma water
potential normal
Decreases
Detected by
receptors in the
hypothalamus
(i)
Secretion of ADH
by pituitary gland
Permeability of
collecting duct
Reabsorption of
water
Write “increases” in one appropriate box and “decreases” in one other appropriate
box.
(1)
(ii)
Give evidence from the flowchart which shows that the control of plasma water
potential involves negative feedback.
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(1)
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(b)
Describe the role of the loop of Henle in the reabsorption of water from the collecting
ducts. You may draw a diagram if it helps your answer.
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(3)
(Total 5 marks)
12.
Mammals and fish remove nitrogenous waste from their bodies in different forms.
S
(a)
Name two polymers present in mammals and fish that contain nitrogen.
1 ........................................................................................................................................
2 ........................................................................................................................................
(2)
(b)
The diagram shows a reaction that occurs in the liver of both mammals and fish.
R
H
N
H
C
C
H
amino acid
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NAD reduced NAD
O
+
H2 O
NH 3
OH
R
+
O
C
C
O
ammonia
OH
X
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(i)
Describe what may happen to molecule X.
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(1)
(ii)
In a mammal the ammonia is converted into urea. Give one advantage of this.
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(1)
(iii)
Describe how the ammonia is removed from the body of a fish.
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(1)
(c)
In a mammal urea is removed from the blood by the kidneys and concentrated in the
filtrate.
(i)
Describe how urea is removed from the blood.
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(2)
(ii)
Explain how urea is concentrated in the filtrate.
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(4)
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(d)
The diagram shows one way in which a person who has kidney disease can have the
condition managed. In the process a fluid is put into the abdominal cavity. Exchange of
materials takes place across the membrane that surrounds the abdominal cavity. This
removes waste products from the blood. After five hours the fluid is drained out of the
cavity and discarded. The cavity is then refilled with fresh fluid.
The table shows the concentration of solutes in the fresh fluid.
Solute
Sodium ions (Na+)
132
Chloride ions (Cl–)
96
Calcium ions (Ca2+)
1.25
Magnesium ions (Mg2+)
0.25
Glucose
Urea
(i)
Concentration/
mmol dm–3
76
0
By what process does urea enter the fluid in the abdominal cavity from the blood?
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(1)
(ii)
Explain why the fluid is changed every five hours.
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(1)
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(iii)
Fluid of the composition shown in the table is used instead of distilled water.
Explain why.
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(2)
(Total 15 marks)
13.
The desert rat is a small mammal which can survive long periods in a hot desert without drinking
any water. In an investigation, groups of desert rats were given either unlimited water or no
water at all. They were fed either a high protein diet or a high carbohydrate diet.
This is summarised in the table.
Group
(a)
Water allowed
Diet
A
Unlimited
High protein
B
None
High protein
C
None
High carbohydrate
All the rats that were fed the high protein diet produced a glomerular filtrate in their
nephrons with a higher concentration of urea than those fed the high carbohydrate diet.
Explain why.
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(1)
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(b)
The mean percentage change in body mass of each group was calculated over sixteen
days. The results are summarised in the graph.
+20
Unlimited
water,
high protein
diet
A
+10
0
Mean
percentage
change in –10
body mass
C
No water,
high
carbohydrate
diet
–20
–30
B
No water,
high
protein diet
–40
–50
0
2
4
6
8
10
Time / days
12
14
16
Explain the difference in change in mass over the sixteen days between
(i)
groups A and B;
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(1)
(ii)
groups B and C.
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(2)
S
(c)
Explain how a more concentrated blood plasma could cause dehydration of cells in the
thirst centre of the hypothalamus.
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(1)
(Total 5 marks)
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