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IB Biology II – Botany Data Practice #1
1.
Name: _______________________________
The unicellular green alga Phaeodactylum tricornutum is photosynthetic. Cell biologists genetically
modified this organism by adding a glucose transporter gene. The modified and unmodified algae were
grown in a nutrient medium under a series of different conditions and the growth rate of the cells was
measured.
12
10
Unmodified algae
Key:
= light and no glucose
in the medium
Modified algae
8
Cell density /
10 7 cells cm –3 6
= light and glucose
in the medium
4
= dark and glucose
in the medium
2
0
0
(a)
2
4
6
8 0
2
Time / days
4
6
8
State the role of glucose in the metabolism of cells.
.....................................................................................................................................
(1)
(b)
Deduce where you would expect to find the glucose transporter protein in the modified algae cells.
.....................................................................................................................................
.....................................................................................................................................
(2)
(c)
Compare the effect of light on the modified and the unmodified cells.
.....................................................................................................................................
.....................................................................................................................................
(2)
Commercially, unmodified algae are grown in shallow sunlit ponds or illuminated containers. The cells
only grow in the top few centimetres. However, the modified algae can grow at any depth.
(d)
Explain why the modified algae can grow at any depth whereas the unmodified algae can only grow
at the surface.
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
(3)
1
2.
A study was carried out to determine the relationship between the diameter of a molecule and its movement
through a membrane. The graph below shows the results of the study.
High
Relative ability to move
through a membrane
Low
0
0.4
0.8
1.2
1.6
Molecular diameter / nm
(a)
From the information in the graph alone, describe the relationship between the diameter of a
molecule and its movement through a membrane.
.....................................................................................................................................
.....................................................................................................................................
(2)
A second study was carried out to investigate the effect of passive protein channels on the movement of
glucose into cells. The graph below shows the rate of uptake of glucose into erythrocytes by simple
diffusion and facilitated diffusion.
Rate of glucose uptake /
mmol cm –3 cells hr –1
500
450
400
350
300
250
200
150
100
50
0
facilitated diffusion
simple diffusion
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
External concentration of glucose / mmol dm –3
(b)
Identify the rate of glucose uptake at an external glucose concentration of 4 mmol dm–3 by
(i)
simple diffusion;
.........................................
(1)
(ii)
facilitated diffusion.
.........................................
(1)
2
(c)
(i)
Compare the effect of increasing the external glucose concentration on glucose uptake by
facilitated diffusion and by simple diffusion.
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
(3)
(ii)
Predict, with a reason, the effect on glucose uptake by facilitated diffusion of increasing the
external concentration of glucose to 30 mmol dm–3.
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
(2)
(Total 9 marks)
3
IB Biology II –Botany Data Practice #2
The Kluane boreal forest ecosystem project was a large scale ten year experimental manipulation of food and
predators on arctic ground squirrel population (Spermophilus parryii plesius).
Three areas were set up:



a food addition area
a predator exclusion area
a food addition area enclosed within a predator exclusion area.
The areas were monitored from 1986 to 1996. In spring 1996 all fences were dismantled and food addition was
stopped.
As a further experiment, spring and summer mark-recapture population estimates of the squirrels were conducted
from spring 1996 to spring 1998. The results for these two years are shown below. The areas are labelled according to
the conditions imposed during the previous ten years.
30
25
20
15
10
Squirrels hectare –1
1.
Name: _________________________________
Control
Predator exclusion
Food addition
Food addition plus
predator exclusion
5
0
Spring
1996
(a)
Summer
1996
Spring
1997
Summer
1997
Spring
1998
State the squirrel population in the food addition plus predator exclusion area in spring 1996.
.....................................................................................................................................
(1)
(b)
Describe the effect of ending food addition on the squirrel population.
.....................................................................................................................................
.....................................................................................................................................
(2)
(c)
Scientists believed that the number of ground squirrels in the boreal forests was limited by an
interaction between food and predators that acted primarily through changes in reproduction. Using
the data, discuss this hypothesis.
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
(3)
4
Research into how the lungs perform during general anesthetics has increased because there are so many pulmonary
complications during operations. It is believed that many inhaled anesthetics affect pulmonary epithelial permeability.
Pulmonary clearing is an indication of whether the alveolar-capillary barrier has been damaged. It can be measured as
the rate at which radioactivity decreases in lungs after inhalation of a radioactive aerosol. The greater the clearing rate,
the greater the damage to the alveolar-capillary barrier. Smoking and lung diseases (such as cancers and asthma) also
significantly increase the clearing rate of radioactive aerosols.
In an experiment, doctors wanted to test the effect of inhaled anesthetics on the permeability between the alveoli and
capillaries. Patients were tested by inhaling a radioactive aerosol one day before their operation and one hour after
their operation.
Three groups of patients each received a different type of anesthetic.
Group 1 :1% halotane

 inhaled anesthetics
Group 2 :1.5% isoflurane 
Group 3: intravenous anesthetic (phentanyl and propofol)
Deviation
Mean
2.0
Clearing rate / % min –1
2.0
1.5
1.0
0.5
0.0
before
anesthetic
after
anesthetic
Group 1
(a)
2.0
Clearing rate / % min –1
Key:
Clearing rate / % min –1
2.
1.5
1.0
0.5
0.0
before
anesthetic
after
anesthetic
1.5
1.0
0.5
0.0
before
anesthetic
Group 2
after
anesthetic
Group 3
Compare the effect of each inhaled anesthetic on the permeability of the alveoli.
.....................................................................................................................................
.....................................................................................................................................
(2)
(b)
Using the data from the graphs, explain whether or not inhaled anesthetics are more dangerous than
intravenous anesthetics.
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
(3)
(c)
Suggest one reason why asthmatic patients were not used in this experiment.
....................................................................................................................................
(1)
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IB Biology II –Botany Data Practice #3
1.
Name: _________________________________
It had always been assumed that eukaryotic genes were similar in organization to prokaryotic genes.
However, modern techniques of molecular analysis indicated that there are additional DNA sequences that
lie within the coding region of genes. Exons are the DNA sequences that code for proteins while introns are
the intervening sequences that have to be removed. The graph shows the number of exons found in genes
for three different groups of eukaryotes.
Percentage of genes
100
80
Saccharomyces cerevisiae (a yeast)
60
40
20
0
40
30
Drosophila melanogaster (fruit fly)
20
10
0
20
15
Mammals
10
5
0
(a)
1
2
3
4
5
6
7
8
9 10 11 12 13 14 15 16 17 18 19 20 <30<40<60>60
Number of exons
Calculate the percentage of genes that have five or less exons in mammals.
.....................................................................................................................................
(1)
(b)
Describe the distribution of the number of exons and the percentage of genes in
D. melanogaster.
.....................................................................................................................................
.....................................................................................................................................
(2)
(c)
(i)
Compare the distributions of the number of exons found in genes of S. cerevisiae and
mammals.
...........................................................................................................................
(2)
6
(ii)
Suggest one reason for the differences in the numbers of exons found in genes of
S. cerevisiae and mammals.
...........................................................................................................................
(1)
Human DNA has been analysed and details of certain genes are shown in the table below.
Gene
Insulin
Gene size / kb*
mRNA size / kb
Number of introns
1.7
0.4
2
Collagen
38.0
5.0
50
Albumin
25.0
2.1
14
Phenylalanine hydroxylase
90.0
2.4
12
2 000.0
17.0
50
Dystrophin
*kilobase pairs
(d)
Calculate the average size of the introns for the albumin gene.
.....................................................................................................................................
.....................................................................................................................................
(2)
(e)
Analyse the relationship between gene size and the number of introns.
.....................................................................................................................................
.....................................................................................................................................
(2)
(f)
Determine the maximum number of amino acids that could be produced by translating the
phenylalanine hydroxylase mRNA.
.....................................................................................................................................
(1)
Hemoglobin is a protein composed of two pairs of globin molecules. During the process of development
from conception to adulthood, human hemoglobin changes in composition. Adult hemoglobin consists of
two alpha- and two beta-globin molecules. Two globin genes occur on chromosome 16: alpha- and
zeta-globin. Four other globin genes are found on chromosome 11: beta, delta, epsilon and gamma. The
graph below illustrates the changes in expression of the globin genes over time.
7
50
Key:
alpha-globin
gamma-globin
beta-globin
delta-globin
epsilon-globin
zeta-globin
40
Percentage of
hemoglobin 30
20
10
0
(g)
10
20
30
weeks of gestation
40
birth
2
6
4
month of age
State which globin genes are the first to be expressed after fertilization.
.....................................................................................................................................
(1)
(h)
Compare the expression of the gamma-globin gene with the beta-globin gene.
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
(3)
(i)
Deduce the composition of the hemoglobin molecules
at 10 weeks of gestation;
.........................................................................................
..........................................................................................
2 months after birth.
..........................................................................................
..........................................................................................
(2)
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IB Biology II –Botany Data Practice #4
The mold Penicillium expansum causes serious crop loss of apples and pears during storage and packing.
The graphs below show the density of P. expansum spores on normal skin, damaged skin and lenticels
(small air pores allowing gaseous exchange) of apples. The experiments show the results at two humidity
levels over a four day period.
High humidity
Low humidity
1000
Log10spores / cm2
1000
Log10spores / cm2
1.
Name: _________________________________
100
10
1
4
12
24
48
96
Time / h
(a)
10
1
0
Key:
100
0
4
12
24
48
96
Time / h
density of spore on normal skin
density of spore on damaged skin
density of spore on lenticels
State the time taken for spores on the skin to reach a density of 100 spores/cm2 at high humidity.
......................................................................................................................................
(1)
(b)
Compare the density of spores on normal skin with spores on lenticels for apples stored at high
humidity.
......................................................................................................................................
......................................................................................................................................
(2)
(c)
Discuss whether apples should be stored at high humidity or low humidity.
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......................................................................................................................................
(3)
2.
Biosphere 2, an enormous greenhouse built in the Arizona desert in the USA, has been used to study five
different ecosystems. It is a closed system so measurements can be made under controlled conditions. The
effects of different factors, including changes in carbon dioxide concentration in the greenhouse, were
studied. The data shown below were collected over the course of one day in January 1996.
1200
1600
1400
1000
1200
CO 2 / 800
ppm
light /
1000 mol m–2 s –1
600
800
600
400
400
Key:
200
0
(a)
(i)
CO 2
Light
200
0
0 1.5 3 4.5 6 7.5 9 10.5 12 13.5 15 16.5 18 19.5 21 22.5 24
Time / hours
Identify the time of day when the sun rose.
...........................................................................................................................
(1)
(ii)
Identify the time of minimal CO2 concentration.
...........................................................................................................................
(1)
(b)
Determine the maximum difference in the concentration of CO2 over the 24-hour period.
.....................................................................................................................................
(1)
(c)
Suggest reasons for changes in CO2 concentration during the 24-hour period.
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
(2)
10
IB Biology II –Botany Data Practice #5
1.
Name: _________________________________
Studies have been carried out on the consumption of oxygen by mammals under different conditions. The
graph below shows the oxygen consumption of a pigmy possum (Cercaertus nanus) at various air
temperatures.
5
4
Oxygen
consumption
/ ml O 2 g - 1 h - 1
3
2
1
0
0
(a)
5
10
15
20
25
Air temperature / C
30
35
40
Estimate the oxygen consumption at 20C.
....................................................................................................................................
(1)
(b)
Outline the effect of temperature on oxygen consumption.
....................................................................................................................................
....................................................................................................................................
(2)
(c)
An air temperature of 32C is the lower critical temperature for a pigmy possum. Deduce the
meaning of lower critical temperature.
....................................................................................................................................
(1)
The graph below shows the metabolic rates of various mammals in relation to air temperatures. Metabolic
rate refers to the rate of use of energy by an animal and may be measured indirectly by oxygen consumption.
The basal metabolic rate (BMR) for each animal, in the absence of stress, is given the value of 100. This
occurs at different temperatures for different animals. The changes in metabolic rate as the temperature falls
are expressed as a percentage of the BMR for each animal.
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(d)
(i)
Identify the mammal with the greatest increase in metabolic rate per degree of temperature, as
the temperature decreases.
..........................................................................................................................
(1)
(ii)
Calculate the average change in the metabolic rate, per degree of temperature, of a weasel as
the temperature decreases from 17C to –20C. Show your workings.
..........................................................................................................................
..........................................................................................................................
(2)
(e)
Suggest one reason for the increase in metabolic rate of mammals at lower temperatures.
....................................................................................................................................
(1)
(f)
Discuss the differences between tropical and arctic mammals regarding the changes in their
metabolic rates as the temperature decreases.
....................................................................................................................................
....................................................................................................................................
....................................................................................................................................
....................................................................................................................................
....................................................................................................................................
(3)
Other studies were carried out to see the relationship of the thickness of fur in mammals to its value as an
insulator. A good insulator prevents heat loss from the animals to their environment. The graph below
shows the relationship of the fur thickness to fur insulation in different mammals. Symbols represent the
transfer of heat from fur to air or fur to ice water.
12
4.0
3.5
white fox
3.0
reindeer
2.5
rabbit
Fur insulation
value / W dm–2
2.0
dall sheep
wolf
grizzly bear
polar bear
red fox
dog
beaver
1.5
seal
lemming
squirrel
1.0
seal in ice water
weasel
0.5
shrew
beaver in ice water
polar bear in ice water
0.0
0
10
20
30
40
50
60
70
80
Fur thickness / mm
Key : ♦ = the rate of transfer of heat from fur to air
◊ = the rate of transfer of heat from fur to ice water
(g)
(i)
Calculate the difference in fur thickness between a reindeer and a grizzly bear.
..........................................................................................................................
(1)
(ii)
Suggest a reason for their similar fur insulation value.
..........................................................................................................................
(1)
(h)
(i)
Calculate the change in the insulation value for a beaver when it enters ice water.
..........................................................................................................................
(1)
(ii)
Suggest an adaptation that allows a mammal to maintain its body temperature while in ice
water.
..........................................................................................................................
(1)
(i)
Discuss the relationship between metabolic rate and fur thickness in mammals.
....................................................................................................................................
....................................................................................................................................
....................................................................................................................................
(3)
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