14: Growth and Development - Problems

Question 1

State the roles of water in seed germination. (4)

Answer 1

Water softens the seed coat and makes the cotyledons swell. This breaks the seed coat so that the radicle can emerge.
Water activates the enzymes in the seed. The enzymes catalyse the conversion of insoluble stored food into soluble food substances.
Water acts as a reactant in the hydrolysis of stored food.
Water acts as a medium for transporting soluble food substances to the growing regions of the embryo.

Question 2

Explain why healthy seeds did not germinate if they are watered and kept at 10ºC. (2)

Answer 2

Warmth is necessary for the enzymes in the seeds to work efficiently.
At low temperatures, enzymes are inactive and insufficient stored food inside the cotyledons are converted into soluble food substances for the growth of the embryo.

Question 3

(CE 2010 I Q8b)
A study was conducted to find out the change of amylase activity and the amount of reducing sugars in seeds during germination. The seeds were washed with water and soaked in dilute sterilising solution for 10 minutes, then rinsed with distilled water and germinated under suitable conditions. The table shows the results.
With reference to the data from day 1 to day 5, describe and explain the relationship between the amylase activity and the amount of reducing sugars in seeds during germination. (2)

Answer 3

When the amylase activity increases, the amount of reducing sugars in seeds increases.
There is increase in amylase activity that catalyse the breakdown of more starch into maltose, which is a reducing sugar.

Question 4

(CE 2010 I Q8b)
A study was conducted to find out the change of amylase activity and the amount of reducing sugars in seeds during germination. The seeds were washed with water and soaked in dilute sterilising solution for 10 minutes, then rinsed with distilled water and germinated under suitable conditions. The table shows the results.
Explain why the amount of reducing sugars in day 9 is lower than day 3 even though the amylase activity in seeds in these two days are the same. (1)

Answer 4

More reducing sugars are consumed in day 9 / less starch is available for amylase activity.

Question 5

(CE 2010 I Q8b)
State two uses of reducing sugars in the seeds during germination. (2)

Answer 5

For energy release for the growth of the embryo.
For forming structural components of new cells like cellulose for growth.

Question 6

A student used the set-up below to investigate the conditions necessary for seed germination.
Give two controlled variables in the experiment. (2)

Answer 6

Species of seeds,
mass of seeds

Question 7

A student used the set-up below to investigate the conditions necessary for seed germination.
State the purpose of boiling the water. (1)

Answer 7

To remove dissolved oxygen in the water.

Question 8

A student used the set-up below to investigate the conditions necessary for seed germination.
Explain why it is necessary to cool the water before the experiment. (1)

Answer 8

To prevent high temperatures of water that may denature the enzymes in the seeds needed for germination.

Question 9

A student used the set-up below to investigate the conditions necessary for seed germination.
State and explain which seed is most likely to germinate. (3)

Answer 9

Seed R,
it can obtain water under the oil layer and oxygen in the air.
Both water and oxygen are necessary for seed germination.

Question 10

A student cut a seed longitudinally. The diagram below shows what they observed.
State the role of structure Y during seed germination, and explain the change in dry mass of it during seed germination. (1+2)

Answer 10

Structure Y contains food reserves for the growth of the plumule and the radicle during germination.
The dry mass of structure Y decreases,
as the food reserves in structure Y are broken down to release energy for growth of the embryo during germination.

Question 11

The photomicrograph below shows a transverse section through the stem of a woody plant.
Name a layer of tissues (X, Y, or Z) which is actively dividing. (1)

Answer 11

Tissue Z

Question 12

The photomicrograph below shows a transverse section through the stem of a woody plant.
Explain if layer P or layer Q is younger with reference to the process of secondary growth in plants. (3)

Answer 12

Tissue Q is younger.
Both tissues P and Q are secondary xylem.
When Z divides, new cells are produced and differentiated to form secondary xylem towards the inside of the stem. The outer layer Q is newly formed.

Question 13

(Edexcel GCE(AS) 2017)
An investigation was carried out into the effects of gamma radiation on the growth of coleoptiles. The mean number of cells in coleoptiles, and the mean cell length in the coleoptiles were determined at regular intervals and presented in the graphs.
Describe the effect of gamma radiation on the mean number of cells in each coleoptile and the mean cell length. (2)

Answer 13

The mean number of cells in the coleoptile decreases when the coleoptile is exposed to gamma radiation.
Yet, gamma radiation has very little effect on the mean cell length.

Question 14

(Edexcel GCE(AS) 2017)
An investigation was carried out into the effects of gamma radiation on the growth of coleoptiles. The mean number of cells in coleoptiles, and the mean cell length in the coleoptiles were determined at regular intervals and presented in the graphs.
The coleoptiles with and without gamma radiation both increased in height but those with gamma radiation were shorter. Explain the effect of gamma radiation on the height of the coleoptiles. (2)

Answer 14

As gamma radiation slows down cell division, the height of the coleoptiles with gamma radiation did not increase as much as the coleoptiles without gamma radiation.
However, since cell elongation is not affected by gamma radiation, the height of the coleoptiles still increases.

Question 15

(Edexcel GCE(AS) 2017)
Give practical details to explain how to compare the number of cells undergoing mitosis in two batches of coleoptiles. (7)

Answer 15

Cut out the tips of the two batches of coleoptiles. Put them on two microscopic slides separately.
Add aceto-orcein stain to each of the coleoptile tips.
Use needles or fine forceps to tear the tips into small pieces. Then cover the torn tips with cover slips.
Press the slides gently with the rubber end of a pencil.
Warm the slides over a hot water bath to intensify the stain.
Observe the slides under a microscope. Under a suitable field of view, count the number of cells undergoing mitosis for each tip.
Calculate and compare the percentage of cells undergoing mitosis of the two tips.

Question 16

The diagram below shows the height of a man at birth and at 18 years old.
Describe and explain the difference in the head-to-height ratio of the man at birth and at 18 years old. (3)

Answer 16

The head-to-height ratio of the man at 18 years old (1:9) is smaller than that at birth (1:5).
This is because the head grows rapidly at the early stage,
and then it grows more slowly than the rest of the body as the person ages.

Question 17

Discuss the similarities and differences between the growth in annual plants and humans. (7+3)

Answer 17

Similarities:
Both annual plants and humans start growing from a single cell (zygote).
Growth of both annual plants and humans involves cell division.
In both annual plants and humans, different body parts grow at different rates.
Differences:
In annual plants, growth occurs in regions containing meristems only.
In humans, growth can occur in all parts of the body.
In annual plants, seeds remain dormant until favourable conditions exist.
In humans, embryos start growing right after fertilisation.