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Biology exam practise > Module 6- Module 6 - Stimuli, both internal and external, are detected and lead to a response > Flashcards

Module 6- Module 6 - Stimuli, both internal and external, are detected and lead to a response Flashcards

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1
Q

What is a tropism?

A

A response to a plant of a directional stimulus

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2
Q

What are 2 examples of tropism?

A
  • phototropism is the growth of a plant in response to light and gravitropism is the growth of a plant in response to gravity
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3
Q

What are growth factors and where are they produced?

A

They stimulate growth of a plant by cell elongation and are produced in growth regions of the plant and then move to were they are needed in a plant.
E.G of a growth factor is an auxin

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4
Q

How does IAA work?

A

IAA moves via diffusion or active transport over short distances and then moves through the phloem over long distances which creates uneven distribution of the auxin IAA creating uneven growth.

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5
Q

What is the distribution of IAA in a plant?

A

Phototropism = IAA may be concentrated in the shaded regions of the plant to stimulate cell elongation so that the shaded areas reach sunlight
Gravitropism = Increased concentration on the lower side which increases cell elongation moving it upwards

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6
Q

Explain phototropism shoot

A
  1. IAA produced in the tip of the shoot
  2. IAA diffuses evenly
  3. IAA moves to the shaded side
  4. IAA stimulates elongation of cells in a shoot
  5. The shoot bends towards light
  6. Positive phototropism
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7
Q

Explain gravitropism shoot

A
  1. IAA produced in the tip of the shoot
  2. IAA diffuses evenly
  3. IAA moves to the lower side
  4. IAA stimulates elongation of cells in a shoot
  5. The shoot will bend against gravity
  6. Negative gravitropism
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8
Q

Explain phototropism root

A
  1. IAA produced in the tip of the root
  2. IAA diffuses evenly
  3. IAA moves to the shaded side
  4. IAA inhibits elongation of the cells in the shoot
  5. The shoot will bend away from the light
  6. Negative phototropism
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9
Q

Explain gravitropism root

A
  1. IAA produced in the tip of the root
  2. IAA diffuses evenly
  3. IAA moved to the lower side
  4. IAA inhibits elongation of cells in a hoot
  5. The shoot will bend in the direction of gravity
  6. Postive gravitropism
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10
Q

Explain why the student removed the shoot tip from each seedling?

A

The shoot tip of the seedlings is where IAA is produced and so it affects the shoot length elongation

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11
Q

How can choice chambers be used to investigate animal responses - light?

A
  1. A petri dish is divided in half and has a lid placed on top
  2. Place damp filter paper on both sides and then cover one half of the lid with black paper including the sides
  3. Place a thin piece of mesh over the petri dish and then place woodlice over this mesh
  4. After 10 minutes, record the number of lice on each side of the chamber
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12
Q

Investigating humidity using choice chambers?

A

Add damp filter paper to one side of the dish and then a desiccating drying agent to the other side.

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13
Q

What are tactic responses (taxes)?

A

This is when an organism moves towards or away from a directional stimulus

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14
Q

What are kinetic responses (kinesis)

A

Movement is affected by a non-directional stimulus

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15
Q

Suggest why the student added glucose solution to each petri dish

A
  • This allows the plant to respire so that it can produced ATP which provides energy for growth
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16
Q

2 marks

When a young shoot is illuminated from one side, IAA stimulates growth on the shaded
side. Explain why growth on the shaded side helps to maintain the leaves in a favourable
environment.

A
  1. The shoot will bend towards the light
  2. Light required for photosynthesis;
17
Q

Give one similarity and one difference between a taxis and tropism

A

Similarity: Movement toward/ away from a directional stimulus
Difference: Taxis –> whole organism moves
Tropism–> A growth response

18
Q

The woodlice were left for 15 minutes before their movement was recorded. Give 2 reasons for this

A

Time for required humidity to be established/ Time for substance to absorb water
So that behaviour typical to humidity
Woodlice is no longer affected by handling

19
Q

3 marks

The fovea of the eye of an eagle has a high density of cones. An eagle focuses the image of its prey onto the fovea.
Explain how the fovea enables an eagle to see its prey in detail.

A
  1. High visual acuity
  2. Each cone is connected to a single bipolar nerve
  3. Cones send separate sets of impulses to brain
20
Q

3 marks

The retina of an owl has a high density of rod cells.
Explain how this enables an owl to hunt its prey at night.

A
  1. High visual sensitivity (retinal convergence)
  2. Several rods connected to a bipolar nerve
  3. Spatial summation to overcome threshold
21
Q

3 marks

A student had stared at a purple square, he saw a green afterimage.
Suggest why.

A
  1. When staring at purple red sensitive and blue sensitive cones are stimulated / green sensitive cones are not stimulated
  2. Red and blue cone cells become exhausted
  3. Afterimage due to green sensitive cone cells working
22
Q

What causes vision using the fovea to have a high visual acuity

A

Impulses along separate optic nerve from each receptor cell / each receptor cell connects to separate bipolar nerve

23
Q

3 marks

What causes vision using the fovea to be in colour

A

Colour detected by cone cells
Fovea contains mainly cone cells
Three types of cone cells: red/green/ blue light sensitive

24
Q

3 marks

Explain why vision using other parts of the retina has high sensitivity to light.

A

Many rods in other parts of retina
Rhodopsin / rod cells works in low light
Rods connected in groups to bipolar nerve
Summation
If enough light above threshold hits any cells in the group, then get nerve impulses to brain/along optic nerve

25
Q

4 marks

Exercise causes an increase in heart rate.
Describe the role of receptors and of the nervous system in this process

A
  1. Chemoreceptors detect rise in CO2
  2. Send impulses to medulla
  3. More impulses to SAN
  4. By sympathetic
26
Q

4 marks

What would happen if CO2 increased (chemoreceptor action)

A
  1. Increase CO2
  2. More impulses to medulla from carotid and aorta receptors
  3. More impulses along sympathetic
  4. Fewer impulses along parasympathetic
  5. To SAN
  6. Heart rate increases
27
Q

4 marks

What would happen if pressure increased (baroreceptor action)

A
  1. Increase in blood pressure
  2. Two types of barorecptor- one detects too low, one detects too high
  3. More impulses sent to medulla from the ‘too high’ detecting recptor in carotid and aorta
  4. More impulses sent on parasympathetic pathway to SAN
  5. Heart rate decreases
28
Q

6 marks

Explain what causes the heart rate to increase when we exercise

A
  1. Rate of respiration increases in muscle cells
  2. Carbon dioxide concentration increases
  3. In blood plasma
  4. Chemoreceptors
  5. In aorta and carotid
  6. Send impulses to medulla oblongata
  7. Increased frequency of impulses
  8. Along sympathetic pathway
  9. To SAN
29
Q

Where is SAN located in the heart

A

Right atrium

30
Q

6 marks

Explain how a rise in blood pressure results in a decrease in the rate of heartbeat.

A
  1. Increase in blood pressure detected by baroreceptors
  2. Two types of barorecptor- one detects too low, one detects too high
  3. More impulses sent to medulla from the ‘too high’ detecting recptor in carotid and aorta
  4. More impulses sent on parasympathetic pathway to SAN
  5. Decreases impulses from SAN
  6. Decreases impulses from AVN
31
Q

6 marks

Explain how a rise in blood pressure results in a decrease in the rate of heartbeat.

A
  1. Increase in blood pressure detected by baroreceptors
  2. More impulses send to medulla oblongata from carotid and aorta receptors
  3. More impulses along parasympathetic, fewer impulses along sympathetic
  4. to SAN
  5. Decreases impulses from SAN
  6. Decreases impulses to AVN

Biology exam practise (23 decks)

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