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Biology A-level > Mass transport in animals > Flashcards

Mass transport in animals Flashcards

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1
Q

Definition of double circulatory system

A

Mammals have a closed (blood in vessels) double circulatory system (blood passes through heart twice for each circuit of the body).

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2
Q

Definition of pulmonary circulation

A

Carries deoxygenated blood to the lungs to take up oxygen and then return it to the heart.

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3
Q

Definition of systematic circulation

A

Carries oxygenated blood to the rest of the body to release oxygen and return to the heart.

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4
Q

What is the main advantage of a double circulatory system for mammals?

A

Increases the blood pressure, therefore increases the rate of blood flow to the tissues.

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5
Q

Explain the role of coronary arteries, and why a blockage in one of these causes a heart attack (myocardial infarction).

A

Coronary arteries carry oxygen to the heart muscle. A blockage, such as: blood clot or fatty acid build-up, can restrict the flow of blood/oxygen to the heart muscle. If the clot blocks the coronary artery, no oxygen is reaching the heart. The heart muscle can no longer respire so muscle cells may die resulting in a heart attack.

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6
Q

Definition of systole

A

Contraction of the heart muscle.

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7
Q

Definition of diastole

A

Relaxation of the heart muscle.

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8
Q

Describe diastole

A

Muscle is relaxed in atria walls.
Pressure increases in atria (blood is moving into atria from veins).
Valves: bicuspid and tricuspid valves open slightly.
Direction: blood starts to enter ventricles.

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9
Q

Describe atrial systole

A

Muscles in atrial walls contract.
Pressure in atria increases.
Valves: Atrio-ventricular valves are opening more as the pressure in atria is higher than in ventricles.
Direction: bloof flows into ventricles.

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10
Q

Describe ventricular diastole

A

Muscles in ventricle walls contract.
Pressure in ventricles increases.
Valves: atrio-ventricular valves are forced to close when the pressure in ventricles is higher than in the atria.
Direction: valves prevent blood flowing back into atria.
Pressure in ventricles is also greater than in arteries.
Valves: semi-lunar valves are therefore forced open.
Direction: blood flows out of ventricles into arteries.

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11
Q

Describe diastole

A

Muscle relaxes.
Pressure in all chambers decreases.
Pressure in arteries is now greater than in ventricles.
Valves: semi-lunar valves are therefore forced to close.
Direction: valves prevent blood flowing back into the ventricles.
Direction: blood enters left and right atria from veins.

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12
Q

What is the equation for heart rate?

A

Heart rate = 60 / (time taken for one heart beat)
1 cardiac cycle = 1 heart beat *

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13
Q

What is the equation to calculate stroke volume?

A

Stroke volume = cardiac output / heart rate

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14
Q

Definition of stroke volume

A

Volume of blood pumped by the left ventricle each time it contracts.

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15
Q

Definition of cardiac output

A

Total volume of blood pumped by the left ventricle in 1 minute.

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16
Q

Why does training lower the resting heart rate of an athlete?

A

Size of the heart/ventricles increase.
Muscle size in the walls increase.
Strength of contraction increase, which increases pressure.
Stroke volume increases.
Cardiac output is the same.

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17
Q

Describe the role of capillaries

A

Tiny vessels that link arterioles to veins; site of exchange between blood and cells.

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18
Q

Describe the role of arteries.

A

Carry blood away from the heart at high pressure, and into arterioles, blood travels in pulses.

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19
Q

Describe the role of veins.

A

Carry blood, at low pressure, from capillaries int the heart.

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20
Q

Describe the role of arterioles.

A

Smaller arteries that control blood flow from arteries to capillaries.

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21
Q

Describe the structure of an artery.

A

Thicker, muscular layer.
Thicker, elastic tissue layer.
Thinner wall overall.
Lumen is relatively narrower.

22
Q

Describe the structure of veins.

A

Tinner, muscular layer.
Thinner, elastic tissue layer.
Thinner wall overall.
Lumen is relatively wider.

23
Q

Describe the structure of capillaries.

A

Onlt 1 cell-thick.
Made of flattened endothelial cells.
Spaces between the endothelial cells allow white blood cells to escape to get to site of infection.

24
Q

How is the structure of the aorta adapted to its function?

A

Elastic tissue allows stretch and recoil to maintain blood pressure.
Elastic tissue stretched when ventricles contract.
Muscle allows contraction, vasoconstriction.
Thick walls stop bursting.
Smooth endothelium reduces friction.
Semi-lunar valves prevent backflow.

25
Q

How is tissue fluid formed at the arteriole end?

A

There is a high hydrostatic pressure in the capillaries created by contraction of the ventricles.
This is greater than the hydrostatic pressure of tissue fluid outside the capilaries.
Therefore, water to move out of the capillaries and into the spaces around cells, forming the tissue fluid.
Small molecules are also forced out.
Cells and the larger plasma proteins are too large to cross the membrane = ultrafiltration.

26
Q

How is tissue fluid formed at the venule end?

A

The loss of water from the capillaries recuces the hydrosatic pressure inside.
So, the hydrostatic pressure is now lower than the tissue fluid outside.
The water potential is lower inside the capillary than outside in the tissue fluid due to the water loss and the larger proteins still in the capillary.
Therefore, water re-eters the capillaries from the tissue fluid by osmosis down a water potential gradient.

27
Q

Describe the structure of haemoglobin.

A

A globular protein with a quaternary structure. Each polypeptide is associated with a haem group, which contains an Fe^2+ ion.

28
Q

Describe how oxygen binds to haemoglobin in humans.

A

Each haem group combines temporarily with 1 oxygen molecule (2 atoms), therefore 4 oxygen molecules are carried by a single haemoglobin molecule in humans.

29
Q

What is the role of haemoglobin?

A
  1. To load oxygen in a high CO2 concentration, where it has a high affinity for oxygen, to form oxyhaemoglobin.
  2. This occurs in the lungs.
  3. Unload oxygen in a low pO2, where it is has a low affinity for oxygen.
  4. This occurs in the repsiring cells/tissues.
30
Q

Why is the curve sigmoid shape?

A

Describe- Bigger increase in pO2 smaller increases in % saturation (quote vlaues from the graph).
Explain- The shape of the haemoglobin molecules makes it difficult for the 1st oxygen molecule to bind.
Describe- Smaller increase in pO2 bigger increases in % saturation (quote values from the graph). Steeper gradient.
Explain- Binding of the 1st oxygen changes the quaternary structure shape of haemoglobin molecule which makes it much easier for other oxygen molecules to bind.
Describe- Bigger increase in pO2 smaller increases in % saturation (quote vlaues from the graph). % saturation flattens out. Explain- Binding of the 4th is harder because it is less likely that a single oxygen molecule will find an empty binding site.

31
Q

What is the relevance of the curve being sigmoid in the lungs?

A

Blood leaving the lungs is almost always completely saturated with oxygen. Even if pO2 in the lungs drops a little below normal, haemoglobin can still pick up a lot of oxygen.

32
Q

What is the relevance of the curve being sigmoid in the body tissues, where respiring cells are using up oxygen?

A

A slight decrease in pO2 of tissues produces a large increase in oxygen release from haemoglobin. O2 is effectively delivered to the tissues that need it.

33
Q

Explain the effect of a higher pCO2 on the ability of haemoglobin to unload oxygen to respiring cells in the body.

A

The higher the pCO2, the more readily the oxygen is unloaded from haemoglobin.
So more oxygen is releases for cells during activity.
So the disassociation curve shifts to the right.
Haemoglobin has a lower affinity for O2.

34
Q

What causes the Bohr shift when a higher pCO2?

A

In the tissues, CO2 is released by respiring cells.
Forms carbonic acid which dissociates into H+, lowering the pH.
Acidity changes the shape of haemoglobin, so now has a lower affinity for oxygen.
O2 is more readily released to respiring tissues.

35
Q

Why are there different versions of haemoglobin?

A

The haemoglobin in different organisms have different affinities for oxygen.
Due to different shapes of the haemoglobin molecule.
Due to amino acid sequences being slightly different.
Resulting in slightly different tertiary and quaternary structures.
Resulting in different binding properties.
Therefore, affinities of different haemoglobin molecules ranges from higher to lower affinity for oxygen.

36
Q

Explain the Bohr shift in terms of a low oxygen environment, low pO2.

A

pO2 is lower for organisms that live in low oxygen environments.
So, the haemoglobin has a higher affinity for O2.
So, it can load more of the available oxygen.
Be saturated at a lower pO2.
Dissociation curve is shifted to the left.

37
Q

Describe the Bohr shift for a foetus.

A

In the uterus, there is a low pO2.
The haemoglobin of the foetus has a higher affinity for O2 than the mother in the same pO2.
Enables oxygen to move from mother to foetus.
Dissociation curve is shifted to the left.

38
Q

Describe the Bohr shift for a lugworm.

A

Live in burrows beneath the sand- low pO2.
Haemoglobin of a lugworm has a high affinity for O2.
Dissociation curve is shifted left.

39
Q

Describe the Bohr shift for a llama.

A

Lives at high altitudes, low atmospheric pO2.
So has to be able to load more oxygen, therefore a llama’s haemoglobin has a high affinty for O2. Dissociation curves shifts to the left.

40
Q

State an advantage of haemoglobin having a higher affinity for O2 in a low pO2.

A

Haemoglobin can be fully saturated with O2 in a low pO2.

41
Q

State a disadvantage of haemoglobin having a high affinity for O2 in a low pO2.

A

Does not readily unload O2 tomtissues, so limits repsiration rates and prevents organism being highly active.

42
Q

Describe the Bohr shift in terms of animals having different activity levels.

A

Organisms that are very active and small, have a higher O2 demand.
They need their haemoglobin to unload oxygen more readily.
To supply more oxygen to tissues.
For faster repsiration.
Therefore, have haemoglobin with a lower affinity.
In all these cases, the dissociation curve is shifted to the right.

43
Q

Describe a Bohr shift for a hawk.

A

A hawk has a very high respiratory rate and lives where there is plenty of oxygen.
So, they have a higher metabolic rate to keep warm.
Therefore, have a higher oxygen demand.
So, need the haemoglobin to unload oxygen more easily.
Dissociation curve shifts to the right.

44
Q

Explain the Bohr shift in terms of small animals.

A

Small animals have a higher SA:VOL ratio.
Therefore, they lose heat more quickly.
So, they have a higher metabolic rate to keep warm.
Therefore, have a higher oxygen demand.
They need the haemoglobin to unload oxygen more easily. Dissociation curve shifts to the right the smaller the animal.

45
Q

State an advantage of the haemoglobin of small animals that has low affinity for O2.

A

O2 is more readily unloaded to tissues; allows for high respiration rate, so organism can be more active.

46
Q

State a disadvantage of the haemoglobin of small animals that has a low affinity for O2.

A

Requires a high pO2 in atmosphere to fully saturate haemoglobin. Organisms can become unconscious quickly at high altitudes.

47
Q

Explain the Bohr shift in terms of large animals.

A

The larger the animal, the more left the curve will shift.
The curve will also be steeper as haemoglobin has a higher affinity for O2.

48
Q

Why does the haemoglobin of small animals have a lower affinity for oxygen?

A

SA:VOL ratio is larger.
Heat loss is quicker.
Rate of respiration is higher.
Requirement of O2 for respiration is higher.
Haemoglobin has a lower affinity for oxygen / it releases O2 more readily.

49
Q

Define cardiovascular disease

A

Covers a range of different problems that can develop int he blood vessesls and the heart, usually when the vessels become blocked.

50
Q

Give 10 examples of risk factors that could cause cardiovascular disease.

A
  1. High fat diet.
  2. High levels of LDL cholesterol.
  3. Obesity.
  4. Smoking.
  5. Diabetes.
  6. High-salt diet.
  7. Age.
  8. Genetics.
  9. Level of exercise.
  10. Gender

Biology A-level (27 decks)

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