Pearson Edexcel International A Level Biology: Homeostasis, Kidney, Osmoregulation and Gene Switching (YBI11)

This guide teaches homeostasis, the kidney, osmoregulation and gene switching from scratch for the Pearson Edexcel International Advanced Subsidiary/Advanced Level Biology specification (YBI11), Issue 2. It covers statements 7.16 to 7.22 of Topic 7. Topics 1 to 4 are International AS (IAS) and Topics 5 to 8 are International A2 (IA2), so this material is A2. There are no tiers. Use it with the revision notes and the practice questions. The course hub, printable checklist and free diagnostics show the rest of the course.

What this guide covers

Statement	What you must be able to do
7.16 (i)	Explain negative feedback and positive feedback control
7.16 (ii)	Explain how negative feedback keeps systems within narrow limits
7.17	Define homeostasis, link it to dynamic equilibrium during exercise, and describe the hypothalamus in thermoregulation
7.18	Describe the gross and microscopic structure of the mammalian kidney
7.19	Explain urea production from excess amino acids and its removal by ultrafiltration
7.20	Explain selective reabsorption in the proximal tubule and the loop of Henle as a countercurrent multiplier
7.21	Explain how the pituitary, hypothalamic osmoreceptors and ADH control plasma concentration and blood volume
7.22	Explain gene switching by transcription factors, with peptide hormones acting extracellularly and steroid hormones intracellularly

7.16 Negative and positive feedback

A control system has a receptor, a coordinator and an effector. In negative feedback, a change away from the set point is detected and the effector response reverses it. The output switches the stimulus off, so the variable stays within narrow limits and oscillates around the set point.

In positive feedback, a change is detected and the response makes the change larger. The system moves away from its starting point until something outside the loop stops it. During childbirth, stretching of the cervix triggers release of oxytocin, which strengthens contractions, which stretch the cervix further. The loop ends when the baby is born. The rising phase of an action potential is another example, because depolarisation opens more sodium ion channels. Positive feedback does not maintain a steady state, so it is not used for homeostasis.

7.17 Homeostasis and thermoregulation

Homeostasis is the maintenance of a stable internal environment, such as core temperature, blood glucose concentration, blood pH and plasma water potential, despite changes outside and inside the body. The state is a dynamic equilibrium: the values are always moving a little either side of the set point, and corrective responses are always being switched on and off.

During exercise, muscle cells respire faster. They release more heat and carbon dioxide and use more glucose and oxygen. Homeostatic mechanisms stop these changes becoming large enough to disrupt enzyme activity. Core temperature may rise by a degree or so and then be held. Without sweating and skin blood flow it would keep climbing.

Thermoregulation is coordinated by the hypothalamus. It contains central thermoreceptors that monitor the temperature of blood flowing through it. It also receives impulses from thermoreceptors in the skin. It acts as the coordinator, and its output goes to effectors.

Condition	Response	Effect
Too warm	Vasodilation of arterioles supplying skin capillaries	More blood near the surface, more heat lost by radiation
Too warm	Sweat glands secrete sweat	Evaporation takes latent heat from the skin
Too cold	Vasoconstriction of skin arterioles	Less blood near the surface, less heat lost
Too cold	Shivering	Rapid muscle contraction releases heat from respiration
Too cold	Raised metabolic rate	More heat released in liver and muscle

The responses return the temperature to the set point, and the receptors then stop stimulating them. This is negative feedback.

Worked example 1. During a 30 minute run, core temperature rises from 37.0 °C to 38.4 °C. The runner loses 1.2 kg of sweat per hour, and evaporating 1 g of sweat takes about 2.4 kJ. Calculate the mean rate of temperature rise and the heat removed by evaporation each hour.

Rate = (38.4 − 37.0) ÷ 30 = 1.4 ÷ 30 = 0.047 °C per minute. Heat removed = 1200 g × 2.4 kJ g⁻¹ = 2880 kJ = 2.9 MJ per hour, about 800 W. Sweat only cools the body if it evaporates.

7.18 Structure of the mammalian kidney

Gross structure. Blood enters through the renal artery and leaves through the renal vein. Urine leaves through the ureter. A section shows an outer cortex, an inner medulla containing pyramids, and a central cavity, the renal pelvis, which collects urine and drains into the ureter. A tough fibrous capsule surrounds the whole organ.

Microscopic structure. The functional unit is the nephron. In order, it contains:

1. Bowman’s capsule around a knot of capillaries, the glomerulus, both in the cortex.
2. The proximal convoluted tubule, in the cortex, lined by cuboidal cells with microvilli and many mitochondria.
3. The loop of Henle, descending into the medulla and returning to the cortex.
4. The distal convoluted tubule, in the cortex.
5. The collecting duct, which runs through the medulla to the renal pelvis and is shared by many nephrons.

The afferent arteriole brings blood to the glomerulus and is wider than the efferent arteriole, which takes blood away. The efferent arteriole leads to capillaries around the tubules, including the vasa recta alongside the loop of Henle.

Worked example 2. A micrograph of a glomerulus is 45 mm wide and the scale shows it is really 150 µm across. Magnification = image size ÷ actual size = 45 mm ÷ 0.150 mm = ×300. Keep both lengths in the same unit.

7.19 Urea and ultrafiltration

Amino acids in excess of need cannot be stored. In the liver, deamination removes the amino group. It is converted to ammonia, which is toxic, and the ammonia is combined with carbon dioxide to make urea. The ornithine cycle details are not required. The remaining carbon skeleton can be respired or converted to carbohydrate or fat. Urea is soluble and much less toxic, so it travels in blood plasma to the kidneys.

Ultrafiltration occurs in the glomerulus. Because the afferent arteriole is wider than the efferent one, hydrostatic pressure in the glomerular capillaries is high. Fluid is forced out through three layers: the capillary endothelium, which has gaps (fenestrations); the basement membrane, a mesh of glycoproteins; and the podocytes of Bowman’s capsule, which leave slits. Water, urea, glucose, amino acids and ions pass into the capsule. Blood cells and most plasma proteins, such as albumin, are too large and stay in the blood. Filtration is by size, so useful solutes such as glucose are filtered and must be reabsorbed.

Worked example 3. The values below are supplied as data. Hydrostatic pressure in the glomerular capillary is 7.5 kPa, hydrostatic pressure in the capsule is 1.5 kPa and the plasma proteins produce an opposing pressure of 3.5 kPa. Net filtration pressure = 7.5 − 1.5 − 3.5 = 2.5 kPa, so fluid moves out. Protein in urine therefore suggests the filter is damaged.

Worked example 4. The filtrate forms at 118 cm³ min⁻¹ and urine is produced at 0.9 cm³ min⁻¹. Reabsorbed = 118 − 0.9 = 117.1 cm³ min⁻¹. Percentage = 117.1 ÷ 118 × 100 = 99.2%. Filtrate per day = 118 × 60 × 24 = about 170 dm³, and urine about 1.3 dm³.

7.20 Reabsorption and the countercurrent multiplier

Proximal tubule. Most reabsorption happens here, about two-thirds of the filtrate. The cells have microvilli for a large surface area and many mitochondria for ATP. Sodium–potassium pumps in the basal membrane move sodium ions out of the cell into the blood, which lowers sodium concentration inside the cell. Sodium ions then enter from the filtrate by co-transporter proteins that also carry glucose or amino acids, so these are absorbed against their concentration gradient. They leave into the blood by facilitated diffusion. Water follows by osmosis. Glucose is normally all reabsorbed, so urine has none.

Loop of Henle. The loop makes the medulla very concentrated, so water can be reabsorbed later.

1. The ascending limb is impermeable to water. Ions (sodium and chloride) are pumped out of the filtrate, by active transport in the thick segment, into the medulla tissue fluid. This lowers the water potential of the medulla.
2. The descending limb is permeable to water. Water leaves by osmosis into the concentrated medulla, so the filtrate becomes more concentrated towards the bend.
3. Fluid in the two limbs flows in opposite directions, so at each level the ascending limb raises the concentration around the descending limb. A small difference at each level is multiplied into a steep gradient from cortex to medulla. This is the countercurrent multiplier.
4. The vasa recta carry away the water and reabsorbed solutes without removing the gradient.
5. The collecting duct passes back down through the concentrated medulla, so water can leave it by osmosis.

Animals with long loops, such as desert rodents, make very concentrated urine.

Worked example 5. Plasma glucose is 5.0 mmol dm⁻³ and filtrate forms at 0.118 dm³ min⁻¹, so glucose filtered = 5.0 × 0.118 = 0.59 mmol min⁻¹. Urine contains none, so all 0.59 mmol is reabsorbed each minute. If plasma urea is 5.0 mmol dm⁻³ and urine urea is 300 mmol dm⁻³, urea is 60 times more concentrated in urine because water is reabsorbed more than urea.

7.21 ADH and osmoregulation

Osmoreceptors in the hypothalamus detect the water potential of plasma. If plasma becomes more concentrated, osmoreceptor cells lose water and shrink, and this stimulates neurosecretory cells in the hypothalamus. They make antidiuretic hormone (ADH), which the posterior pituitary gland secretes into the blood.

ADH binds to receptors on collecting duct cells (and the distal tubule). Vesicles carrying aquaporins fuse with the membrane, so the walls become more permeable to water. More water leaves the filtrate by osmosis into the concentrated medulla, giving a small volume of concentrated urine. The plasma becomes more dilute, blood volume is restored and the osmoreceptors stop signalling. If plasma is too dilute, less ADH is released, aquaporins are removed from the membrane and a large volume of dilute urine is produced. This is negative feedback on both plasma concentration and blood volume.

Worked example 6. After a hot day, plasma concentration rises from 285 to 295 mOsm kg⁻¹ and urine flow falls from 1.9 to 0.6 cm³ min⁻¹. Percentage change in urine flow = (0.6 − 1.9) ÷ 1.9 × 100 = −68.4%. Plasma concentration rose only 3.5%, yet urine flow fell by almost 70%.

7.22 Switching genes on and off

A transcription factor is a protein that binds to a specific DNA sequence near a gene, often in its promoter region. An activator helps RNA polymerase bind and start transcription, so the gene is switched on. A repressor blocks binding, so the gene is switched off. Cells of one organism hold the same genes but make different proteins because different transcription factors are active.

Steroid hormones (such as oestrogen) are lipid-soluble. They diffuse through the phospholipid bilayer and bind a receptor protein inside the cell. The hormone–receptor complex enters the nucleus, acts as a transcription factor, binds the DNA and switches on a gene. Transcription then gives mRNA, which is translated to a protein.

Peptide hormones (such as insulin and ADH) are water-soluble and cannot cross the membrane. They bind receptors on the cell surface membrane, acting extracellularly. This triggers a chain of reactions involving a second messenger such as cyclic AMP, which activates transcription factors in the cell and so changes which genes are expressed.

Common errors

• Saying negative feedback “stops” a change instead of reversing it.
• Writing that the hypothalamus is the effector in thermoregulation.
• Saying the loop of Henle reabsorbs water in both limbs.
• Stating that ADH makes the kidney “make” more water.
• Saying filtration is selective for useful solutes.
• Writing that steroid hormones bind to cell surface receptors.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level Biology (YBI11) specification, Issue 2, published by Pearson Education Limited.