Biology Paper 1 flashcards

All 20 cards

1. Card 1

   Trace the full pathway of a deoxygenated red blood cell from the body to the lungs, naming all chambers and valves it passes through.

   Deoxygenated blood from the body enters the right atrium via the vena cava. It then passes through the tricuspid valve into the right ventricle. From the right ventricle, it goes through the pulmonary valve into the pulmonary artery, which carries it to the lungs.

2. Card 2

   Define the primary purpose of a potometer experiment and explain how it indirectly measures the rate of transpiration.

   A potometer measures the rate of water uptake by a plant shoot, which is an indirect measure of transpiration. As the plant transpires, it loses water, creating a tension that pulls water up the stem from the potometer, causing an air bubble in the capillary tube to move. The distance the bubble moves over a set time indicates the rate of water uptake.

3. Card 3

   A researcher wants to investigate the effect of different light intensities on the rate of transpiration in a plant using a potometer. Describe the experimental setup, including the key variable to manipulate and at least two variables to control.

   The researcher should place a plant shoot in a potometer and introduce an air bubble. The key variable to manipulate is light intensity, which can be varied using a lamp placed at different distances or with different wattage bulbs. Controlled variables include temperature (using a water bath), humidity (by avoiding mist or fan), and ensuring the plant has acclimatized before measurements begin. The distance the air bubble moves over a set time is measured for each light intensity.

4. Card 4

   Explain why a potometer measures water uptake and not the actual rate of transpiration.

   A potometer measures water uptake, which is not exactly the same as transpiration. While most water uptake replaces water lost through transpiration, a small amount of the absorbed water is utilized by the plant for processes like photosynthesis. Therefore, the measurement is an indirect estimation rather than a direct measure of transpired water.

5. Card 5

   A group of scientists is designing an experiment to study the effects of dietary vitamin D supplementation on bone density in adolescents living at high latitudes. Outline the critical elements they must include in their experimental design, beyond just administering supplements.

   The experiment must include a control group receiving a placebo. Key variables to standardize across all participants include age, baseline diet, and the amount of time spent outdoors. The study should ideally involve a large sample size and be conducted over a long term. Vitamin D levels in the blood should be measured at regular intervals, and bone density assessments should be performed to evaluate the outcome.

6. Card 6

   Compare the structure and formation of maltose and sucrose, highlighting both their similarities and differences in terms of constituent monosaccharides and glycosidic bonds.

   Both maltose and sucrose are disaccharides formed by a condensation reaction, releasing a water molecule. Maltose is composed of two alpha-glucose molecules linked by a 1-4 glycosidic bond. Sucrose, on the other hand, is formed from one glucose molecule and one fructose molecule, joined by a 1-2 glycosidic bond. Their primary difference lies in their monosaccharide components and the specific carbon atoms involved in the glycosidic linkage.

7. Card 7

   Describe the formation of a peptide bond between two amino acids, specifying the functional groups involved and the type of reaction.

   A peptide bond forms between two amino acids via a condensation reaction. The amine group (NH₂) of one amino acid reacts with the carboxyl group (COOH) of another amino acid. During this reaction, a water molecule (H₂O) is released, and a covalent peptide bond (CO-NH) forms, linking the two amino acids.

8. Card 8

   A protein's tertiary structure is crucial for its function. Explain how hydrogen bonds and ionic bonds contribute to maintaining this 3D shape, and what conditions can disrupt them.

   Hydrogen bonds, though individually weak, are numerous and form between polar R groups, collectively stabilizing the tertiary structure. Ionic bonds are stronger and occur between oppositely charged R groups. High temperatures can break hydrogen bonds and ionic bonds, leading to protein denaturation. Changes in pH specifically alter the charges on R groups, disrupting ionic bonds and also causing denaturation.

9. Card 9

   An experiment uses beetroot discs to study membrane permeability. Explain how temperature affects the permeability of the beetroot cell membrane and how this change can be quantitatively measured.

   At low temperatures, the cell membrane is less fluid, reducing permeability. As temperature increases, phospholipids move faster, making the membrane more fluid and permeable, causing more red pigment (betalain) to leak. At very high temperatures, membrane proteins denature, causing the membrane to fall apart and release a large amount of pigment. The amount of leaked pigment can be quantitatively measured using a colorimeter, which detects the absorbance of light by the red solution; higher absorbance indicates greater membrane permeability.

10. Card 10

    Explain how ethanol affects cell membrane permeability and how this effect can be quantified using a colorimeter in a beetroot experiment.

    Ethanol is a solvent that dissolves the phospholipid bilayer of the cell membrane. As the concentration of ethanol increases, more of the membrane is dissolved, leading to increased permeability and a greater leakage of the red betalain pigment from the beetroot cells. This leakage can be quantified using a colorimeter, where a higher absorbance reading of the surrounding solution indicates more pigment has leaked, thus reflecting greater membrane permeability.

11. Card 11

    A patient is diagnosed with emphysema, a condition where the alveoli merge and are destroyed. Using Fick's Law of Diffusion, explain why this condition leads to breathlessness.

    Fick's Law states that the rate of diffusion is directly proportional to surface area and concentration difference, and inversely proportional to thickness. In emphysema, the merging and destruction of alveoli significantly reduce the total surface area available for gas exchange. This decrease in surface area directly lowers the rate of oxygen diffusion into the blood, leading to less oxygen being transported and consequently causing breathlessness.

12. Card 12

    Explain the role of the concentration difference and membrane thickness in maximizing the rate of diffusion according to Fick's Law, providing a biological example for each.

    A larger concentration difference creates a steeper gradient, leading to a faster rate of diffusion. For instance, in the alveoli, continuous blood flow removes oxygen, maintaining a steep concentration gradient between the alveolar air and the blood. A thinner membrane reduces the distance substances need to travel, increasing diffusion rate. Alveolar walls are only one cell thick, providing a very short diffusion path for gases.

13. Card 13

    Distinguish between simple diffusion and facilitated diffusion, including the types of solutes they transport and whether they require membrane proteins.

    Simple diffusion involves the movement of small, lipid-soluble solutes directly across the cell membrane down their concentration gradient, without the aid of membrane proteins. Facilitated diffusion, however, transports larger or polar solutes down their concentration gradient, but it requires specific membrane proteins (channels or carriers) to assist their passage across the lipid bilayer.

14. Card 14

    Describe the four main stages of DNA replication, ensuring you mention the key enzymes involved and the concept of semi-conservative replication.

    DNA replication begins with unwinding, where helicase breaks hydrogen bonds, separating the double helix into two template strands. Next, free DNA nucleotides align complementarily to each template strand (A with T, C with G). Then, DNA polymerase joins these nucleotides, forming phosphodiester bonds to create new strands in the 5' to 3' direction. Finally, this process results in two identical DNA molecules, each containing one original and one new strand, demonstrating semi-conservative replication.

15. Card 15

    Outline the process of transcription, specifying where it occurs within the cell and the role of RNA polymerase.

    Transcription occurs in the nucleus. It begins with helicase unwinding a segment of DNA. RNA polymerase then moves along the template strand, aligning free RNA nucleotides opposite their complementary bases. RNA polymerase subsequently joins these nucleotides together with phosphodiester bonds, forming an mRNA strand. This mRNA strand, which is a copy of the coding strand (with U instead of T), then leaves the nucleus through a nuclear pore.

16. Card 16

    Explain the process of translation, detailing the roles of mRNA, tRNA, and ribosomes, and where it takes place in the cell.

    Translation occurs at the ribosomes in the cytoplasm. The mRNA strand attaches to a ribosome, which then reads its codons one by one. tRNA molecules, each carrying a specific amino acid, have anticodons that complementarily pair with the mRNA codons. As the ribosome moves along the mRNA, peptide bonds form between the amino acids brought by successive tRNA molecules, building a polypeptide chain until a stop codon is reached, at which point the chain is released.

17. Card 17

    The Meselson-Stahl experiment was pivotal in proving DNA replication is semi-conservative. Describe the key observation made after one generation of bacterial growth in light nitrogen (¹⁴N) following initial growth in heavy nitrogen (¹⁵N), and explain what this implied.

    After one generation of growth in light nitrogen (¹⁴N) following initial growth in heavy nitrogen (¹⁵N), Meselson and Stahl observed that all the extracted DNA settled at a medium density when centrifuged. This single band of intermediate density implied that each new DNA molecule contained one heavy ¹⁵N strand and one light ¹⁴N strand, definitively supporting the semi-conservative model of replication and ruling out conservative replication.

18. Card 18

    A couple is concerned about the inheritance of cystic fibrosis, an autosomal recessive condition (f). The male is a carrier (Ff), and the female is also a carrier (Ff). Draw a genetic diagram to predict the genotypes and phenotypes of their offspring, including the ratios.

    Parents' genotypes: Male = Ff, Female = Ff. Gametes: Male = F, f; Female = F, f. Punnett Square: FF, Ff, Ff, ff. Offspring genotypes: 1 FF : 2 Ff : 1 ff. Offspring phenotypes: 3 Normal (FF, Ff) : 1 Sufferer (ff).

19. Card 19

    A male with normal vision (X^B Y) and a female carrier for colour blindness (X^B X^b) are planning to have children. Construct a genetic diagram to predict the genotypes and phenotypes of their potential offspring, specifying ratios for both males and females.

    Parents' genotypes: Male = X^B Y, Female = X^B X^b. Gametes: Male = X^B, Y; Female = X^B, X^b. Punnett Square: X^B X^B, X^B X^b, X^B Y, X^b Y. Offspring genotypes: 1 X^B X^B : 1 X^B X^b : 1 X^B Y : 1 X^b Y. Offspring phenotypes: 1 Normal female (X^B X^B) : 1 Carrier female (X^B X^b) : 1 Normal male (X^B Y) : 1 Colour blind male (X^b Y). The probability is 50% for normal vision and 50% for colour blindness in sons, and 50% normal and 50% carrier in daughters.

20. Card 20

    Evaluate the advantages and disadvantages of using a large sample size in an experiment designed to study the effect of a new fertilizer on plant growth.

    A large sample size increases the statistical power of the experiment, making the results more reliable and representative of the general population of plants. It also helps to minimize the impact of individual variations or anomalies, increasing the confidence that any observed effects are due to the fertilizer and not random chance. However, a large sample size can be more time-consuming, resource-intensive, and costly to manage, potentially leading to practical limitations in experimental setup and data collection.

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