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Hardest BMAT Section 1 Questions and Answers

Advice & Insight From BMAT Specialists

The second section of the BMAT can be challenging for candidates, especially if they have not taken all subjects to A Level. Here, we present 10 difficult questions from our question bank. We’ve removed the multiple choice options to make these questions even more difficult – all questions in our bank include four options to choose from. 

Biology

Explain the process of alternative splicing and its significance in gene expression regulation.

Alternative splicing is a post-transcriptional process in eukaryotic cells where exons and introns of pre-mRNA are differently arranged, leading to multiple mRNA transcripts from a single gene. This process allows a single gene to code for various protein isoforms, enhancing the complexity of the proteome. Different combinations of exons can be included or excluded, resulting in distinct functional proteins. This enables cells to perform diverse functions despite having a limited number of genes. The regulation of alternative splicing is critical for tissue-specific expression and development.

What is the role of the sodium-potassium pump in maintaining the resting membrane potential of a neuron?

The sodium-potassium pump is a critical cellular mechanism found in the plasma membrane of neurons and other cells. Its primary role is to maintain the resting membrane potential, which is the electrical charge difference across the cell membrane when the cell is at rest. This pump actively transports three sodium ions out of the cell for every two potassium ions it brings into the cell. This results in a net loss of positive charges from the inside of the cell, creating a negative resting potential.

Compare and contrast the structure and function of smooth endoplasmic reticulum and rough endoplasmic reticulum in eukaryotic cells.

Smooth endoplasmic reticulum (SER) and rough endoplasmic reticulum (RER) are both essential components of eukaryotic cells, but they differ in structure and function. RER is studded with ribosomes on its cytosolic surface, giving it a “rough” appearance. It plays a key role in protein synthesis, particularly for proteins targeted to be exported from the cell or localised in the cell membrane. In contrast, SER lacks ribosomes and is involved in lipid metabolism, detoxification, and calcium ion storage. It synthesises lipids, metabolises carbohydrates, and detoxifies drugs and poisons. Additionally, SER plays a role in muscle cell contraction by storing and releasing calcium ions.

 
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Chemistry

Explain the principles behind chromatography and how it can be used to separate and identify different components in a mixture.

Chromatography is a versatile separation technique used to separate and identify different components in a mixture based on their differing affinities for a stationary phase and a mobile phase. The stationary phase could be a solid or a liquid fixed on a solid support, while the mobile phase is a liquid or gas that carries the mixture through the stationary phase. As the mixture travels through the stationary phase, the components with higher affinity for the mobile phase move faster, leading to their separation from the rest of the mixture. The separated components form distinct bands or spots, which can be analyzed to identify the substances present. Chromatography finds applications in various fields, from forensics and pharmaceuticals to environmental analysis, providing valuable insights into complex mixtures.

Discuss the greenhouse effect and its impact on global climate change. What are some measures that can be taken to reduce greenhouse gas emissions?

The greenhouse effect is a natural phenomenon essential for maintaining Earth’s temperature. It involves the trapping of some of the Sun’s energy by greenhouse gases (GHGs) in the atmosphere, such as carbon dioxide, methane, and water vapor. However, human activities, particularly the burning of fossil fuels and deforestation, have significantly increased GHG concentrations, leading to enhanced greenhouse effect and global warming. This contributes to climate change, causing more frequent and severe heatwaves, extreme weather events, rising sea levels, and disruptions to ecosystems.

To reduce greenhouse gas emissions, measures must be taken at individual, community, and governmental levels. These include transitioning to renewable energy sources, improving energy efficiency, promoting sustainable transportation, adopting greener agricultural practices, and implementing international agreements and policies to limit emissions.

What are acids and bases? Provide examples of strong acids and bases and explain their behaviour in aqueous solutions.

Acids are substances that release hydrogen ions (H⁺) in aqueous solutions, while bases are substances that release hydroxide ions (OH⁻). Strong acids and bases completely dissociate in water, meaning they release a high concentration of H⁺ or OH⁻ ions. An example of a strong acid is hydrochloric acid (HCl), which dissociates into H⁺ and chloride ions (Cl⁻) in water. On the other hand, sodium hydroxide (NaOH) is a strong base, dissociating into Na⁺ and OH⁻ ions.

In aqueous solutions, strong acids aggressively donate H⁺ ions, leading to a low pH (highly acidic). Strong bases are proactive acceptors of H⁺ ions, resulting in a high pH (highly basic). It’s important to note that strong acids and bases can be highly corrosive and hazardous, requiring careful handling and dilution to safe levels for various applications.

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Physics & Mathematics

A triangle has side lengths of 8 cm, 12 cm, and 15 cm. Determine whether the triangle is a right-angled triangle using the Pythagorean theorem and explain your reasoning.

To determine if the triangle with side lengths 8 cm, 12 cm, and 15 cm is a right-angled triangle, we can use the Pythagorean theorem. The theorem states that in a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. In this case, 15^2 = 8^2 + 12^2. After evaluating both sides of the equation, we find that 225 ≠ 208, meaning the triangle is not a right-angled triangle.

A car is traveling at a constant speed of 30 m/s. Suddenly, the driver slams the brakes, and the car comes to a stop in 5 seconds. Calculate the car's acceleration and the distance it traveled during the braking.

The car’s acceleration can be calculated using the equation a = (v_f – v_i) / t, where v_f is the final velocity (0 m/s, as the car comes to a stop), v_i is the initial velocity (30 m/s), and t is the time taken (5 seconds). Plugging the values into the equation, we get a = (0 – 30) / 5 = -6 m/s². The negative sign indicates that the acceleration is in the opposite direction to the initial motion.

To find the distance traveled during braking, we can use the equation d = v_i * t + 0.5 * a * t^2. Substituting the values, d = 30 * 5 + 0.5 * (-6) * 5^2 = 75 – 75 = 0 meters. Therefore, the car traveled 0 meters during braking.

Two point charges, q1 = +2 μC and q2 = -3 μC, are placed 10 cm apart. Calculate the electric potential energy of the system.

The electric potential energy (EPE) of two point charges is given by the equation EPE = k * (q1 * q2) / r, where k is Coulomb’s constant (8.99 × 10^9 N·m²/C²), q1 and q2 are the magnitudes of the point charges (2 μC and 3 μC), and r is the distance between the charges (10 cm = 0.1 m). Substituting the values, EPE = 8.99 × 10^9 * (2 × 10^-6 * -3 × 10^-6) / 0.1 = -5394 J. The negative sign indicates that the charges have opposite signs and their potential energy is attractive.

An object is thrown vertically upwards with an initial velocity of 20 m/s. Calculate the maximum height it reaches and the time it takes to reach the highest point. Also, determine the time it takes for the object to return to the ground.

To determine the maximum height reached by the object, we can use the kinematic equation vf^2 = vi^2 + 2aΔy, where vf is the final velocity (0 m/s at the maximum height), vi is the initial velocity (20 m/s), a is the acceleration due to gravity (-9.81 m/s²), and Δy is the change in height (maximum height – initial height = H – 0 = H). Solving for H, we get H = (vf^2 – vi^2) / (2a) = (0 – 20^2) / (2 * -9.81) ≈ 20.4 meters.

To calculate the time it takes to reach the highest point, we can use the kinematic equation vf = vi + at, where vf is 0 m/s, vi is 20 m/s, a is -9.81 m/s², and t is the time. Solving for t, we get t = (vf – vi) / a = (0 – 20) / -9.81 ≈ 2.04 seconds.

Lastly, to find the time it takes for the object to return to the ground, we double the time it took to reach the highest point. So, the total time of flight is approximately 2.04 * 2 = 4.08 seconds.

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