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© UCLES 2015 BioMedical Admissions Test Section 2 solutions : Specimen

2 © UCLES 2015 [Turn over 1 Process 1 is nitrogen fixation which is the process of turning nitrogen gas from the air into nitrogen compounds in the soil. Process 2 is denitrification which is the release of nitrogen back into the air. Process 3 is the breakdown of nitrogen -rich dead plant material such as plant protein. There fore the correct answer is H. 2 Addition polymerisation takes place between unsaturated molecules, such as those containing the C = C double bond ( alkenes ) which can join with another C = C bond to form a C – C linkage in a polymer. Fully saturated molecule s, such as alkanes, cannot undergo addition polymerisation. Alkenes have the general formula C nH2n whilst alkanes have the general formula C nH2n+2 . Both formulae can have other atoms in them replacing one or more H atoms e.g. C l, Br, I. In this question co mpound 1 is fully saturated, it is an alkane; there are no double (or triple) bonds present. Compound 2 is an alkene as is compound 4 (with 2 C l atoms). Compound 3 matches the general formula of an alkane (with 1 Br atom) and therefore is fully saturated. Compound 5 matches the general formula of an alkene (with 4 Cl atoms). Therefore the correct answer is E (2, 4 and 5 only) 3 Statement 1: Aluminium is a goo d conductor of heat and would increase, not decrease, the r ate of heat loss by conduction. Statement 1 is therefore not correct. Statement 2: Th e aluminium sheet does trap air and the fact that air is trapped re duces heat loss by convection. Statement 2 is therefore correct. Statement 3: Aluminium is shiny and therefore a poor emitter of therma l radiation. It therefore does reduce heat loss by radiation. Statement 3 is correct. The correct option is that statements 2 and 3 only are correct. Therefore the correct answer is F. 4 The area of a triangle is given by The height is the vertical height which is given as √ . In this case the area is . Because then this simplifies to . Therefore the correct answer is C 5 The SAN pacemaker is found within the wall of the right atrium which leaves the answers of B or F. Whilst the left side of the heart pumps oxygenated blood to the body, the right pumps d eoxygenated blood to the lungs. T herefore the correct answer F. . 2 1 height base      2 2 2 2 2 2 4 8 2 2 4 2 2       2 2 2  2 3 2 2 2 6   

3 © UCLES 2015 [Turn over 6 Thi s type of question relies on studying the equation and the numbers of atoms on both sides of the equation. In this question the Na atoms give the starting point as on the right there are c × 5 and on the left are a × 1 and b × 2. The first effort would be would be to make c = 1, a = 1 and b = 2 in order to balance the Na atoms and then use them to find the value of d. Using the original values ( c, a and b ) the P atoms are also balance d so it is then a matter of adding u p the O and H atoms on the left. This gives d = 2 as a value to balance them. Therefore the correct answer is D (1 , 2 , 1 , 2) 7 The mass of the two cars is the same which means that kinetic energy is proportional to speed squared. Therefore, as car Q has twice the speed of car P, it will hav e fou r times the kinetic energy. Gravitational potential energy is proportional to height, and so the car which rises 50 m will have double the potential energy of the car that has only risen 25 m. The correct answer is therefore that car Q will have twice as much potential energy but four times a s much kinetic energy as car P. Therefore the correct answer is C. 8 In the expression every term inside the bracket has to be squared. The easiest way is to consider each term separately. . Remember that when raising one power to another, you multiply them, so : . In the same way we h ave to square √ which leaves us with z. The square of is . The expression simplifies to . Therefore the correct answer is B, . None of the other expressions are equivalent to this. 9 As the genotypes of P and Q are given, the genotype of S will be a combination of one allele from each of P and Q which makes S heterozygous. There is a 50% chance that U will inherit the recessive allele from S. If T is homozygous recessive then individu al U will have a 100% chance of inheriting a recessive allele from T. So the chance of individual U, inheriting both recessive alleles is 50% (answers A and C). However, if T is heterozygous, then individual U will have a 50% chance of inheriting the rece ssive allele from T. Therefore the chance of individual U inheriting both recessive alleles from 2 heterozygous parents will be 25%. Therefore the correct answer is C. 4 22 3 2 2 3 x x                   2 23 2 2 3 x x 3y 6y z y x 6 3 4 z y x 6 3 4

4 © UCLES 2015 [Turn over 10 Bond breaking is an endothermic process whilst bond making is an exothermic process. The equation is exothermic overall, which means that more energy is released when bonds are made than is needed for bonds being broken. The bonds to be broken are on the left (N 2 and 3 × H2) and made on the right (2 × 3 × N – H) of the equation. This means that the 6 × N – H bond energies must be greater than the N 2 and 3 × H2 bond energies in total. Looking at the number of bonds from the mole ratios in the equation (1 : 2 : 3) the correct answer is D: . 11 When travelling at terminal velocity, the force of air resistance (drag) on the parachutist will be eq ual and opposite to his weight. This is true when travelling at a high terminal velocity before opening the parachute and when travelling at a lower terminal velocity after opening the parachute. Terminal velocity means zero acceleration which means zero resultant force. During the act of opening the parachute there is a rapid deceleration which means that momentarily there must be a resultant upwards f orce acting on the parachutist. During that instant there must therefore be air resistance which is much greater than the weight. The correct graph is therefore one that shows the same air resistanc e force, in the same direction throughout , apart from during the mo ment of o pening the parachute. And during that moment there must be a much larger air resistance force, ag ain all in the same direction. The only graph that shows all this is graph A . B and D are incorrect because they show the air resistance force changing direction, which it cannot do. It will always be upwards. C and D are incorrect because they show different air resistance forces at terminal velocity before and after opening the parachute. Therefore the correct answer is A. 12 The probability of an event , if all events are equally likely, is the number of ways for something to happen divided by the total number of possible results . The balls are identical except for colour, and are well mixed, so picking any of the colours is equally likely. The probabil ity that the first ball taken out was red is . The chosen ball is now replaced, irrespective of colour. There are still balls in the bag. The probability that the s econd ball taken out was blue is . Therefore the correct answer is E. 13 The transfer between one neuron and the next is via a transmitter substance that diffuses across the synaptic gap hence statement 4 is correct. The arrival of the signal at the end of the neuron next to the synapse stimulates the release of the transmitter, he nce statement 3 is also correct. Therefore the correct answer is E. y x z 3 6   z y x x   z y x   z y x y  

5 © UCLES 2015 [Turn over 14 From the information provided it is clear that the compo und is ionically bonded. Remember that the overall charge of any compound is zero, so t he total sum of the charges of the individual ions in the compounds A – F must equal 0 if the compound exists . The method is to work out for each compound the amount of positive ( +) charges (Mg and H) and the amount of negative ( –) charge s (PO 4). Only Mg(H 2PO 4)2 correctly balances the + and – charges. Therefore the correct answer is B: Mg(H 2PO 4)2 15 Source X: 24 hours is 5 half -lives. So in that time the activity of source X will halve 5 times. 320 to 160 to 80 to 40 to 20 to 10. So after 24 hours the activity of X will have fallen to 10 counts per minute. Source Y: 8 hours is 3 half -lives. So in that time the activity of Y will halve 3 times. 480 to 240 to 120 to 60. So after 24 hours the activity of Y will have fallen to 60 counts per minute. The combined count rate after 24 hours is therefore 10 + 60 = 70 count s per minute. Therefore the correct answer is D. 16 To rearrange a formula the ‘subject’ has to be isolated. In the formula the term has to be isolated. There are several different ways of doing this but these all include the same steps. In the first instance add to both sides of the equation. This gives . Then subtract r from both sides of the equation. This gives . Now divide b oth sides of the equa tion by 6 to give . Finally multiply both sides of the equation by to give . This is not exactly the same as any of the given answers but and so the answer arrived at is equivalent to response E. The refore the correct answer is E: .  1 6 1 2     n n d r  2d   2 1 6   n n d   1 1 6 2     n n d r   r nn d     1 1 6 2 2   6 1 1 2 2 r nn d     )1 ( 2 n n   1 6 1 2 2     n n r d   n n nn    3 2 1  n n r d     3 2 6 ) 1(

6 © UCLES 2015 [Turn over 17 During breathing out, the diaphragm muscles relax so statement 2 is incorrect. The ribcage moves down and in during breathing out, which is statement 1. This causes a decrease in the volume of the thorax leading to an increase in pressure within the lungs, which is statement 3. As 1 and 3 are right , the correct answer is E. 18 Different methods can be used to calculate the answer and one is shown below: Maximum m ass of PbS from the or e = kg Relative formula mass of PbS = 207 + 32 = 239 Number of moles of PbS = × = = 1.4 Number of m oles of Pb = 1.4 × 207 , so the correct option must be more than 207 ( = 1 × 207), and less than 414 (= 2 × 207). There is only one option that matches. Therefore the correct answer is D, 289.80 kg 19 The trough to peak height is 16 – 10 = 6 m. But amplitude is defined as the maximum displacement from the equilibrium position, i .e. from the middle to the trough or peak. Amplitude is therefore = 3 m. Th e period of the wave is 12 hours (from the g raph). 12 hours = 12 × 60 minutes = 12 × 3600 seconds. Frequency = eriod = Hz. Therefore the correct answer is A (amplitude = 3 m, frequency = Hz .

7 © UCLES 2015 [Turn over 20 One way to approach this is to think about which quadrants of the graph the equa tions cover, the and intercepts, and any symmetrical shapes . Quick sketches will help. and are quadratic functions – they have a symmetrical U shape. For the – sign means it is upside down, and the 1 means at so the top of the upside -down U is above the –axis , at . So, 2 and 3 could intersect. is a straight line, with a gradient of +3, a –intercept at and an – intercept at . It is in the same quadrants as either or , or both of them, and so could intersect. is a straight line, with a gradient of +1 a –intercept at and an – intercept at . For it is in the same quadrants as , so these graphs could intersect. intersects the –axis , and the graph of intersects the – axis , so these two graphs may not intersect. intersects the –axis at –6, and intersects the –axis at , so these two lines do not intersect. This is option E. It is clear from the plots that and do not intersect. An alternative approach is to solve each pair of equations. The correct answer is E (3 and 4 ). 2x y 2 1 x y   2 1 x y   2x y 2 1 x y   2x y 2 1 x y   2 1 x y   6  x y 2 1 x y  

8 © UCLES 2015 [Turn over 21 Oxygen is required for aerobic respiration which releases energy. The energy released is used for active transport which moves substances up the concentration gradient. The only answer that shows movement against the concentration gradient is C. Therefore the correct answer is C. 22 The Periodic Table is structured so that: the number of the group shows the number of electrons in the outer shell of the atom; the number of the period shows the total number of shells of the atom. In this example Y has gained a 3– charge i.e. 3 electrons have been added to the atom to make the configuration of a noble gas of 2, 8, 8. The atom Y must have the electron configuration 2,8,5. This means it is in Group 15 and Period 3. Therefore the correc t answer is C (Group is 15, P eriod is 3). 23 Output power = 0.50 kW = 500 W. Therefore output voltage = = = 50 V. Number of turns on the secondary is therefore = = 300 turns. Therefore the correct answer is B. (Note the fact that the transformer is 100% efficient is not information that you need to answer this question, as you are already given the output power.)

9 © UCLES 2015 [Turn over 24 The volume of a cylinder (a circular prism) is the base area times the height , . As the cylinder’s height and internal diameter are the same as the diameter of the sphere then both of these are equal to 2 r. This means th at the volume of the cylinder is . The volume of the cylinder is . The volume of the sphere is . The fraction of space inside the cylinder taken up by the sphere is . Simplify by cancelling the terms to leave . The refore the correct answer is D, . 25 The process of genetically engineering this bacterium requires a restriction enzyme to cut out the fluorescent gene from jellyfish DNA (answer B). This gene needs to be put into an appropriate vector (answer A) which uses a ligase (answer D). However, it is the gene that is being added, not the protein that it codes for, so answer C is correct because it is not needed . 26 There are several methods that will give the correc t answer and one is shown below : No. of moles of I = . = = 0.5 No. of m oles of O = . = = 1.25 Ratio of I : O is 0.5 : 1.25 i.e. 2:5 Therefore the correct answer is E ( I2O5) h r2  r r 2 2  3 2 2 2 r r r     3 3 4 r 3 3 2 3 4 r r   3 2 2 1 3 4   3 2

10 © UCLES 2015 [Turn over 27 First, consider these microwaves travelling in air. We are told they have a wavelength of 12 cm (= 0.12 m) and a speed of 3.0 × 10 8 m/s. This means we can work out their frequency, from speed = frequency wavelength, = f λ. f = = . = = × 10 10 = 0.25 × 10 10 = 2.5 × 10 9 Hz . Now consider the microwaves passing through the plastic. We know the plastic will slow the waves down, therefore reducing the wavelength, but that the frequency remains unchanged. We are therefore looking for a frequency of 2.5 × 10 9 Hz but a wavelength shorter than 12 cm by a factor of 2/3 (i .e. 8.0 cm). Therefore the correct answer is B.