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Midterm

Animal Adapatations Midterm 2013.pdf

6 Pages
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Department
Biology
Course Code
BIO4120
Professor
Tom Moon

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BIO 4120 - Animal Adaptations Mid-term Examination Fe2r0a,r;y M.oo.n Write answers in the examination booklet provided. Make certain you clearly number your answers. Note: total marks possible being 40 (+2 for bonus); answers will be judged on content and thought, NOT length!!! Use figures/cartoons where possible and I encourage point form where this does address the question asked. 1. Paper-in-lieu of 50% of midterm mark – note in the exam booklet either the reference of the paper you will critique or the topic you will discuss; if you fail to put anything here, you will not be able to accept this option at a later date. 2. Answer 4 of the following 5 questions: (5 marks each – you need 5 marks of information) a. The terms adaptation and natural selection are often discussed together; define these terms and discuss the interrelationship between them. Can define adaptation according to Hochachka as “modification of the characteristics (physiological, biochemical) of organisms that facilitate an enhanced ability to survive and reproduce in a particular environment   fitness” (1.5 marks). Natural selection according to Darwin is a process of molding an organism so that favorable mutations tend to be retained in a population based upon the existing environment as those individuals with these mutations survive and produce the next generation (1.5 marks). These two are interrelated whenever we discuss the relationship between the environment and reproduction, as it is natural selection that selects those individuals that survive and reproduce and thus those traits that facilitate this are adaptive. Would also accept that natural selection is the driving force for the adaptation of specific characteristics in an environment (2 marks). Could also show the cartoon (not necessary): b. Allozymes and isozymes arise by different processes and have different functional characteristics. Using an example, distinguish between these two variant types. Allozymes – arise from allelic variants (the two alleles of a particular gene are not identical) and such changes are often SNP (single nucleotide polymorphisms); these do not breed-true (i.e. they may not be transmitted to the next generation); given that the changes leading to allozymes can be very few, they generally have little effect on enzyme/protein function (1.5 marks). Isozymes – arise from a duplicated locus so that the gene has 2 rather than one copy so that one copy can become ‘silent’, accumulate significant changes in sequence, leading to significant changes in function. These breed-true from one generation to the next (1.5 marks). An example would be LDH (the only one I spoke about; Adrenoceptors and Hb vs Mb could work for isozymes but I didn’t speak about allozymes here?). The isozymes are the M/A or H/B subunits which aggregate in a manner dependent upon the amounts of M and H proteins available in that cell; aggregate according to the polynomial equation giving LDH M and LD4 H which has 4 totally different kinetics (see graph). There are allozymes of the H/B subunit but apparently not the M/A and as such LDH H ’ is 4inetically similar to LDH H (2 mar4s) + LDH5 is anaerobic. Responsible for pyruvate + NADH  lactate + NAD . Keeps glycolysis going. LDH1 is aerobic. It is reversible but acts mostly as lactate oxidase. The resulting pyruvate is used to continue the TCA cycle. There is inhibition at high [pyruvate] because of the abortive tertiary complex (LDH-pyruvate-NAD ). The NAD+ to NADH ratio is ~ 1000. At high [pyruvate], NADH can’t bind. c. Discuss K mnd V max how they are determined, and their significance in an enzyme kinetic response. These are complex ‘constants’ that can be approximated as with the LDH example above by simply estimating Vmax as the maximum velocity that the enzyme achieves independent of [substrate] (1 mark) and Km which is the [substrate] at 50% of Vmax (1 mark). In reality approximation is not appropriate and a specialized plot like the ‘double-reciprocal’ plot or Lineweaver-Burke plot can be used so that Km and Vmax are calculated. Appropriate graphs with Km and Vmax indicated were also acceptable (for example the graph on the right) (1 mark). Lineweaver-Burke plot seen in the left graph below: Their significance is that Km values are an estimate of affinity of the enzyme for its substrate or how effectively it binds its substrate (1 mark), and Vmax given that it is independent of [S]s is an estimate of [enzyme]. It is only by increasing the quantity of the enzyme that Vmax will change (1 mark) d. Enzymes that demonstrate sigmoidal kinetic curves are very different from those showing hyperbolic curves; discuss 3 such differences and their importance. 1) Number of subunits – sigmoidal enzymes have at least 4 subunits, while hyperbolic enzymes have anywhere from 1 to 4 or more subunits (1 mark). 2) Shape of the curve – sigmoidal vs hyperbolic which is based upon the idea of cooperativity, meaning that subunits interact with one another (1 mark). Each subunit can exist in a R (relaxed) and a T (tense) formation and these 2 states are in some state of equilibrium; positive heterotropic effectors (or modulators) push this equilibrium towards the R-state meaning higher binding affinity while a negative heterotropic effectors pushes this equilibrium towards the T-state or lower binding (1 mark). Appropriate graphs indicating sigmoidal and hyperbolic curves were also accepted (see question C – graph on the right). 3) Regulation of hyperbolic enzymes generally occur by increasing or decreasing quantities of the enzyme (altering Vmax), or by feedback inhibition (blocking of binding site); regulation of sigmoidal enzymes can be via quantitative changes but more likely by changes in K0.5 values moving subunit along the R to T conformation (1 mark). The importa
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