1. A population of wild geese has the following frequency foralleles at the alcohol dehydrogenase locus: ADH+ 0.70,ADH null 0.30. In a population of 1,000 birds, what willbe the expected frequencies of each genotype (and thenumber of birds that have each genotype)?*null non-functional protein=2. A major winter storm decimated to this population of wildgeese reducing its number to 100 birds. Their genotypesare displayed below:Genotype +/++/null null/nullNumber 51 42 7Is this an example of genetic drift? Prove this by using the C-squared statistical test where:c2 = (observed expected)²/(expected)-

Approach to solving the question:Question 1:Genotype Frequencies in a Population of 1,000 BirdsWe…

Answered: 1. A population of wild geese has the…

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter11: Genome Alterations: Mutation And Epigenetics

Section: Chapter Questions

Problem 8QP: Bruce Ames and his colleagues have pointed out that although detailed toxicological analysis has...

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Bruce Ames and his colleagues have pointed out that although detailed toxicological analysis has been conducted on synthetic chemicals, almost no information is available about the mutagenic or carcinogenic effects of the toxins produced by plants as a natural defense against fungi, insects, and animal predators. Tens of thousands of such compounds have been discovered, and he estimates that in the United States adults eat about 1.5 g of these compounds each daylevels that are approximately 10,000 times higher than those of the synthetic pesticides present in the diet. For example, cabbage contains 49 natural pesticides and metabolites, and only a few of these have been tested for their carcinogenic and mutagenic effects. a. With the introduction of new foods into the U.S. diet over the last 200 years (mangoes, kiwi fruit, tomatoes, and so forth), has there been enough time for humans to develop resistance to the mutagenic effects of the toxins present in those foods? b. The natural pesticides present in plants constitute more than 99% of the toxins we eat. Should diet planning, especially for vegetarians, take into account the doses of toxins present in the diet?
3 i. Define Biological Value (BV) of Proteinsii. What are the advantages of this protein evaluation method when compared to Nbalance and chemical score?5iii. A bull consumes 7 kg DM of feed containing 60 g N/kg. The bull excretes 120 g N inthe faeces and 50 g N in the urine. The feacal and urinary N had 20 g metabolic feacalnitrogen (MFN) and 6 g endogenous urinary nitrogen (EUN) respectively.Calculate the BV of the protein in the diet.
1. explain why the substitution of Leu > Val and Arg > Lys in protein does not cause a decrease in the biological activity of the enzyme ! 2. explain why the substitution of Ala → Pro and Gly → Val in protein does not cause a decrease in the biological activity of the enzyme !
1. Acid phosphatases are an important group of enzymes that can be detected in human bloodserum. Under slightly acidic conditions (pH 5.0), this group of enzymes can hydrolyzebiological phosphate esters as follows:R-O-P-O3-2 + H 2O R-OH + HO-P-O3-2.Acid phosphatases are produced and can be detected in erythrocytes, kidney, spleen, the liver,and prostrate gland. The enzyme from the prostrate gland is clinically important because anincreased activity in the blood is frequently an indication of cancer of the prostrate gland.Tartrate ion can strongly inhibit the phosphatase from the prostrate gland, but not acidphosphatases from other tissues. How can you use the information above to develop a specificprocedure for measuring the activity of the acid phosphatase of the prostrate gland in humanblood serum?
15. The deficiency of glucose-6-phosphate dehydrogenase (G6PD) is one of the most common inborn errors of metabolism worldwide. This congenital disorder generally results from mutations that are spread throughout the entire gene of G6PD. Three mutations for G6PD, AAAAGAGGG AAACAUGGG, „CACAAUCAC-CACCAUCAC, AUGACUAAA AUGGCUAAA are responsible for severe deerease of the protein activity. What types of mutations are these? To answer the question please: I) explain what the mutation is; 2) describe the types of mutation you know; 3) explain the difference between germline and somatic mutati ymem
1. Hexokinase (HK) catalyses the conversion of glucose to glucose-6-phosphate. There four distinct mammalian HK isoforms [I, II, III, and IV (glucokinase)]. HK I in red blood cells (RBCs) has a Km of 0.05mM, while HK IV, in the liver, has a Km of 5 mM. Explain how the marked differences in Km between HK1 and HKIV supports the function of red blood cells and liver
3. Aspartate aminotransferase (AspAT) catalyzes the following reaction: COO™ C=O CH₂ COO Oxaloacetate The AspAT enzyme has two active-site arginines, Arg 386 and Arg 292, that interact with the a- carboxylate and ß-carboxylate groups on the aspartate substrate, respectively. Investigators stud- ied the mechanism of AspAT in more detail by constructing mutant AspAT enzymes in which either or both of the essential arginines were replaced with a lysine residue. The kinetic parame- ters for the wild-type enzyme and nutant enzymes are shown in the table. NH—CH-COO T CH₂ COO™ Aspartate α-Amino acid Enzyme 1. Wild-type Asp AT(Arg 292 Arg 386) 2. Mutant Asp AT(Lys 292 Arg 386) 3. Mutant Asp AT(Arg 292 Lys 386) 4. Mutant Asp AT(Lys 292 Lys 386) Asp-AT a-Keto acid KM (MM) 4 326 72 300 Kcat (S-1) 530 4.5 9.6 0.055 (a) The Michaelis constant is strictly not equivalent to an equilibrium binding constant; however under certain conditions the Michaelis constant does approach the equilibrium (dissoci-…
6
6. A. List all of the ionizable functional groups that are found in insulin when in aqueous solution. List which amino acid residues have these ionizable groups and list all of the pka and pKb values (including the R groups) that are on both polypeptide chains that make up insulin. (see the table at the end of this HW set; note that tyrosine and cysteine both have unusual pka's, since these side groups ionize above the pKR's given to have a negative charge). B. The isoelectric point of insulin is reported to be around 5.3-5.35. Using the method covered in class, estimate the isoelectric point of insulin and compare your answer to the values above. C. For a polypeptide to be soluble in an aqueous solution, is it good to be near the isoelectric point? Why or why not? Notes: a couple of unusual R group's that ionize (cysteine and tyrosine have R groups that have pka values; histidine has a pkb). The table at the end of this homework set (also in the lecture notes) lists the pka's and…
8. TABLE 3.3 Catalytic Activity of a Variety of Enzymes Enzyme Nonenzymatic t,2' Turnover number? Rate enhancement 78,000,000 yr 130,000 yr 120 yr 69,000 yr 69 yr 2.9 yr 7.3 yr 7 wk 1.4 x 107 5.6 x 1014 2.1 x 1012 6.0 x 1012 1.2 x 1012 2.8 x 10" 1.9 x 10" 3.9 x 10" 1.0 x 10° 1.9 x 10 7.7 x 10 4.6 x 10 39 OMP decarboxylase Staphylococcal nuclease 95 Adenosine deaminase 370 AMP nucleosidase 60 Cylidine deaminase Phosphotriesterase Carboxypeptidase A Ketosteroid isomerase 299 2,100 578 66,000 Triosephosphate isomerase Chorismate mutase 1.9 d 4,300 7.4 hr 5 sec 50 1 x 10 Carbonic anhydrase Cyclophilin, human 23 sec 13,000 "The time that would elapse for half the reactants to be converted to product in the absence of enzyme. The number of reactions catalyzed by a single enzyme molecule per second when operating at a saturating substrate concentration. The increase in reaction rate achieved by the enzyme-catalyzed reaction over the noncatalyzed reaction. Source: A. Radzicka and R. Wolfenden,…
Fill out the concept map. In the middle circle, describe a just law. For the outside circles, describe the purpose of the different kinds of law. Chese laws affect me in a vay thatSeanner
3. Hepatocellular carcinoma (HCC) is among the leading causes of cancer death worldwide. Studies of gene expression in HCC cells demonstrated that the enzyme ornithine aminotransferase (OAT) is upregulated in these cells. Ornithine (structure to the right) is both an intermediate in the urea cycle as well as a precursor for the synthesis of the nonessential amino acid proline. Unlike many aminotransferases, OAT acts on the side chain amine of ornithine rather than the alpha-amino group. wwwwwww (a) If OAT takes ornithine and alpha-ketoglutarate as substrates, draw the structures of the products released after catalysis. (b) What cofactor does OAT use and how is it linked to the enzyme? (c) Devise a strategy to alleviate HCC based on OAT and describe the specific steps required to accomplish this. What is a major challenge in this approach? What pre-clinical effects would you want to demonstrate before attempting to treat patients?

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