Given the shown relationships, which statement is true? A rabbit has a lower mass-specific metabolic rate than a horse O It is impossible to tell from this graph whether a rabbit or a horse has a higher metabolic rate or higher mass-specific metabolic rate OA rabbit has a higher metabolic rate than a horse A rabbit has a higher mass-specific metabolic rate than a horse
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- According to Kleiber’s law, for the vast majority of animals, an animal’s metabolic rate is proportional to m 0.75, where m is the animal’s mass. When we plot the graph of the common logarithm of the mass on the horizontal axis and the common logarithm of the metabolic rate on the vertical axis, we get a straight line. Explain why the graph is a straight line. What is the slope of the line?The average daily metabolic rate for captive animals from weasels to elk can be expressed as a function of mass byr = 140.2m0.75,where m is the mass of the animal (in kilograms) and r is the metabolic rate (in kcal per day).The Nutrition Facts label on most food products shows % daily values based on a food energy requirement of 8,360 kJ (2,000 kcal) at normal resting state. a) Assuming that the efficiency of converting food energy to ATP is 50%, calculate the mass of ATP (in kg) that is harvested daily by the human body from 8,360 kJ of food energy. A total of 30.5 kJ of energy is needed to synthesize one mole of ATP under standard conditions. The molar mass of ATP is 505 g/mol. b) For an average 68-kg (150 lbs.) human adult, calculate the % mass of extracted ATP (from your answer in 2a) relative to body weight.
- Hemoglobin glycation (so named to distinguish it from glycosylation, which is the enzymatic transfer of glucose to a protein) is a non-enzymatic process that involves reaction of the N-terminal amino group of hemoglobin and glucose. The amount of glycated hemoglobin (GHB) is usually about 5% of total hemoglobin (and corresponds to a blood glucose concentration of 120 mg/100 mL). However, in people with untreated diabetes this value may be as high as 13%, which indicates an average blood level of about 300 mg/100 mL -dangerously high. One of the aims of insulin therapy is to maintain GHB values of about 7%. Draw a possible chemical scheme for the glycation of hemoglobin.Write down the energy content in kilo-calories (kcal) from one gram of fat, one gram of carbohydrate and one gram of protein obtained by complete combustion in calorimeter (assume by yourself and show the calculation steps) and by completely broken down in human body in the following table. Please give references. Complete combustion in calorimeter * Completely broken down in body 1 g of Fat 1 g of Protein 1 g of Carbohydrate 1 g of Mixed FoodDrink # Glucose Concentration in mM 1 2.455 mM 2 0.307 mM 3 3.750 mM 4 1.100 mM 5 1.500 mM From here, each group needs to convert the concentration of their drink from mM of the diluted drink into g/100mL of the undiluted drink. This is where you come in! The students are stuck at this point and really need your help! Main Goal: Convert the experimental glucose concentration (provided in the table below) from mM (millimolar) to g/100 mL. Hints & Tips Take the 1/100 dilution factor into consideration Molarity will open the doors for you to get to grams. Once at M assume 1L of solution. (Hint: this will help you when converting from M to moles) You will also need to calculate the molar mass of glucose which has the formula C6H12O6 There are 5 steps in total Task 1: Write a step-by-step guide with key words for the Chem 2 students on how to do these calculations. You can use diagrams, a table or just type the steps out. Ensure the steps…
- Drink # Glucose Concentration in mM 1 2.455 mM 2 0.307 mM 3 3.750 mM 4 1.100 mM 5 1.500 mM From here, each group needs to convert the concentration of their drink from mM of the diluted drink into g/100mL of the undiluted drink. This is where you come in! The students are stuck at this point and really need your help! Main Goal: Convert the experimental glucose concentration (provided in the table below) from mM (millimolar) to g/100 mL. Hints & Tips Take the 1/100 dilution factor into consideration Molarity will open the doors for you to get to grams. Once at M assume 1L of solution. (Hint: this will help you when converting from M to moles) You will also need to calculate the molar mass of glucose which has the formula C6H12O6 There are 5 steps in total Task 1: Write a step-by-step guide with key words for the Chem 2 students on how to do these calculations. You can use diagrams, a table or just type the steps out. Ensure the steps…In biological systems, enzymes are used to accelerate the rates of certain bio- logical reactions. Glucoamylase is an enzyme that aids in the conversion of starch to glucose (a sugar that cells use for energy). Experiments show that 1 pg mol of glucoamylase in a 4% starch solution results in a production rate of glucose of 0.6 ug mol/(mL)(min). Determine the production rate of glucose for this system in the units of Ib mol/(ft³)(day). Ibmol Answer: 0.0639 ft day True FalseBriefly discuss why the mass specific metabolic rate of an elephant is much lower than that of a mouse.
- Calculate all the glucose data into cmol. What is the number of cmol glucose at timepont 10.42 (19.65 g/L)?Twenty-three milligrams of glucose were eaten by the bacteria Sanacoccus pumasareus. Calculate the hypothetical amount of ATP your patient can generate under fermentative metabolism with this amount of glucose. (Note: Glucose MW-180.16 g/mole; 1 mole= 6.02 x 1023 molecules (Avogadro's number)). O2.8 x 10^21 ATPs 01.5 x 10 20 ATPs 01.5 x 10^21 ATPs O2.8 x 10^20 ATPs O No ATP produced since it's fermentation O Lacks information, cannot be determined Nexte PreviousA bull consumes 7 kg DM of feed containing 60 g N/kg. The bull excretes 120 g N in the faeces and 50 g N in the urine. The faecal and urinary N had 20 g metabolic faecal nitrogen (MFN) and 6 g endogenous urinary nitrogen (EUN) respectively. Calculate the biological value (BV) of the protein in the diet.