In a Nitrogen metabolism study, the following data were obtained over the test period:

Nitrogen consumed 175 grams
Total Fecal N 35 grams
Metabolic Fecal N 10 grams
Total Urinary N 80 grams
Endogenous Urinary N 10 grams

Calculate:
a. Apparent Nitrogen digestibility (%)
b. True Nitrogen digestibility (%)
c. Nitrogen Balance (g)
d. Apparent Nitrogen Retention (%)

Answers

Answer 1

Answer:

a. 80%

b. 86%

c. 60 g

d. 34.29%

Explanation:

From the given information:

The apparent nitrogen digestibility (AD) is the ratio of the difference between the consumed nitrogen and fecal nitrogen to the consumed nitrogen.

Mathematically:

[tex]AD = \dfrac{consumed \ nitrogen - fecal \ nitrogen }{consumed \ nitrogen }[/tex]

[tex]AD = \dfrac{175 -35 }{175}[/tex]

AD = 0.8

To percentage, the apparent Nitrogen digestibility = 80%

b. True Nitrogen digestibility is calculated by the formula:

[tex]=\dfrac{consumed \ nitrogen - (fecal \ nitrogen -metabolic \ fecal)}{consumed \ nitrogen}[/tex]

[tex]= \dfrac{175 -(35-10)}{175}[/tex]

[tex]= \dfrac{175 -25}{175}[/tex]

= 0.86

= 86%

c. Nitrogen Balance = consumed Nitrogen - Fecal nitrogen - total uninary

Nitrogen Balance = (175 - 35 - 80) grams

Nitrogen Balance = 60 g

d. Apparent Nitrogen Retention (ANR) is computed as follows:

[tex]ANR = \dfrac{(consumed \ nitrogen-fecal\ nitrogen - total \ uninary )}{nitrogen \consumed }\times 100\%[/tex]

[tex]ANR = \dfrac{(175-35- 80 )}{175 }\times 100\%[/tex]

[tex]ANR = \dfrac{(60 )}{175 }\times 100\%[/tex]

ANR = 34.29%


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Two different enzymes catalyze the same reaction and both exhibit the same Vmax. When Enzyme A was run with a 40 uM substrate, the initial rate (Vo) was 10 uM/min, and when g it was run with a 4 mM substrate, the Vo was 20 uM/min. Estimate the approximate Vmax and Km of Enzyme A. When Enzyme B was run with 80 uM substrate, the initial rate (Vo) was 10 uM per minute. Estimate the Km of Enzyme B.

Answers

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Explanation:

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equating equation (1) and (2):

[tex]40V_{max} -10k_m = 400 --- (1)[/tex]

[tex]4000V_{max} -20k_m = 8000 --- (2)[/tex]

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Since V_{max} of A ie equivalent to that of B; then:

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[S] = 80  uM

V = 10 uM/min

[tex]10 = \dfrac{11.02 \times 80}{k_m + 80 }[/tex]

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10k_m = 881.6  - 800

10k_m = 81.6

k_m = 81.6/10

k_m = 8.16 uM

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