HUBS2206 Human Biochemistry and Cell Biology Assignment Brief 2026 | UON Australia

HUBS2206 Assignment Brief 

Learning Outcomes

Question 1: 

 

Figure a): Initial velocity rate of Alcohol Dehydrogenase (V0) against increasing Ethanol concentration ([S]). Each cuvette (Ethanol, Phosphate Buffer, NAD+ and ADH) absorbance was measured at 340 nm and at 5-second intervals. Ethanol substrate concentration increases in each cuvette from 0.625, 1.25, 2.5,  5  , to  10  mM.  The trendline is drawn logarithmically,  indicating the saturation of alcohol dehydrogenase, as it plateaus out.

Question 2:  

As the ethanol substrate concentration of each cuvette increases, the initial velocity of each reaction increases. The rate of reaction and initial velocity continue to increase until the enzyme alcohol dehydrogenase becomes fully saturated with a specific concentration of substrate (10 mM of Ethanol), resulting in the graph plateauing. The individual results were not accurate; however, the logarithmic trendline gives an indication of where the reaction plateaus. In this case, when the substrate reaches 10 mM, the initial velocity starts to plateau at 0.0981 mM/min, indicating it has reached its full capacity for the substrate.

Questions 3:

 

Figure b): Double reciprocal linear plot for initial velocity (V₀) of Alcohol Dehydrogenase and Ethanol substrate concentration ([S]). The reciprocal values for V₀ and [S] from the saturation curve are taken to give a linear graph. The individual plot points were not expected, however, from the linear trend line, the equation of the line is derived (y = 62.271x − 2.1072), and Vmax and Km can be calculated by finding the x and y intercepts of the line.

Question 4.

Vmax: the reciprocal of the y-axis
Substitute x = 0 into the equation of the line Y = 62.271x − 2.1072
When x = 0
Y = 62.271(0) − 2.1071
Y = − 2.1071
Vmax is the reciprocal of this value
Y = − 0.47456 mM/min
Vmax = − 0.47456 mM/min

Km: is the negative reciprocal of the X intercept

Substitute y = 0 into the equation Y = 62.271x − 2.1072
When y = 0
0 = 62.271x − 2.1071
2.1071 = 62.271x

x = 2.1071 / 62.271

Multiply by -1, and find the reciprocal of this value

x = − 0.3383758 x^-1
x = − 0.3383758
x = -2.95529 mM
Km = -2.95529 mM 

Therefore; KM = -2.95529 mM and VMax = -0.47456 mM/min 

These results were not expected, as they are both negative values. The expected Vmax value should have been positive

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