A biologist has placed three strains of bacteria (denoted I, II, and III) in a t

Question

A biologist has placed three strains of bacteria (denoted I, II, and III) in a test tube, where they will feed on three different food sources (A, B, and C). Suppose that 400 units of food A, 600 units of B, and 440 units of C are placed in the test tube each day, and the data on daily food consumption by the bacteria (in units per day) are as shown in the table. How many bacteria of each strain can coexist in the test tube and consume all of the food?

strain I

strain II

strain III

Bacteria Strain I

Bacteria Strain II

Bacteria Strain III

Food A 1 2 0

Food B 1 1 2

Food C 0 1 2

Explanation / Answer

Units of food A =400

units of food B = 600

Units of food C = 440

since the bacteria can coexist .

Let the bacteria of strain 1 be x

the bacteria of strain 2 be y

bacteria of strain 3 be z

x+ 2y = 400 -i)

x + y + 2z =600 -ii)

y + 2z =440 -iii)

adding i and iii and subtracting ii we get

2y = 240 i.e. y= 120

so putting in iii) we get z = 160

putting y in i) we get x=160

so strain I is 160 ; strain 2 is 120 and strain 3 is 160

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