Question

What is the growth rate (h-1) of microbial culture, if cells divide every six hour?

A 1.38 h-1

B . 0.23 h-1

C 0.35 h-1

D 0.115 h-1

E 30 h-1

Suppose you decided to repeat Frederick Griffith experiment, and you added a DNAse I (an exonuclease which digests single and double-stranded DNA) to the mixture of dead R and dead S bacterial strains. After incubation, the mixture was used to infect mice. What would be the outcome of the infection? Why did it occur?

A

The mice will die. Transformation was terminated by the DNAse I treatment.

B

The mice will die. Transformation was not required.

C

The mice will live. Transformation was not required.

D

The mice will die. Transformation can’t be terminated by the DNAse I treatment.

E

The mice will live. Transformation was terminated by the DNAse I treatment.

Which of the following is(are) an example(s) of anabolic metabolism? Mark all that apply. Include the metabolic pathway(s) with dual functions.

Multiple answers:You can select more than one option

A

Wood–Ljungdahl pathway

B

Iron oxidation

C

Anammox

D

Gluconeogenesis

E

Calvin cycle

F

Hydrogen oxidation

Answer 1

1. Growth rate (h-1) is calculated by the number of individuals divided by the time taken for each division. If a cell is taking 6 hrs to divide from one cell to two cells then growth rate will be = 2/6 = 0.35h-1

Hence the correct option is: C. 0.35h-1.

2. As the DNAse I treatment will digest the single and double stranded DNA strands, the transformation wont be occuring between the R and S strain and hence the mice will live. Hence the correct option is: C. the mice will live. Transformation was not required.

3. All oxidation reactions are catabolic reactions. And the Wood–Ljungdahl pathway is both anabolic and catabolic. The other two pathways, gluconeogenesis ( formation of glucose) and calvin cycle (formation of glucose) both are anabolic pathway as both produce glucose.

Hence the correct options are: D. gluconeogenesis and E. calvin cycle.

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