TLS Online TPP Program

#Question id: 314


Which statement explains the cleaning action of soap on greasy dishes?

#SCPH28 | Zoology
  1. The soap changes the water-solubility of the grease so that it is easily dissolved by the water.

  2. The grease is trapped inside the hydrophobic interior of micelles made of soap molecules.

  3. The soap chemically breaks down the grease into smaller, more water-soluble molecules.

  4. The soap hydrates the grease with its polar head groups and holds it in suspension.

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TLS Online TPP Program

#Question id: 4808

#SCPH12 I Genetics

When F1 female Drosophila of the genotype AaBbCc is test crossed, ( Gene A and B located at 10 map unit in transe configuration but gene C are independently assort for both gene. Which of the following test cross progeny that is that is more frequent?

TLS Online TPP Program

#Question id: 11964

#SCPH28 | Zoology

 Animals cannot produce enzymes to digest cellulose, yet many termite species consume cellulose from plant material as a main part of their diet. How do termites access the nutrients contained in cellulose?

TLS Online TPP Program

#Question id: 15638

#SCPH28 | Zoology

You have isolated two mutations that show decreased expression of the Lac operon. However, unlike like the promoter mutations, these mutations don’t respond to the inducer IPTG. These mutations, designated Lac3– and Lac4–, are evaluated for the quantity of ß-galactosidase and permease activity expressed with or without IPTG:
 
Mapping experiments reveal that Lac3– and Lac4– are different short deletions located in the region before the start of the LacZ gene. Given the data shown above suggest which genetic element(s) in addition to part of the promoter has been deleted in each mutant. 

TLS Online TPP Program

#Question id: 2620

#SCPH05 I Biotechnology

Which option is correct about catabolite activator protein binds as

TLS Online TPP Program

#Question id: 11708

#SCPH28 | Zoology

Use the following clinical laboratory test results:

Urine flow rate = 1 ml/min

Urine inulin concentration = 100 mg/ml

Plasma inulin concentration = 2 mg/ml

Urine urea concentration = 50 mg/ml

Plasma urea concentration = 2.5 mg/ml

What is the glomerular filtration rate (GFR)?