Bioanalytical tools and techniques

Compare and contrast parsimony, maximum likelihood, UPGMA and neighbor-joining methods.

Maximum parsimony: – This is a character based method. In it, the topology that requires the smallest number of substitutions is the best tree….

Describe technical and theoretical challenges associated with building phylogenetic trees.

Phylogenetic trees are constructed by choosing an optimal tree, which is itself based on some model of evolutionary change. This approach is known…

What are the similarities and differences between x-ray and MRI (magnetic resonance imaging) images??

Differences between X-ray and MRI images of head, knee, and chest are: An X-ray will give an image of bones present in these body…

Name some of the stationary phases of gas chromatography?

a) Cyanopropylphenyl dimethyl polysiloxane, b) Carbowax polyethylene-glycol, c) Biscyanopropyl cyanopropylphenyl polysiloxane d) Diphenyl dimethyl polysiloxane

Which of the following is not a detector of gas chromatography?

a) Electron capture detector b) Radium ion detector c) Flame ionization detector d) Thermal conductivity detector. Ans. b) Radium ion detector is not the…

Name the scientists who got noble prize for invention of chromatography?

Martin and Synge got noble prize in Chemistry in the year 1952 for the invention of partition chromatography. Their full names are Archer John…

Lecture-1 : Chromatography

Lecture-1 : Chromatography “Chroma” means colour and “graphy” means graphical representation. Separation of components of a mixture on the basis of colour and then identifying…

Lecture 3: Ion- Exchange chromatography

Lecture 3: Bioanalytical tools Important points: This is the most common method for separation of proteins from a mixture. This is so because first…

Lecture 2: Gel filtration chromatography

Lecture 2: Bioanalytical tools Important points It is also called as molecular sieve chromatography or sieve exclusion chromatography. A column made up of porous…

A mixture is having 5 components A, B, C, D and E of size 2, 4, 6, 8 and 10 µm. This mixture is applied to a column having beads of size, 2, 4, 6, and 8 µm. After this, mobile phase is applied. The value of Partition coefficients (Kd) of sample components in stationary and mobile phase will be in the ratio: a) E>D>C>B>A b) A>B>C>D>E c) D>C>B>A>E d) A>D>B>C>E

Correct Answer is b. Explanation: Distribution coefficient = (concentration of component in stationary phase)/ (concentration of component in mobile phase). It is represented by…