Question

Compare the wild type protein to the mutant protein. Then classify the mutation.

	DNA	mRNA	Protein
Wild type DNA	‘3 TAC CCC GAT ACT 5’	AUG GGG CAU UGA	Met Gly His
Mutant DNA	‘3 TAC CCC GAG ACT 5’	AUG GGG CAC UGA	Met Gly His

1. Nonnsense
2. Silent
3. Missense
4. Neutral

Let E in the following reaction stand for some enzyme: E + H₂O₂ --à E H₂O₂ --à E + H₂O + O₂ ( 2 nanoseconds). This reaction shows that E_____.

1. Does not change permanently in the chemical reaction
2. It probably has a role that is related to time.

1. Only 1
2. Only 2
3. Both 1 and 2
4. Neither 1 or 2

How many grams of sodium chloride (NaCl) should you add to 99 .9% grams of water to make 0.04% sodium chloride?

1. 4 grams
2. 99.96 grams
3. 0.04 grams
4. 0.4 grams

. Lactase+ Lactose-----> Lactase+ Lactose----->Lactase+ Glucose+ Galactose

       Which of the following are final products of an enzymatic reaction?

1. Lactase lactose complex
2. Lactose
3. Glucose, galactose
4. Only lactase

Answer 1

c)missense: mutation in which point mutation does not affect the amino acid and it position. thus it can be in seen in protein Sequence no changes in amino acid

C) BOTH 1 AND 2

by looking at reaction scheme

E does not change permanently because E is catalyzing the decomposition H202 into H20 AND 02 and at the final step we are getting E back as same ( thus first statement is true)

since E is changing it structure and forming an adduct with H202 so we can infer that E probably has a role that is related to time.

c ) 0.04 g

99.9 gm of water stand for pure water so by calculating w/w (weight / weight) = 0.04% we need to have 0.04 gram of NaCl in 100gm of water

c) glucose, galactose

lactose is a disaccharide of glucose + galactose and having linkage O-β-d-galactopyranosyl-(1–4)-β-d-glucopyranose