TLS Online TPP Program

#Question id: 72


For the reaction catalyzed by the enzyme hexokinase: Glucose + ATP → glucose 6-phosphate + ADP the equilibrium constant, Keq, is 7.8 × 102. In living E. coli cells, [ATP] = 5 mM, [ADP] = 0.5 mM, [glucose] = 2 mM, and [glucose 6-phosphate] = 1 mM. Which of the following conclusions made on the basis of the above calculations is incorrect?

#Unit 1. Molecules and their Interaction Relevant to Biology
  1. The reaction is therefore far from equilibrium.

  2. The cellular concentrations of the products are much lower and those of the reactants are much higher.

  3. The reaction therefore tends strongly to go to the right.

  4. The cellular concentrations of the products are much higher and those of the reactants are much lower.

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TLS Online TPP Program

#Question id: 29105

#Unit 8. Inheritance Biology

If two set of chromosome is considered in G1 phase, then at which phase of the cell cycle, the no of chromosome as well as no of DNA cell both are doubled?

TLS Online TPP Program

#Question id: 29106

#Unit 8. Inheritance Biology

For the proper segregation of the chromosome at anaphase-I chiasma formation take place. At which stage of the cell cycle crossing over takes place?

TLS Online TPP Program

#Question id: 29107

#Unit 8. Inheritance Biology

Aneuploids may arise through nondisjunction, the failure of homologous chromosomes or sister chromatids to separate in meiosis. which of the following statements is correct?

a) Meiosis I non-disjunction leads to produce all gamete become aneuploidy

b) Meiosis II non-disjunction leads to produce half of gametes n+1 and n-1 while other half normal haploid

c) Either meiosis I or meiosis II non-disjunction leads to produce all aneuploidy gamete

d) Meiosis II non-disjunction leads to produce all gamete become n+1 and n-1

TLS Online TPP Program

#Question id: 29108

#Unit 8. Inheritance Biology

If 3 genes are unlinked which is X, Y and Z, the probability of progeny being X+ X-Y+Y- Z- Z- from a cross between  X+X+Y+Y-Z+Z- and X-X-Y+Y+Z-Z- parents will be____

TLS Online TPP Program

#Question id: 29109

#Unit 8. Inheritance Biology

If the probability of being blood types are;
Blood type A = 1/4
Blood type B = 3/4
Blood type O = 1/2
what is the probability of being either blood-type A  or blood-type B ?

TLS Online TPP Program

#Question id: 29110

#Unit 8. Inheritance Biology

When two parents heterozygous for albinism mate (Aa X Aa), the probability
of their having a child with albinism (aa) is 1/4 and the probability of having a child with normal pigmentation (AA or Aa) is 3/4 . What is the probability of this couple having 5 children, two with albinism and three with normal pigmentation?