Question

questions 1-4?

1.

Use the following diagram of the lac I gene and lac operon.   Where would the lac repressor be bound in a (nonmutant) E. coli cell that is growing in no glucose and high lactose?

(I = lac repressor gene; Z, Y, A = lac operon structural genes; P = lac promoter; O = lac operator)

		The repressor would not be bound
		P
		I
		P and O
		O

2.

Euchromatin is associated with more loose and open organization of DNA and genes in these regions are more likely to be active and transcribed.

		False
		True

3.

There are mutations to the lac operon that allow beta galactosidase to be expressed constitutively even when lactose is absent. Which of the lac genotypes below would enable for this constitutive expression?

		I+ P+ Oc Z+ Y+ A+
		I+ P- O+ Z+ Y+ A+
		Is P+ O+ Z+ Y+ A+
		I+ P+ O+ Z- Y+ A+
		I+ P+ O+ Z+ Y+ A+

4.

If a typical somatic cell has 14 chromosomes, how many chromosomes are expected in each gamete of that organism?

		14
		56
		28
		7
		23

Answer 1

Q.1. Where would the lac repressor be bound in a (nonmutant) E. coli cell that is growing in no glucose and high lactose

Answer- The repressor would not be bound

(Note: lactose is inducible operon. When lactose is absent then the repressor protein will bound to the operator site to prevent the transcription. In the presence of lactose, the repressor will bound to the inducer (lactose) and hence, cannot bind to the operator so, transcription proceeds)

Q.2. Euchromatin is associated with more loose and open organization of DNA and genes in these regions are more likely to be active and transcribed.

True.

During Giemsa staining process euchromatin show faint stain while heterochromatin is darkly stained. Heterochromatin is tightly coiled with the protein part (darkly stained) and contains gene-poor regions. While euchromatin is loosely packed (light stained ) and contains gene-rich regions. As euchromatin is loosely arranged, so the DNA of the euchromatin is accessible for the enzymes responsible for the transcription process. therefor, euchromatin contains genes.

Q.3. There are mutations to the lac operon that allow beta galactosidase to be expressed constitutively even when lactose is absent. Which of the lac genotypes below would enable for this constitutive expression?

Answer- I+ P+ Oc Z+ Y+ A+

(In this mutant, the operator has a constitutive mutation, therefore no repressor can boun to the operator in any condition. If the repressor is failed to bind to the operator, then the transcription will always proceed and beta-galactosidase will be expressed constitutively weather the lactose is present or absent).

Q.4. If a typical somatic cell has 14 chromosomes, how many chromosomes are expected in each gamete of that organism?

The gametes are produced by meiotic cell division. Meiosis is also called reduction division as the chromosome numbers become half in the products.

So if the somatic cell contain 14 chromosomes then the gametes will contain the half number of the somatic chromosomes.

Answer: 7.