Question

21 A A A Aa AOE E ADA Eta In class Prokaryotic Gene Reg: lac operon Name Styles Styles Pane Dictate When: → Is the Z gene expressed? in Units Glucose present Lactose present I* PO*Z*Y* 100 IPO*ZY+ I* PO*Z*Y* 200 IP-OCZY 1* P*O*Z-Y IP-O*Z*Y POZY* I'POCZY IP-O*Z*Y* POZY IPOCZY IPOZY I* POZY I* PO*ZY IPOCZY English (United States) Focus - - 803 S %

Answer 1

S.No.	Glucose	Lactose	Expression of Z	Explanation
I+P+O+Z+Y+	1	100	Expressed	As glucose is in low amount and lactose is in high amount, therefore lac Z is expressed.
I+P+O+Z+Y+ I+P+O+Z+Y+	2	200	Expressed	As glucose is in low amount and lactose is in high amount, therefore lac Z is expressed.
I-P-OcZ+Y+ I+P+O+Z-Y+	0	0	Expressed	As Oc is dominant, cis-acting and constitutive in nature and due to mutation in lac operator doesn’t allow the binding of normal repressor with operator, therefor continuous expression takes place.
I+P-O+Z+Y+ I-P+O+Z+Y+	1	100	Expressed	I+ and P+ are dominant thus the expression of the lac Z takes place.
I+P+OcZ+Y+ I+P-O+Z+Y+	1	100	Expressed	As Oc is dominant, cis-acting and constitutive in nature and due to mutation in lac operator doesn’t allow the binding of normal repressor with an operator, therefor continuous expression takes place.
IsP+O+Z+Y+ I-P+OcZ-Y-	0		Expressed	Lac Is is dominant and thus inducer gets changed and cannot bind to operator but as there is no inducer thus no expression can be there but Oc is dominant, cis-acting and constitutive in nature and due to mutation in lac operator doesn’t allow the binding of normal repressor with operator, therefor continuous expression takes place (thus expression takes place even in absence of lactose).
I+P+O+Z+Y- I+P+O+Z+Y-			Not expressed	No expression, as no inducer is present.
I+P+O+Z+Y- I+P+OcZ+Y-			Expressed	As Oc is dominant, cis-acting and constitutive in nature and due to mutation in lac operator doesn’t allow the binding of normal repressor with operator, therefor continuous expression takes place.