Lec 4 The lac operon - Nayak

O mutants have a mutation in the operator region of the operon. This mutation prevents the repressor protein from binding to operator

• The + means that gene is present, and - means gene is absent. Having O means that O is a genetic element that controls expression of Z and Y. 
• C1 = No it cannot produce B-galactosidase without being induced, beucase, for Z to activate, Lactose must be present (inducer).
• C2 = Z is produced when induced (such as waen lactose is present (inducer) Z is produced)
• C3 =No it cannot produce permease/Y without being induced, 
• C4 =   Y  is produced when induced
• C5 =  what conclusion can we draw about O+Z+Y+ :: its wildtype which works only when induced
• 

• Now we see O+ and F'O+, F' measn the bacteria contains a plasmid. The First genotype is the chromosomal geneotype, and the second is the Plasmid genotype. THe F' indicates that this genotype is of the plasmid. 
• We have O+Z+Y+ and F Plasmid (O+Z - Y+)
• C1 =No it cannot produce B-galactosidase without being induced, beucase, for Z to activate, Lactose must be present (inducer).
• C2 = Z is produced when induced (such as when lactose is present (inducer) Z is produced)
•  
• Since Only Z has different in both chromosome and plasmid, we can make a statement about that: Z+ is DOMINANT to Z-. THIS IS NOT ALWAYS THE CASE. But after looking at what will happen, if the inducer is present or not, we can say which is Dominant.

• C3 =No it cannot produce permease/Y without being induced, 
• C4 =   Y  is produced when induced
• C5 =  Z+ is DOMINANT to Z-.
• 

• O^c means Constitutively ON, meaning its always on, even when uninduced. 
• Oc tells me that Z+ and Y+ will always be on whether they are induced or not.  So all C1,C2,C3,C4 will have + sign. Because, non of them need an inducer to be on, if there is an inducer, meh, ok, its still on. 
• C5=  Oc is constitutive 
• 

We have O+ Z- Y+ /  F' Oc Z+ Y-

Now for C1 and C2: we know that we are not giving any inducer, so will Z be on or off. Since, Z+ is connected to Oc, and Z- is connected to O+. and we know that Z- is recessive to Z+, we can say that Z+ will prevail. Now that Z+ is going to be constitutively ON. meaning both C1 and C2 will be +. Meaning, in noninduced and in induced conditions, B-galactosidase enzyme will be produced. 

• C3: Y+ is dominant to Y-. So, leave Y- and Oc and Look at O+ and Y+. so, O+ we know is wildtype, so it only turns on Y when inducer is present, as seen in strain 1. So C3 is going to be -. b/c Y will only turn on when inducer is present. 
• C4 =  +, b/c inducer is present so will turn on.

• C5: Since O+ controls the activity of Z- and Y+
• and  Oc controls the activity of Z+ and Y-, meaning Oc and O= control the expression of genes on the same molecule of DNA, we can say that O mutant is Cis acting.
• O mutant regulate expression of genes on the same molecule of DNA - we say O is cis acting
• 

• C1 & C2 =  Since Z+ is dominant to Z-, its activity will be superior, and since Z+ is linked to Oc, meaning Z+ will be expressed whether inducer is present or not
• C3 & C4 =  Since Y+ is dominant to Y-, its activity will be superior, and since Y+ is linked to Oc, meaning Y+ will be expressed whether inducer is present or not
• 

• Z+ is dominant to Z-, and Z+ is linked to O+. 
• C1: When inducer is not present, Z will not be active, so -.
• C2:  When inducer is present, Z is active so +
• Y+ is dominant to Y-
• C3: -
• C4:  +
• 

• Cis-acting regulatory elements can only regulate expression of genes on the same molecule of DNA as the element
• (Location dependent)
• 

• Trans-acting regulatory elements can regulate expression of genes that are located on the same or different DNA molecule
• (Location independent)
• 

I mutants in the Lac Operon model have a mutation in the repressor gene of the operon.

• Strain 1:  Only + in place of Inducible, so we can say I+ is inducible
• Strain 2:  All columns have +, so we can say I- remains ON always, so I- is constitutive
• Strain 3: We see the + and - sequence is same as strain 1, s this means I= is dominant to I-
• Strain 4:  The Z C1 and C2 seem normal, but Y's last 2 columns are a problem.
• So In the noninduced Y column: We see that Y+ is dominant to Y-, and Y+ is bound to I-, which would keep the Y+ always ON (constitutive)/ Also, we see that Y columns display an inducible function, and we know that only I+ is inducible, and we know that I+ in strain 4 has Y-, so by itself, it cannot exhibit the induced activity of Y, so I+ has to effect the expression of Y+ which is connected to I-. This tells us that I+ is trans acting.  
• 

Repressor is a protein that prevents transcription by binding the promoter or binding downstream

• Strain 1:  Only Induced columns have +, so I+ is inducible.
• Strain 2: ALL columns are never expressed, even when induced, so this means I^S is always repressed.
• Strain 3:  All columns are never expressed, we have both I^S and I+, this means I^S is dominant to I+
• 

• 1) LacI binds to Inducer (IPTG) 
• 2) LacI binds to DNA (operator) using an independent operator binding site  
• 3) LacI bound to inducer cannot bind to DNA
• 4) LacI^s cannot bind inducer but can still bind the operator and, thus acts as a constitutive repressor

 

LacI is a sensor and a regulator!