My professor was talking about double mutant analysis with null mutants, and how double mutant analysis won't work with hypomorphs. I really don't understand the concept of double mutant analysis.

He was talking about pathways - if A and B are in the same pathway, then you would mutate A and he said something about the phenotype being identical. If A and B are in different pathways, you could mutate both of them and there would be a "stronger" phenotype.

He also said it would only work with null mutants, not with hypomorphs, because with hypomorphs some of the pathway would still be active, not completely shut off.

I really don't understand any of this - what a double mutant is, the whole pathway thing, why the phenotype is "identical" or "stronger", or why you use hypomorphs. I would really appreciate a rundown on the idea of double mutant analysis because my professor's explanation was so unclear. Thanks in advance.

  • $\begingroup$ Do you mean why you DO NOT use hymomorphs ? $\endgroup$
    – biogirl
    Commented Dec 9, 2013 at 11:45
  • $\begingroup$ you may also want to see this $\endgroup$
    – biogirl
    Commented Dec 9, 2013 at 12:12
  • $\begingroup$ Basically your professor was saying some hypomorphs (weakly expressing phenotypes) do not necessarily show their phenotype since they still express some of the gene products and there are other pathways that might compensate for their lack of expression of that gene to an extent. So if you want to see a phenotypic effect then you got to have a null in those situations since that will be harder to compensate for specially in organisms that do not have that many redundancies such as Drosophila. But even nulls are not exempt to the above rule if there is another pathway compensating for them! $\endgroup$ Commented Nov 6, 2014 at 12:01

2 Answers 2

  • What is a double mutant ?

See this image.

In this mutant 1 has one mutation, mutant 2 has another mutation while the double mutant has "double" or 2 mutations.

Key : A,B means double mutation , A means mutation in A and B means mutation in B.

  • Types of double mutations

a)Additive phenotype: A, B = A + B Implies that A and B function independently and on different processes.
Eg. A = short roots, B = no root hairs A, B = short roots with no root hairs. Conclusion: A and B function independently on different aspects of root development (A is required for normal root growth and B is required for root hair development.

b) Epistatic phenotype: A, B = either A or B

i) If A = B = A, B then epistasis implies that both genes function in the same pathway and that each gene is essential to the process. Eg. A = short roots, B = short roots A, B = short roots. Both A and B are required for normal growth of roots.

ii) If A is opposite in phenotype to B, and A, B = A, then epistasis implies that A and B are in the same pathway and B is a negative regulator of A. Eg. A = long roots, B = short roots A, B = long roots. A is required to suppress root growth and B is a negative regulator of A.

c) Synergistic phenotype: A, B >>> A or B.

Implies that A and B contribute independently to the same process. Eg. A = long roots (wt = 2 cm, A = 4 cm) B = long roots (B = 4 cm) A, B = very long roots (20 cm). A and B negatively control root growth independently of one another and are performing partially redundant functions.

When your professor was saying that if A and B are in same pathway and you mutate A and phenotype comes as if both A and B have been mutated, then he was probably referring to the epistatic phenotype.

When he was talking about having a stronger effect by mutating them both, he was talking about Synergistic phenotype.

Source : Double mutant analysis

  • $\begingroup$ I do not know about the hypomorph thing. Maybe someone else will help you ! $\endgroup$
    – biogirl
    Commented Dec 9, 2013 at 12:06
  • $\begingroup$ Feel free to ask anything u do not understand. These things are tricky ! $\endgroup$
    – biogirl
    Commented Dec 9, 2013 at 12:07

Hypomorphs (now commonly termed 'dominant negative') have a reduced (hypo) phenotype as compared the wildtype. Null (knockout) mutations, however, have the phenotype removed, so to say.

The reason that you would use null rather than hypomorph mutations to do double mutant analysis is because with the hypomorph, you would still see some effect of the phenotype you are trying to have eliminated.

Why do you want to eliminate a phenotype? Let's say you want to try to find the pathway for the genetic interactions between A and B. You have 2 animals you want to cross (I work with C. elegans worms, and use double mutants often). If we look at the wildtype homozygous of both animals, they will be the same. Animal 1: +;+/+;+. Animal 2: +;+/+;+. (where + denotes wildtype, and ';' means they are not linked, so the genes are independent of each other).

We will use 'a' to denote null mutation A, and 'b' to denote null mutation B.

Animal 1 will have mutation A. Animal 2 will have mutation B. So now Animal 1 will look like a;+/a;+, and Animal 2 will look like +;b/+;b. Notice that each locus is corresponding with itself - ie: in the region where Animal 1 is still +, Animal 2 is b, and vice versa (for null mutation 'a').

So then we cross Animal 1 with Animal 2 and get a;+/a;+ x +;b/+;b which will segregate in a 9:3:3:1 ratio (according to independent assortment of 2 independent genes - just draw a 4x4 punnet square to convince yourself).

We are interested in achieving the 1/16 a;b/a;b animal (also can be noted aabb, btw). And THAT is our double mutant.

Because we use a null allele, there is no chance of the wildtype phenotype "leaking" within our cross, giving us a false ratio or false assessment, as would be the case with a hypomorph (reduced, but not gone).

Now, when talking about genes in a pathway: I think this is what your professor was referring to. When we are mapping how genes interact, and in what order, we can either have the genes interacting in redundant or sequential pathways.

Redundant pathways: A -> C and B -> C. So either A or B can have an effect on C, meaning that having both is redundant, because one can do the job.

Sequential pathway: A -> B -> C OR B -> A -> C. So both are existing in the pathway in sequence, and both are needed.

Just as a side note, whenever you are using double mutants to find out if your pathway is redundant or sequential, you MUST compare it to a single mutant!!! Otherwise you WILL NOT be able to tell the difference between your results.

In a single mutant (A, for example), when using in a redundant pathway, you will still get phenotype C, since B is available. Just note that your C phenotype will be weaker. When you then use a double mutant, knocking out both A and B, you will see NO expression of phenotype C. The same will apply if you knock out B instead of A.

Now, for a sequential pathway. If you use a single mutant (again, let's use A as the example) you will have no phenotype C, since A is needed to interact with B (whether upstream of downstream) in order to have an effect on C. And, if you use a double mutant (so knock out both A and B), then you will also have no expression of C, since both A and B are gone. (again, same applies if you use a B single mutant first)

So, for a redundant pathway: single mutant will have C phenotype, double mutant will not have C phenotype. There IS a change!

For a sequential pathway: single mutant will have no C phenotype, and double mutant will have no C phenotype. There is NO change!

If you would have used a hypomorph to do these experiments, then you would probably see some percentage of C expression in your double mutant experiments, making your results inconclusive.


I hope this helps!

[source: you can find this information in any genetics textbook (in my case, since i study worms, WormBook is a great source); I have given this information from my own experience. I have an honours degree in Genetics (Biochemistry and Medical Genetics) and am currently continuing my education in grad school (biochemistry and molecular biology).

  • $\begingroup$ Null mutants are not necessarily KO! they can for example be substitutions causing a premature stop codon. Hence it would be great if you made that clear in your response! $\endgroup$ Commented Nov 6, 2014 at 0:29

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