Expert: Dana Krempels, Ph.D. - 6/3/2007

Question

Hi
This is not a homework question-really.The answer is there in the book but I can't figure how they arrived at it.The text book says when a cross takes place between AB and ab we get 4 progeny types,the 2 parental genotypes AB and ab and the 2 recombinants Ab and aB...First of I don't understand how a cross can yield the parental genotype.Then it also says that the rarest genotype is a product of multiple crossovers,so does it mean that most progeny will have parental genotype?If so then in the following question,it doesn't seem so.Can you please explain this phenomenon to me?
In drosophilia,the allele b gives black body (wild type is brown),at a different gene locus,wx gives waxy wings(wild type is non waxy),and at a third locus,the allene cn gives cinnabar eyes(wild type eyes are red)A female which is heterozygous for all three genes is test crossed.In this case ,the fly is a male homozygous for black ,waxy and cinnabar.
The following progeny are observed
5 normal
6 black,waxy,cinnabar
69 waxy,cinnabar
67 black
382 cinnabar
379 black,waxy
48 waxy
44 black,cinnabar
A.explain which of these progeny represent the parental chromosomes and which are results of recombination?
B.Draw the alleles in their proper position on the chromosme of the triple heterozygote.  

Answer
Hi, Geetha,

The question you've asked would need an entire Genetics lecture to adequately answer, but I'll try to do my best in an email.  Let me start with the stuff you probably already  know.

Usually, when we say "parental" genotype, we refer to the P generation that has given rise to an F1 under study.  In "traditional" crosses that means two parents that are *true breeding* (homozygous) for one or more traits are mated to produce offspring that are hybrid for all the traits in question.  

In the examples you're studying, you're considering *linked* genes--those located on the same chromosomes.  That means they will be inherited together in their original configuration on the P generation's chromosomes unless crossing over separates them.

If you start with a P generation in which one parent is AABB and the other is aabb, then 100% of the offspring will be AaBb.  This is your F1.  Every somatic cell of the F1 individual will have one chromosome with AB and one with ab.

When the F1 individuals manufacture gametes of their own, this is the point where some of the chromosomes may undergo crossing over and recombine so that the result--in some cases--will be a chromosome with Ab and a reciprocal one (the other partner in the crossover) with aB.  These are the *recombinant* chromosomes.  If there has been no crossover between the A and B loci, then the chromosomes will remain in the parental configuration (AB and ab).

Crossovers that separate linked genes are not all that common, and most of the gametes produced by meiosis will not have crossover between the loci you happen to be studying.  This is why MOST of the gametes resulting from meiosis will remain in the parental configuration (AB and ab).

If one crossover is rare, then multiple crossovers are even more rare, and so it doesn't often happen that ONE crossover creates a recombinant configuration (Ab and aB) only to have a second one occur at a different spot between your loci of interest to restore the original parental configuration (AB and ab).  The two crossovers must occur between the A and B loci to have this effect, and the closer the two loci are together, the less likely this is to happen.

This movie might help you visualize what's going on:

http://www.bio.miami.edu/dana/mov/crossing_over.mov

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The second part of your question, regarding a three-point cross used for linkage mapping, is related to the first part, but will still require some work.

In the problem you state above, you say that you have a heterozygous female who is test-crossed to generate the numbers of offspring you listed.  Note that in this case "heterozygous" does not mean "trihybrid."  In order to get the proportions of offspring in the cohort you itemize, the parental chromosomes cannot be from true-breeding parents.

The first step is to pair up the reciprocal classes of traits, meaning pairing up the chromosomes that were produced by a crossover event in the same location, relative to the three loci.  You've already listed them in reciprocal class, so I'll assume you know how to do that.  

Your next question is the good one:  Didn't the book say that the most common result should be parental configuration?  If so, then why are "wild type" and "fully mutant" the *least* common configuration, with the most common reciprocal group being "cinnabar" vs. "black, waxy."

The correct answer requires you to think outside the box, using the fact that the most common configuration *is* the parental configuration!  This means that the parents of the heterozygous female in your breeding experiment were:

cinnabar cncn (chromosomes:  b+b+ cncn wx+wx+)
black waxy bb wxwx (chromosomes:  bb cn+cn+ wxwx)

Since the *least* common configurations are the wild type and fully mutant (which are reciprocals of each other), that must mean that a *double crossover* was required to create that configuration.  And that means that the cinnabar locus is the *central marker*.  There had to be two crossovers--one between b an cn *and another* between cn and wx--to make the resulting rare chromosomes:

b+b+ cn+cn+ wx+wx+ (wild type)  
and bb cncn wx wx (black, waxy, cinnabar)

The parents of the heterozygous female, then, were:

b+b+ cncn wx+wx+ and bb cn+cn+ wxwx

NOT b+b+ cn+cn+ wx+wx+  and bb cncn wx wx as you might automatically think if your heterozygous female was an F1 trihybrid progeny of two true-breeding strains for all three traits, which she was NOT.

So to answer your question A:  

b+b+ cncn wx+wx+ and bb cn+cn+ wxwx are your parental chromosomes, and all the other configurations are recombinants.

I'm not sure I can *draw* the chromosomes of the heterozygote in a line using the text email here, but the order of the three loci on her two chromosomes would look like this:

b+b+ cncn wx+wx+/bb cn+cn+ wxwx

Does that make sense?

I know linkage and mapping are tricky, and I hope this helped a little bit.  If not, please write back and I'll try to clarify any rough spots.

Good luck!

Dana