• Gene product interaction
• Complementation 

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• Recombination by Crossover of Chromosomes

In real living organisms, all traits result from the interaction of many gene products.  Suppose two enzymes are required to produce a trait.  What may be the result?  What will happen to Mendelian inheritance ratios?
 

• Recessive epistasis.  The presence of recessive alleles at one locus makes useless the presence of dominant alleles at another locus.  This happens if two enzymes are needed in series; "the chain breaks" if either link fails.  The Mendelian ratio of a dihybrid cross is 9:7.
• Dominant epistasis.  The presence of one dominant allele at one locus compensates for the lack of dominant alleles at the other locus.  If it works in both directions, the ratio from a dihybrid cross is 15:1.

1. In sweet peas, purple is dominant; white is recessive.  Suppose two recessive white flowers from different breeders are cross-bred together.  The flowers--surprise--show the dominant purple color.  When self-crossed, the ratio is 9 purple: 7 white.  Explain these results.

2.  In fancy fowl, feathers are dominant to non-feathered.  Two pure-breeding feathered birds from two different fowl fanciers are bred, producing lovely feathered offspring.  The offspring are intercrossed, and a few of their offspring have no feathers.  The ratio is:
15 feathered: 1 non-feathered.  Explain these results.

3. Beetle colors. 

Suppose the color of beetles is determined by this enzyme pathway.  Enzymes A and B are needed in series, but enzyme C represses enzyme A.

Figure out what dihybrid crosses would produce what ratios.  Remember that you have to specify the entire genotype, including the purebreeding locus; for instance:
AABbCc would be a dihybrid cross for B and C, but AA would be the same for all offspring.

•   9   : 4  : 3 
• 12   : 3  : 1 
• 13  : 3 

Complementation refers to the fact that mutations producing null alleles in different genes can be complemented, or compensated in effect, by the gene product provided elsewhere.  An example of complementation is two-way recessive epistasis: The two white strains of sweet pea complement each other:

P1 P1 p2 p2   X   p1 p1 P2 P2  -->  P1 p1 P2 p2  (purple flowers)

You can use complementation analysis to intercross a large number of independently isolated null phenotypes, and figure out how many different genes are required for the enzyme pathway to produce the phenotype.

See the MIT site on Complementation. 

Two DNA helices may "cross over" or recombine, a process requiring breakage of each strand at the phosphodiester backbone and ligation to another strand.  There are two different classes of recombination:

• Homologous recombination occurs when two homologues (regions of nearly identical gene sequence) associate and exchange.  This type of recombination can occur with many different sequences, but the two sequences which actually recombine must be nearly identical.
• Site-specific recombination occurs when a gene (or phage genome, or plasmid) needs to insert itself into (or excise itself out of) a larger genome.  This recombination involves little or no homology, but involves enzymatic recognition of a particular short DNA  sequence.

Problem:  Is hin recombinase involved in homologous recombination, or site-specific recombination?  Explain your answer.

Homologous recombination

This image series shows how two homologous chromosomes within a tetrad cross over and exchange portions of their arms. (Problem: What phase of meiosis?)  The blue and red arms designate homologues from different parents.  (Why are they each double already?)Click the image:

What happens to Mendelian ratios when two genes are linked on the same chromosome?
 

• Random reassortment.  If the map positions, or loci, of the two genes are far enough apart so that many crossovers occur between them, then they may appear to reassort independently, with Mendelian ratios.  This is actually what happened with several of Gregor Mendel's pea traits.
• Deviation of ratios--"parental" vs. "recombinant."  If the loci are near enough so that recombination is less than 50%, then linkage will be favored for the pattern of alleles that was the same for each parent.

                A    B            a    b                             A    b            a    B
Parents   --------     X    -------                            --------     X    -------
               A    B            a    b                              A    b            a    B

                                     A    B                                          A    b
F1                               --------    < Same phenotype! >    --------
                                    a    b                                            a    B
 

Test Cross                       45%          A    B           5%
with  a    b                                         ---------
         --------                                         a    b
          a    b
                                        45%            a    b     5%
                                                            ---------
                                                            a    b

                                               5%          A b             45%
                                                             --------
                                                              a    b

                                               5%        a    B       45%
                                                             ---------
                                                              a    b

Which classes are "parental"?  "Recombinant"?

Map Distances.
For distances <10%, where the proportion of double crossovers is small, the rate of recombination is approximately proportional to the physical distance along the DNA helix.   So one can map the order of genes on a chromosome by intercrossing strains with closely linked markers.
1% recombination = 1 map unit

Problem: What happens when there are double crossovers?  How does this interfere with map calculations?

LOD scores and pedigrees.
Today we use statistical analysis of human pedigrees to calculate the linkage between particular diseases and regions of DNA that may contain the gene for the disease.  The "log of the odds" (LOD) score yields the probability of a given recombination frequence for a given set of pedigree data.