ZOOL 304

End-of-Chapter Questions

Chapter 5

• Question 5.1, about haplotypes.
• Questions 5.2, 5.3, relating allele frequencies to mutation-selection balance.
• Question 5.4, relating allele frequencies to heterosis.

Question 5.1 (text page 111)

The answer to this question should be fairly obvious.  If not, read on.

First, translate the jargon.  

Haplotype means "haploid multi-locus genotype".  "Haplotype" may refer either to the allele combination in an entire haploid set of chomosomes or to the allele combination along a particular chromosome.  The term is more strictly used with the latter meaning, to describe individual chromosomes.  Just as alleles are different versions of the same gene, so haplotypes are different versions of the same (homologous) chromosome.  

Two homologous chromosomes differ in haplotype if they differ at any gene (any locus) along the entire length of the chromosome.  Remember the metaphor of genes on chromosomes as "bead on a string"?  That metaphor is helpful here.  A locus is a bead position.  An allele is a bead of a particular color.  A haplotype is an entire string of beads, with beads of particular color at each position.

A novel haplotype is a particular string of alleles that is different from any other string that has gone before.  Such strings can form by mutation (of course), but also by recombination of pre-existing strings.

Imagine a pair of chromosomes as a two strings of beads lying side by side. Each bead position corresponds to a gene, and each bead is a particular allele.  At any given position, the pair of chromosomes is homozygous if the beads are the same color and heterozygous if they are different.  

Now cut each string at the same position and exchange the pieces.  Repeat if desired.  This is equivalent to recombination by crossing-over.  If the bead patterns were mostly the same before the cut, the exchanged parts might match completely and you couldn't tell that any exchange had taken place.  No new haplotypes would be formed.  But if there were several differences in the original strings, the recombined strings which result from the exchange would each be different from the original.  Each would represent a novel haplotype.

A "population with high genetic diversity" is one in which many genes (loci) are polymorphic (i.e, with more than one common allele).  If many genes are polymorphic, many individuals will be heterozygous at one or more genes.  Many strings (chromosomes) will be different, each with different-colored beads at many different positions.

Now, how do you get "novel haplotypes"?  

Let's explain this first for a simple, Mendelian, two-factor cross.

Actually, let's begin with a one-factor cross, with two alleles at the single locus.  We start with two homozygous parents (AA and aa), each bearing one of the two different alleles.  Together they can form two kinds of gametes, A and a, each of which is a different haplotype.  In each subsequent generation, we again get the same two haplotypes, A and a.  Recombination plays no role in this single-locus example.

Now let's consider a two-factor cross, with two loci on the same chromosome and with two alleles at each locus.  As in the single-locus example above, let's presume that our parents are homozygotes coming from different inbred lines, AB/AB and ab/ab.  Their offspring will be heterozygotes, AB/ab.  Without meiotic recombination, these heterozygous offspring (like those above) can form only the same two types of gametes, AB and ab, as their parents.  

However, with meiotic recombination by chromosomal crossing-over, these same heterozygotes can form four different haplotypes, AB and Ab and aB and ab.  Two of these, Ab and aB, are novel haplotypes (i.e., novel because they differ from either of the two parental haplotypes).

Now let's make the situation a bit more interesting by increasing the allelic diversity.  This time let's start with parents that are both heterozygous and different from one another, AB/ab and AB/ab so that there are four different alleles at each of the two loci.  Now there are four different parental haplotypes (i.e., one haplotype for each parental chromosome, AB, ab, AB, and ab).  With crossing over, as in the heterozygotes of the two-allele example above, each of these parents can form four haplotypes.  Together the gametes from these two parents include eight different haplotypes, AB, Ab, aB, ab, AB, Ab, aB and ab.  

After two more generations of recombination and fertilization, we can still recover the original set of four parental haplotypes AB, ab, AB, and ab as well as the four additional first-generation haplotypes Ab, aB, Ab, and, aB.  But by now we can also find eight more novel haplotypes, AB, Ab, aB, ab, AB, Ab, aB, and ab.  Thus, with meiotic recombination, we have gone in three generations from four haplotypes to sixteen haplotypes.  That's all, though.  That is as many haplotype combinations as are possible from two loci with four alleles apiece.

From these examples, we see that the number of possible chromosomal haplotypes equals the product that we can compute by multiplying together the numbers of alleles at all of the polymorphic loci.  

• 1 locus with two alleles = 1 x 2 = 2 haplotypes
• 2 loci with 2 alleles each = 2 x 2 = 4 haplotypes
• 2 loci with 4 alleles each = 4 x 4 = 16 haplotypes
• 4 loci with 4 alleles each = 4 x 4 x 4 x 4 = 256 haplotypes.  

In the examples above, with only a few polymorphic loci and only a few alleles at each polymorphic locus, all of the possible haploid allele combinations can appear within a few generations (as long as crossing-over is fairly frequent between loci).  

But with more loci and/or with more alleles, the number of possible haplotypes rapidly increases to astronomical size.  With only four alleles at each of fifteen loci, the number of possible haplotypes would exceed one billion.  With hundreds to thousands of loci per chromosome, it doesn't take a very high proportion of polymorphic loci before the total number of possible haplotypes vastly exceeds the population size.  Then every new generation produces haplotypes that have never before existed.  With intragenic crossing-over, recombination can even create novel alleles.

Now you should be set to answer the question.  Actually, its already been answered above, if you can sort through the verbiage.

Questions 5.2 and 5.3

You cannot just answer these questions on the basis of intuition.  You must perform an appropriate calculation using the information provided.  

There are three steps to answer these questions.  First, given the disease prevalence provided, you need to calculate the frequency of the recessive allele in the population.  Second, given the range of mutation rates provided, you need to calculate a range of expected allele frequency at mutation-selection balance (using the equation provided in the text).  Third, you need to compare these two results to determine whether "these diseases can be explained as an equilibrium between mutation and selection".

The biggest "catch" in these questions is the realization that disease prevalence is NOT the same as allele frequency.  The disease is a phenotype.  For a disease caused by a recessive allele, the prevalence of the disease is the frequency of the homozygous recessive genotype.  

To figure the allele frequency, you need to refer to the Hardy-Weinberg frequencies:

• [freq AA] = p²
• [freq Aa] = 2pq  
• [freq aa] = q²

From these Hardy-Weinberg frequencies, we know that a homozygote genotype frequency is the square of the allele frequency.  If we know the genotype frequency for the disease, we find the allele frequency q by taking the square-root.   

The square root of 1/10,000 is 1/100.  The square root of 1/100,000 is about 1/316.  The square root of 1/4000 is about 1/63.

Then compare this frequency with the expectation for mutation-selection balance for recessive mutations.  

The equations for mutation selection balance may be found on page 102.  These equations may be derived from the population genetic model of selection, simply by introducing a term for mutation rate.)  Note that the expectation differs depending on dominance.

q = equilibrium frequency of the mutant allele
u = mutation rate
s = selection coefficient against the mutant (i.e., fitness = 1-s )
       (For a lethal phenotype, s = 1 so fitness = 0.)

q = u / s  for dominant alleles.

q = square root ( u / s )  for recessive alleles.

Note that the square root of a fraction is larger than the fraction.  For example, the square root of 1% is 10%.  Thus the expected frequency for a recessive mutant allele is much larger than that for a dominant mutant allele, at mutation-selection balance.  And for recessive mutations, the prevalence of carriers (i.e., heterozygous individuals) is much, much higher at mutation-selection equilibrium than the prevalence of the disease.

One final note.  Don't interpret the quoted mutation rates (10^-4) to 10^-6) as rates for a specific mutation at a particular base pair.  Such per-base-pair rates would be much lower.  Efforts to measure deleterious mutation rates predate any practical means for determining the DNA sequence of particular alleles.  Thus mutation rates have traditionally been measured by counting any mutation which renders the gene non-functional as equivalent to any other such mutation.  Thus "the mutant allele" is really a set of mutant alleles, all with similar phenotypic consequences.  So the quoted range of mutation rates is approximately that for any major deleterious mutation within an entire gene.

Question 5.4

This question invites you to apply the result from Question 5.3 in the equation for heterosis, and calculate the selection coefficient for the heterozygote which would yield the observed allele frequency.

The relevant equation, along with explanation, is given on page 104.  

 

Notes for chapter 1 / 2 / 3 / 4 / 5 / 6 / 7 / 8 / 9 / 10 / 11 / 12 / 13 / 14 / 15 / 16 / 17

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