You’d know from the previous post that regular body cells such as kidney cells or liver cells undergo mitosis to make identical 2n copies of themselves that regenerate the organs and body as older cells die.
Meiosis has a different purpose. It is a specialised and more complicated form of cell division restricted to the sex organs. Through meiosis, ovaries make egg cells and testes make sperm cells containing half the number of chromosomes (half the genetic content of the parent). Meiosis makes n cells from 2n cells, and this occurs only in the sex organs.
When an n sperm fertilises an n egg, its genetic material enters the egg’s nucleus and the egg acquires a full number (n + n = 2n) of chromosomes. With this complete set of genetic material, half from each parent, the fertilised egg then divides via regular mitosis, and again, and again, until after many, many divisions and a lot of complicated processes, a new and complete organism forms. In time that organism will produce its own eggs or sperm via meiosis to continue the cycle.
We’ll now step through the entire process of meiosis — and knowing how cells divide through mitosis will make this much easier to follow, as you will see!
Meiosis comprises two stages of division, Phase 1 and Phase 2. Each phase is broken down into interphase, prophase, metaphase, anaphase and telophase as with mitosis.
Phase 1 The first division. At the end of this stage the cell has divided, but the chromosomes have not separated. The following images will help explain.
Interphase 1 The chromosomes replicate themselves as they would in mitosis.
Prophase 1 Again as in mitosis, the chromosomes are visible if stained, and appear as duplicates joined at a centromere.
In the 2n cell above, ignore the colours and focus on the two pairs of chromosomes that are similar in shape: the long pair and the short pair. Each of these pairs are called homologous pairs (from the Greek homos, “same” and logos, “relation”). A 2n cell has n pairs of homologous chromosomes. In the diagram below one of each homologous pair is dark grey and the other light grey to make it easier to follow the movements of these pairs during meiosis:
Metaphase 1 The homologous pairs line up along the equatorial plane of the cell. The nuclear membrane disappears at the end of this stage as it does during mitosis.
Anaphase 1 There is a crucial distinction here between mitosis and meiosis. The chromatids separate at this stage in mitosis, but in meiosis the chromatids stay together and it is the homologous pairs that separate. it is one of each pair that goes to either pole of the cell in anaphase 1.
Telophase 1 As in mitosis, a nuclear membrane forms around each group of chromosomes, the cell membrane indents, and two new cells result.
But as you can see from the colour-coding, these cells are not identical as they would be after mitosis. This is another key difference with meiosis at this stage. Homologous chromosomes look the same and carry the same genes, but they do not necessarily carry the same versions of those genes.
Let’s assign some values to these homologous chromosomes to make this clearer. Going back to Mendel’s peas, let’s have the long chromosomes carry the genes for seed colour (yellow or green) and the short chromosomes carry the genes for seed shape (round or wrinkled). Let’s further have the light grey chromosomes carry the dominant gene (’Y’ or ‘R’) and the dark grey chromosomes carry the recessive gene (’y’ or ‘r’). (Please note this is very simplified and purely for illustration — pea plants have more chromosomes and genes than this.)
Phase 2 The second division. At the end of this stage the chromosomes have separated but have not replicated (they did that in Phase 1).
Interphase 2 This is a very brief stage in Phase 2.
Prophase 2 The chromosomes appear as the original pair in telophase 1. Here we’ll continue to use our illustrative pea chromosomes — can you see Mendel’s law of segregation in effect? The ‘Y’ and ‘y’ genes have separated into different cells, as have the ‘R’ and ‘R’ genes.
Metaphase 2 The chromosomes again line up across the middle of each cell:
Anaphase 2 The chromosomes separate and one of each pair goes to opposite ends of the cell.
Telophase 2 The nuclear membrane reforms and the cell divides. The end result is four n cells from one 2n cell.
Here we ended up with two Ry cells and two rY cells.
If you go back to the metaphase 1 diagram, you’d see how both light coloured chromosomes could just have easily been aligned at top. Following through the steps, with that scenario we’d have ended up with two RY cells and two ry cells instead. Can you see how Mendel’s law of independent assortment applies here?
And that is meiosis! The example here with four original chromosomes was a very simple one. Imagine the genetic variation in the gametes of dogs (starting with 39 homologous pairs) or cattle (30 homologous pairs)!
We’ve seen how Mendel’s first two laws apply at the cellular level: next we’ll go deeper still into the chromosome and cover Mendel’s third law, the law of dominance. But with a twist, for not all inheritance is Mendelian!
From the last post you’ll know that meiosis results in four n cells from one 2n cell. These n cells are known as gametes or sex cells — males make sperm sex cells and females make egg sex cells.
Still another term for an n cell is haploid cell. When a haploid sperm fertilises a haploid egg, the two nuclei merge to form a 2n, or diploid, cell. The haploid cells in the diagram below have two distinct chromosomes each (n = 2). All four chromosomes end up in the fertilised egg, but because the four are really two homologous pairs, we say 2n = 4:
Let’s add our pea genes from the previous post to make a very simplified diploid pea cell:
This RrYy diploid cell now carries the genes for round (’R') pea shape, wrinkled pea shape (’r'), yellow seed colour (’Y') and green seed colour (’y'). Should it undergo mitosis and develop into an adult pea plant, it will only ever produce round and yellow seeds despite carrying the other genes, for round is always dominant to wrinkled, and yellow is always dominant to green.
Though this plant will only ever have round yellow seeds, it is heterozygous for both seed shape and seed colour. Thus via Mendel’s first law of segregation and his second law of independent assortment covered here at the cellular level, that plant will produce sex cells of RY, Ry, rY and ry combinations.
Mendel knew from his pea experiments that ‘R’ and ‘Y’ are dominant over ‘r’ and ‘y’ respectively, and derived his third law of dominance from these observations. He knew that a dominant character would express itself whether in the presence of a recessive one or not, but did not know the underlying reasons for this.
To explain what dominance actually is, let’s backtrack a bit.
An organism’s entire genome (its complete set of DNA, or complete set of genetic material) is spread over its full complement of chromosomes. A dog’s genome is spread over 78 chromosomes for example. You’ll know from Meiosis that chromosomes in a diploid cell exist as homologous pairs. A dog would thus have 39 homologous pairs, with one homologue of each pair having come from the mother and the other having come from the father.
Geneticists designate an order and number to the homologous pairs of species’ genomes. The largest chromosome pair is usually ‘1′, the second largest ‘2′ and so on. The very last pair is always the sex chromosomes and not numbered, but rather called X and Y in mammals or Z and W in birds.
Each homologous pair carries specific genes. In dogs, a gene for brown coat colour is found only on their chromosome 11, while a gene linked to blue eye colour is on their chromosome 18. The gene for pea seed colour is on a pea’s chromosome 1, while the gene for seed shape is on chromosome 7.
A pea has two ‘1′ chromosomes, and thus has two copies of the seed colour gene, one on each homolog. If these copies are identical (eg both are ‘Y’ or ‘y’), we say the plant is homozygous for that trait. If the copies are different (eg one is ‘Y’ and one is ‘y’), the plant is heterozygous for that trait.
Not only are specific genes found on specific chromosomes, they are also found at specific locations on those chromosomes. A gene location is called a locus (plural: loci). Bearing in mind that the following is illustrative only, we could represent our pea genes like so:
The locus is named for the dominant form of the gene that resides there. One homologue in the above diagram carries the ‘Y’ version of the gene at the Y locus, while the other carries the ‘y’ version of the same gene. Similarly for ‘R’ and ‘r’ at the R locus. People refer to ‘dominant’ and ‘recessive’ genes — and I’ve done the same in this blog to simplify things where appropriate — but it is more correct to refer to dominant and recessive versions of a particular gene. And like everything in biology (!) these different versions of the same gene have their own name: alleles. In the diagram above, ‘Y’ and ‘y’ are the alleles for seed colour at the Y locus, and ‘R’ and ‘r’ are the alleles for seed shape at the R locus.
Though a pea plant may carry both the ‘Y’ and ‘y’ allele, something about the ‘Y’ allele dominates, as if the ‘y’ were never there. What makes a particular allele dominant?
It may surprise you to learn that a gene is nothing more than a set of instructions for making a particular protein. It may surprise you even more to learn that most enzymes and some hormones are proteins or peptides (protein subunits)! Thus proteins are crucial to the functioning of an organism.
It follows that a dominant allele is simply a piece of genetic code that makes a protein which ‘out-competes’ the protein coded by the recessive allele. In some cases the recessive allele codes for a ‘broken’ (non-functioning) protein and the dominant protein ‘wins’ by default. This is actually the case with the Y allele in pea plants.
A plant with yellow peas produces a protein which switches on a particular set of genes in the peas. That set of genes in turn produces proteins which destroy chlorophyll. Chlorophyll is a green pigment, thus destroying it leaves the peas a yellow colour. A plant with green peas is homozygous recessive — it has two copies of the ‘r’ allele. The ‘r’ allele codes for a broken protein: it is defective and cannot switch on the chlorophyll-destroying genes. With no ‘R’ allele to produce a functioning protein, the chlorophyll is undisturbed in the pea, and the pea stays green.
This example of one allele dominating over another is called complete dominance, and is classic Mendelian inheritance. The heterozygous phenotype is indistinguishable from the homozygous phenotype. Examples abound in the animal world too, with simply-inherited traits typically expressed this way. The polled (hornless) allele in cattle is completely dominant to the horned allele. The suri fleece type in alpacas is completely dominant to the huacaya fleece type. Black feathers in chickens are dominant to red feathers.
Mendel was very lucky to have chosen seven pea traits that not only happened to mostly reside on separate chromosomes, but which also followed complete dominance. More so when you know that a pea plant has 2n = 14, ie he chose seven genes on five out of a possible seven chromosomes!
As the field of genetics advanced, scientists discovered that not all inheritance was as simple or predictable as Mendel’s observations. These forms of inheritance became known as non-Mendelian inheritance, and to know and understand these emphasises not only how random inheritance truly is, but how lucky Mendel was!
This post will discuss non-Mendelian inheritance, but first let’s recap Mendelian inheritance.
Mendelian inheritance relies on three laws:
the law of segregation
the law of independent assortment
the law of dominance
It’s important to stress that Mendel was referring to the segregation and independent assortment of what he termed ‘factors’, what we now call genes. His results suggested that these ‘factors’ of heredity behaved as units, or particles, that segregated and assorted independently of, and which were completely uninfluenced by, each other. One unit was as likely to end up in an organism with any other, unrelated, unit as not.
It wasn’t until 1902 that the concept of chromosomes and their being the carriers of multiple ‘factors’ (genes) was even suspected. In other words, genes did not exist as isolated particles, but instead resided on chromosomes with other genes.
Up until now this discussion has glossed over chromosomes as being the carriers of multiple genes, and focussed instead on tracking four alleles each on its own chromosome during meiosis. Showing these chromosomes independently segregating and assorting inferred the same behaviour on the alleles they carried — if chromosomes assorted independently, it made sense that the alleles did too.
Chromosomes do independently segregate and assort. But because they carry multiple genes, Mendelian inheritance doesn’t always apply at the gene level, so let’s explore the concept of non-Mendelian inheritance further with examples.
In the diagram below you’d expect the ‘Y’, ‘y’, ‘Z’ and ‘z’ alleles to stick with their respective homologues, ie you’d expect ‘Y’ and ‘Z’ to be inherited together and likewise for ‘y’ and ‘z’:
Here we say that the Y and Z loci are linked as they are on the same chromosome. Mendel’s first law of segregation still applies, as the ’Y’ and ‘y’ alleles, and the ‘Z’ and ‘z’ alleles, will still separate when their homologues separate during meiosis. But how do the genes at the Y locus independently assort from the genes at the Z locus if they are inherited together? This linkage of genes is the exception to Mendel’s second law.
Even so, this is an exception and not a complete breakdown of the law, for chromosomes have a tendency to cross over! This is when homologues exchange equivalent pieces of themselves during prophase 1 of meiosis before the chromosomes separate into separate gametes. Here, segments around the Y loci have been exchanged:
As the break occurred between the Y and Z loci, those alleles are now linked in a different combination — ’y’ is now with ‘Z’ and ‘Y’ is now with ‘z’ — and we say they have recombined. The genes at the Y and Z loci have still independently assorted as if they were on separate chromosomes the whole time. But there is a caveat.
Multiple crossovers are actually common during prophase 1, and loci that are far apart are the most likely to recombine often. There is less chance of breakages occurring between loci that are closer together, and thus more chance that closely linked loci are inherited together. This is represented below with the multiple loci bunched together at the tip of each chromosome having the same background colour as the respective Y locus:
Thus closely linked loci are more likely to be the exception to Mendel’s law of independent assortment.
Mendel’s third law of dominance is also subject to exceptions. This law assumes complete dominance, where one inherited allele is expressed and to the exclusion of its recessive counterpart. But as genetics progressed through the 20th century, it became more and more clear that other types of dominance existed.
Partial dominance, also called incomplete dominance, is where the expression of the heterozygote is somewhere between that of the two homozygotes. A classic example in animal breeding is the Andalusian fowl. Black feather colour is partially, not completely, dominant over white feather colour. A homozygous black chicken is fully black in colour and a homozygous white chicken is fully white in colour. Crossing the two produces offspring with slate-blue feathers. The white gene is a dilution gene which partially dilutes (washes out) the black colour. An all black chicken thus has no dilution of feather colour while a white chicken has fully diluted feather colour.
Codominance is where the expression of the heterozygote is exactly midway between that of the two homozygotes. Both genes are expressed equally. A classic example occurs in cattle: crossing a homozygous red bull over a homozygous white cow produces a roan calf, one with a coat of evenly mixed red and white hairs.
Overdominance is where the expression of the heterozygote outperforms that of the homozygous dominant genotype. An example is warfarin resistance in rats. The gene for resistance is dominant, and both heterozygotes and homozygotes for this gene are unaffected by warfarin. Warfarin thins the blood and prevents blood from clotting and in high enough concentrations will cause fatal internal bleeding. Vitamin K is a blood-clotting agent essential in the diet and counters warfarin’s action. Rats homzygous for warfarin resistance need a higher level of vitamin K than they can get naturally. Thus rats homozygously not resistant to warfarin as well as rats homozygous for resistance will both succumb to warfarin. Those that are heterozygous for resistance are unaffected. The warfarin-resistant gene is overdominant with respect to rat survivability: rats heterozygous for warfarin resistance outperform (survive) rats homoozygous for warfarin resistance.
Sex-linked inheritance is another example of non-Mendelian inheritance. The sex chromosomes in mammals are named X and Y. Females are XX and males are XY. Of genes found exclusively on the X chromosome (called X-linked genes), XY males will only have one copy while XX females will have two. A classic example of sex-linkage involves tortoiseshell cats. Tortoiseshell is a mixture of black and orange patches. All tortoiseshells are female*, because the gene for orange colouration is found only on the X chromosome. Possible genotypes in females are OO (orange), Oo (tortoiseshell) and oo (black). Males can only be O (orange) or o (black). * In very rare cases a male may be XXY (this would be called Klinefelter syndrome in humans) and thus tortoiseshell, but also sterile.
And then there is epistasis, still another cause of non-Mendelian inheritance! Epistasis is where genes at one or more loci determine how genes at another locus — and this could be on a different chromosome — are expressed. An example is Labrador Retriever coat colour. ‘Labs’ come in three colours: black, yellow, and chocolate. These colours are determined by genes at the B (black) locus and E (extension of pigmentation) locus.
A phenotypically black Lab could be genotypically BBEE, BBEe, BbEE or BbEe. A yellow Lab could be BBee, Bbee or bbee. From this you can see that the expression of the ‘B’ allele is dependent on the presence of at least one ‘E’ allele. A dog homozygous for ‘e’ will always be yellow regardless of whether it also carries the ‘B’ allele. But dogs with at least one ‘E’ allele but homozygous for ‘b’ will be chocolate — possible genotypes in this case are bbEe and bbEE.
These are just some examples of non-Mendelian inheritance. But from the ones outlined above you’d be getting a good feel for just how complicated patterns of inheritance can be, and how fortuitous Mendel was to have chosen the pea traits he did to have been able to unravel the mysteries of inheritance at all! You may also be getting a feel for just how random genetic inheritance can be, and next week we’ll explore that randomness some more with a bit of maths.
We saw earlier how chromosomes randomly assort to form sex cells during meiosis, and that the combination of chromosomes (and alleles) an organism acquires is down to which particular sperm fertilised which particular egg to make it ‘it’ and no other.
Zygote is the word for ‘fertilised egg’ in the biological/genetic worlds. A zygote is one-celled (unicellular), but through mitosis becomes a multicellular embryo, which through more mitosis and differentiation of those cells into organs and the like, develops further into a foetus until ready to be born. (It’s only natural for people to refer to foetuses so close to birth as babies or calves or puppies, but until the moment of birth they are still technically ‘foetuses’.)
So far we’ve looked at very simple models of one or two loci segregating and assorting— far too simple for the real world, but the intrinsic randomness of gene assortment could still be seen.
While this is a blog about animal breeding, let’s stick with peas a little longer as we explore this randomness further.
A Punnett square is a handy tool for working out the possible outcomes of a one locus mating. You’ve seen this one before, for heterozygous parents, but this time think of ‘R’ and ‘r’ as the possible gametes that can form from meiosis. Those gametes can combine in four ways:
From this we can expect three-quarters of the offspring to express the dominant ’round’ trait and one-quarter to express the recessive ‘wrinkled’ trait. Statistically, the phenotype of round seed to wrinkled seed should appear in a 3:1 ratio. We’d also expect one-quarter of all offspring to be homozygous dominant, half to be heterozygous dominant, and one-quarter to be homozygous recessive. Statistically, these three unique genotypes ‘RR’, ‘Rr’ and ‘rr’ should occur in a 1:2:1 ratio.
Please note the word ’statistically’— statistics calculates the probability of an event occurring, not the certainty of it. While we may expect to see these outcomes there is no guarantee they will occur. For example, there’s a 50% chance that a baby will be a boy or a girl, which implies a family would comprise of equal numbers of boys and girls. Yet, back when larger families were the norm, it wasn’t unusual to know of ones with, say, six boys and no girls, or four girls and no boys.
Statistics work best on very large data samples, and it is usually only after many many repeated events that the observed results begin to resemble the predicted ones. Look again at Mendel’s pea data— his results that best matched the predicted 3:1 ratio were those with literally thousands of progeny. His results with ‘only’ hundreds of progeny weren’t quite as close.
Now let’s calculate the probable outcomes of two-loci matings with a Punnett square as well, again assuming two heterozygous parents. You can see how quickly things get more complicated just by moving from four gametes to eight:
Here the predicted phenotypes round and yellow:round and green:wrinkled and yellow:wrinkled and green are in a 9:3:3:1 ratio. These four phenotypes are from nine unique genotypes, in a 4:2:2:2:2:1:1:1:1 ratio.
A Punnett square could in theory be used to determine the possible gametes and genotypes from any number of loci matings, but as the above example shows, this tool doesn’t scale very well. Feel free to confirm for yourself, but just three loci would increase the number of squares to 64 (eight possible gamete combinations from each parent), and four loci would require 256 squares (sixteen possible gametes from each parent)!
There’s a much easier way to calculate the number of gametes for any number of loci — deriving a formula!
An organism heterozygous at one locus can produce two unique gametes: A and a.
An organism heterozygous at two loci can produce four unique gametes: AB, Ab, aB and ab.
An organism heterozygous at three loci can produce eight unique gametes: ABC, ABc, AbC, Abc, aBC, aBc, abC and abc.
There’s a pattern here: 1 locus, 2 unique gametes per parent if heterozygous at that locus 2 loci, 4 gametes per parent if heterozygous at each of those loci 3 loci, 8 gametes per parent if heterozygous at each of those loci
Mathematically: 21 = 2 22 = 4 23 = 8
Or, 2n = number of unique gametes, where n is the number of loci. If we plug n = 4 into this formula, we get the 24 = 2 × 2 × 2 × 2 = 16 gametes per parent mentioned above.
Now let’s work out how many unique genotypes arise from these unique gametes.
From the one-locus Punnett square above, three unique genotypes are possible if both parents are heterozygous at this locus.
From the two-locus Punnett square above, nine unique genotypes are possible if both parents are heterozygous at both loci.
And a three-locus Punnett square would reveal twenty-seven unique genotypes if both parents are heterozygous at all three loci (colour-coded here to make these slightly easier to find).
Mathematically: 31 = 3 32 = 9 33 = 27
Or, 3n = number of unique genotypes, where n is the number of loci for which both parents are heterozygous.
But what if one of those parents is heterozygous at additional loci? How many unique genotypes then?
Let’s go back to locus C. If one parent is homozygous for this then it has for unique gametes: ABC, AbC, aBC and abC. If the other parent is heterozygous then its unique gametes increase to eight: ABC, AbC, aBC, abC, ABc, Abc, aBc and abc.
Being heterozygous at one locus has enabled the number of unique gametes to double, simply because the additional ‘c’ allele increased the number of unique combinations the alleles could sort into. For every ABC there can be an ABc, and so on.
I was going to delve into some more maths here but it was bogging this discussion down too much. I wrote a new post instead — for now please be satisfied that we can calculate the number of unique genotypes for any number of loci with this formula:
Number of unique genotypes = 3n × 2m where n is the number of loci both parents are heterozygous at, and m is the number of loci one parent is heterozygous at.
Another word for ‘genotype’ is zygote, our fertilised egg at the beginning of this discussion, that will ultimately grow and develop and be born as a living animal.
Thus the number of unique genotypes possible is really the number of unique zygotes possible from any union of any female and any male.
We’ve only looked at three loci so far, but animals are far more complicated that that — they are likely to have thousands or even tens of thousands of heterozygous loci.
Burt let’s assume a sire and a dam each heterozygous for 100 loci just for argument’s sake. Our formula becomes: 3n × 2m = 3100 × 20 = 3100 × 1 (20= 1. Any number to the power of zero is one) ≅ 5 × 1047(’≅’ means ‘approximately equal to’) ≅ 500, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000 ≅ 500 billion billion billion billion billion unique zygotes!
That is an eye-watering number of potential offspring.
And this number is assuming there are only two alleles per locus — there are plenty of genes with three or more alleles. Of course, one parent can only ever have two of these, but it’s conceivable that four different alleles are carried between both parents. This of course just adds to the sheer randomness of it all.
Never mind that the randomness increases with additional loci! Let’s add just two more loci, that only one parent is heterozygous at:
Mating two superior animals together does improve your chance of producing superior offspring, but because of the completely random nature of gene assortment there is no way to predict which particular combination of genes will end up in which gamete. There is no way to control which egg is fertilised by which sperm, and ultimately chance and even luck still factor in the genome you are dealt. Some gamete combinations may be so detrimental that the embryo dies very early on. Two animals who together produced a superb animal may never do so again, but at the same time a disappointing mating outcome doesn’t mean future ones will be.
So yes, there absolutely is a lot of randomness in genetic inheritance — far, far more than you may have thought. It is important to be aware of this randomness, and how, because of it, you do not have as much control over your breeding as you may think. But this is a blog about better breeding, and just knowing what you can’t control is a big step on the way to being a better breeder. We will be going deeper and deeper in our understanding, covering genes, statistics and selection strategies along the way.
But first, for completion, I do want to finish the maths touched on here — and there is now a post dedicated to just that here. After that we’ll begin a discussion on gene frequencies in populations!
This post explains the maths behind the formula 3n × 2m (the number of unique zygotes) mentioned in last week’s post.
Let’s go back to that three-locus Punnett square, where both parents are heterozygous at A, B and C. We determined visually — and not that easily! — that there are 27 unique genotypes (colour-coded here):
Break each of the three loci into three individual squares:
While there are four possible arrangements, there are three unique genotypes from each locus (remember that Aa and aA, Bb and bB, and Cc and cC are the same), so long as both parents are heterozygous at each.
Thus every possible zygote from those two parents would have a genotype comprised of (AA or Aa or aa) and (BB or Bb or bb) and (CC or Cc or cc).
In statistics, where we’d say ‘or’, we’d write ‘+’, and where we’d say ‘and’, we’d write ‘×’. Thus there are (any three of A) × (any three of B) × (any three of C) =3 × 3 × 3 = 33 = 27 unique genotypes possible.
Similarly you could go back to a simpler two-locus Punnett square, where we already know there are nine unique genotypes:
and confirm that there are (any three of AA or Aa or aa) × (any three of BB or Bb or bb) =3 × 3 = 32 = 9 unique genotypes possible.
And without even attempting to draw a four-locus Punnett square (!!), we could calculate the number of unique genotypes for two parents heterozygous at four loci to be (any three of A) × (any three of B) × (any three of C) × (any three of D) =3 × 3 × 3 × 3 = 34 = 81 unique genotypes possible.
Hopefully it’s much clearer now where the 3 in 3n comes from, and why it’s to the power of n loci? (Always remember that this is when both parents are heterozygous at n loci.)
But what about the 2m component of the formula, where m is the number of additional loci only one parent is heterozygous at?
Let’s go back to a two-locus Punnett square, but this time with only one heterozygous parent at one locus:
You can see that there are, in a sense, two duplicate sets, as parent 2 can only ever contribute a ‘B’ allele. It has no ‘b’, and because of this the number of unique genotypes at this locus is determined solely by parent 1. As there is no second ‘b’ to join up with another ‘b’, the three genotypes AAbb, Aabb and aabb can’t be formed.
The number of unique genotypes thus drops from nine: AABB AABb AaBB AaBb AAbb Aabb aaBB aaBb aabb
to six: AABB AABb AaBB AaBb aaBB aaBb
The possible genotypes at the A locus are still (AA or Aa or aa) but the possible genotypes at the B locus are limited to (BB or Bb). ‘bb’ is never possible.
Thus every possible zygote is now (AA or Aa or aa) and (BB or Bb). Or, (any three of A) × (any two of B) =3 × 2 = 6 unique genotypes possible.
Let’s add another locus, C, for which parent 2 this time is heterozygous. As with B, the possible genotypes are CC or Cc. Every possible zygote is now (any three of A) × (any two of B contributed by parent 1) × (any two of C contributed by parent 2) =3 × 2 × 2 = 3 × 22 = 12 unique genotypes possible.
Can you see how it doesn’t matter which parent is heterozygous at which locus? Parent 1 could have been heterozygous at both loci B and C, or parent 2, and the end result would be the same.
Let’s now work through all possible unique genotypes when two parents are heterozygous at two loci (A and B), and one parent is heterozygous at two additional loci (C and D), by combining what we’ve covered above:
The number of possible genotypes is (AA or Aa or aa) and (BB or Bb or bb) and (CC or Cc) and (DD or Dd).
Or, (any three of A) × (any three of B) × (any two of C) × (any two of D) =3 × 3 × 2 × 2 = 32 × 22 = 36 unique genotypes possible.
What about a fifth locus, E, at which both parents are heterozygous, and a sixth one, F, at which just one is heterozygous?
(any three of A) × (any three of B) × (any three of E) × (any two of C) × (any two of D) × (any two of F) =3 × 3 × 3 × 2 × 2 × 2 = 72 unique genotypes possible.
Or, 33 × 23.
The more variations of these combinations of heterozygous loci you construct, the more clear it is to see the pattern, or the formula 3n × 2m, where n is the number of heterozygous loci both parents have, and m is the number of heterozygous loci one parent has.
Please do note that this assumes only two alleles exist for each locus. As mentioned last week, many genes do contain three, four, or even more alleles, which does complicate the maths, but the important thing here is that you can see how 3n × 2m was derived in the first place, and hopefully I’ve done that here for you.