Chickens have 78 chromosomes. They are diploid animals, meaning that each cell in the body carries the chromosomes in pairs. Since there are 78 chromosomes, there are therefore 39 pairs (two of Chromosome 1, two of Chromosome 2 and etc). Each pair of chromosomes is nearly identical in the genes that they carry and the pair of chromosomes are known as 'homologous chromosomes' or 'chromatids' ie, same set of chromosomes. The exception to this rule are the sex-chromosomes, Z and W, where roosters have two Z chromosomes and hens have only one Z chromosome, plus one W chromosome. It is this difference in the sex-chromosomes that result in sex-linked characteristics.
Each chromosome contains many genes. The location or position along a chromosome where a gene may be found, is its locus (Latin for 'location'; plural: loci). "Alleles" are the alternate genes possible on a specific locus. For example, the lavender gene (lav) may only be found on the Lavender locus. As only one mutation has been known to occur at this locus, there is a total of two alleles possible, ie the lav mutation and the wild type allele: Lav+ (wild type alleles are identified by the plus symbol: + ). Therefore, the possible locus allele combinations on the two homologous chromosomes are:
Lav+ / Lav+
Lav+ / lav
lav / Lav+
lav / lav
As mentioned previously, sex chromosomes are different to the above autosome pairs, and different again between roosters (2 Z chromosomes) and hens (W & Z chromosomes). Males may have two alleles at the same locus on the sex chromosomes (ie. homozygous), or two different alleles at the same locus (ie. heterozygous). Females on the other hand can only have one allele of a gene since there is only one Z chromosome present. They are referred to as hemizygous.
Dominant / Recessive Inheritance
This inheritance mode is where a single copy of a dominant gene is needed for expression to occur (ie. the gene will be expressed regardless of whether the bird carries only one copy of the gene or two copies of the gene (heterozygotes & homozygotes respectively). However, two doses are needed for expression of a recessive gene (ie homozygotes only express the gene).
Why? To keep it simple (and to keep it that way, we will bend the truth of the matter just a little), if a bird has one copy of the Black gene and one copy of the Lavender gene, Black being the 'darker' colour will easily mask the presence of the lighter coloured Lavender. Hence, the bird appears black. When referring to the 'outward' physical attribute of an animal, we use the term 'Phenotype'. Therefore in our above example of a bird with both Black and Lavender alleles present, the 'Phenotype' is Black, but the 'Genotype' (the term used to describe its genetic makeup) would be Lav+ / lav (where Lav+ = Black and lav = Lavender).
Now, let's take that bent truth and bring it back to some level of truth. But we will use the word 'Working' to replace the term 'Dominant', and the words 'Not Working' to replace the term 'Recessive'. The truth hence is that if the Black gene is Working well, it will allow the accumulation of melanocytes (the bits of pigment that give the black colour) in the feather. As long as there is a copy of the Black gene that is Working, it will allow the melanocytes to accumulate in the feathers. If both copies of the Black gene are Not Working as well as they could be, it only allows the partial accumulation of melanocytes in the feathers, giving a 'less black' or 'grey' aka 'Lavender' colour.
You can see where we are going now. Hence, if the bird is to show any Lavender colour, it really needs to have two copies of the recessive Lavender gene (lav / lav), which in reality would be two copies of the Black wild type gene that are failing to work correctly. Why it should fail is another encyclopedia of information.
Let's take this example one step further into the breeding environment. There are two types of cells in our bodies: Autosomal cells and Gamete cells. Autosomal cells contain the full compliment of chromosomes, ie. it contains all 78 chromosomes. The gametic cells contain only half that number. Remember the chromosomes are in pairs, so the gametic cells contain one half of each pair, or 38 chromosomes + either the Z or W chromosome.
Both the male and female bird will produce gametic cells (sperm and ovule respectively) that come together during fertilisation to form the full compliment of 78 chromosomes. Coming back to our Black and Lavender genes, a rooster that is homozygous (has two copies) for the Black gene in its autosomal cells will in its gametic cell contain one copy of the Black gene (ie Lav+). A hen that is homozygous for the Lavender gene in its autosomal cells will in its gametic cell contain one copy of the Lavender gene (ie lav). So its really simple to see that the resulting embryo after fertilisation will contain one copy of the Black and one copy of the Lav gene (Lav+ / lav) in its autosomal cells. The offspring would be considered heterozygous for the Lavender trait (it would still be phenotypically black, but genotypically heterozygous for black and lavender).
This is the basic principle of genetics. No doubt the term 'Punnett square' (the term being derived from the inventor, Reginald C. Punnett) would be familiar to some. This is just a simple means to visualise the transference of genes to offspring and calculate the percentage of offspring that will carry a particular genotype. Let us stay with our example of crossing a homozygous Black rooster (Lav+ / Lav+) with a homozygous Lavender hen (lav / lav). The progenitor of a gametic cell contains the full compliment of chromosomes (ie 78 chromosomes). Through a process known as meiosis, a single progenitor cell doubles the number of chrosmosomes (ie to 156 chromosomes) whilst simultaneously dividing, resulting in 4 separate cells or gametes. Each of the 4 gametes will contain a quarter of the 156 chromosomes (ie 39 chromosomes). Through the magic of life, all 4 gametic cells contain different combinations of genes represented on the 39 different chromosomes. But let us just stick to the Lav+ and the lav genes, where we know that each parent bird only contains the one genotype. In the Punnett square below, the rooster and hen are each represented with two gametic cells and not four, because technically, the progenitor cell duplicated before dividing.
The following examples show us how to calculate the probablity of obtaining offspring of a certain colour depending on the parent's genetic makeup. The top row of Punnett square cells contain the genetics for the Male and the cells in the left column contains the genetics for the Female. The remaining inner cells contain the genetics for the offspring.
Example 1: One parent is homozygous Black (Lav+ / Lav+) and the other homozygous Lavender (lav / lav)
Female / Male Lav+ Lav+
lav Lav+/lav Lav+/lav
lav Lav+/lav Lav+/lav
We see from the punnett square that the coming together of the gametes after fertilisation give a result of:
100% Lav+/lav (heterozygous Black) - all visually Black but carry the Lavender gene.
Other dominant / recessive breeding combination examples are:
Example 2: Heterozygous Black (Lav+ / lav) paired to homozygous Black (Lav+ / Lav+).
This produces 100% Black phenotype, but 50% are carrying the Lavender gene:
Female / Male Lav+ Lav+
Lav+ Lav+/Lav+ Lav+/Lav+
lav Lav+/lav Lav+/lav
50% Lav+/ Lav+ (homozygous Black) - visually Black.
50% Lav+/ lav (heterozygous Black) - visually Black but carry the Lavender gene.
From examples 1 and 2, we see that the genotypes vary in percentage depending on the genetics of the parents. But in both cases, all offspring are the same colour, Black. This is a result of the 'dominance' of the Black gene.
Example 3: Heterozygous Black (Lav+ / lav) paired to homozygous Lavender (lav / lav).
Female / Male lav lav
Lav+ Lav+/lav Lav+/lav
lav lav/lav lav/lav
50% Lav+ / lav (heterozygous Black) - visually Black but carry the Lavender gene.
50% lav / lav (homozygous Lavender) - visually Lavender.
In this instance, 50% of the offspring have received two copies of the lavender gene (lav) and hence, without the presence of the Black (Lav+) gene, appear visually Lavender.
Keep in mind that the percentages calculated are but only a probability. It does not mean that if you hatch two chicks from the pair in Example 3, you will definitely get one Black and one Lavender chick. To obtain the predicted ratios of more complicated genetic combinations, it is necessary to hatch literally dozens if not hundrends of chicks from the same pair of birds.
Other forms of inheritance are:
Incomplete Dominant Inheritance
In the dominant / recessive examples, the Black, carrying lavender (heterozygote) bird is not distinguishable from the homozygous Black (not carrying lavender gene), as the black (wild type: Lav+) gene is completely dominant to the lavender gene. With Incomplete Dominance inheritance, the heterozygote gives an intermediate phenotype. For example, a Splash bird with the genotype Bl / Bl (two copies of the Blue gene) appears Whitish / Splash in colour. When paired to a Black bird (no blue genes: bl+ / bl+ ie. wild type), will produce an intermediate shade between the Whitish/Splash colour and Black, ie. Blue (genotype: Bl / bl+ ). Hence, depending on the combination of the alleles Bl and bl+, three different colours are possible: Whitish / Splash (Bl / Bl); Blue (Bl / bl+); and Black (bl+ / bl+). This is in contrast to the above dominant / recessive inheritance that only gives two colours: Black or Lavender. Hence, knowing if a gene is dominant or incompletely dominant helps us understand and predict potential progeny outcomes.
Sex-Linked genes are genes on the Z chromosome. Roosters have two Z chromosomes and hens have only one Z chromosome, plus one W chromosome. As the hens only have one sex-linked allele for any given locus, both dominant and recessive sex-linked genes are expressed with just one gene (as hemizygous).
If the sex-linked recessive allele is homozygous in the male, and the sex-linked dominant allele is found in the female, all offspring males will inherit the sex-linked dominant gene from the mother, and all offspring females will inherit the sex-linked recessive gene from the fathers. The following is a punnett square example of a sex-linked cross with the gene for Gold - fancy term for Brown - (s+), and Silver - fancy term for White - (S) colour found at the same locus and only on the Z chromosome. Silver (S) is the dominant colour (gene). The result is all female offspring will be the colour of the father and all male offspring the colour of the mother.
Silver Hen / Gold Rooster s+ (on Z) s+ (on Z)
S (on Z) s+/S (Silver males) s+/S (Silver males)
- (no chromosome) s+/- (Gold females) s+/- (Gold females)
50% s+/S (heterozygous Silver) - visually all Silver and all Males (remember the two copies of the Z chromosome).
50% s+/- (hemizygous Gold) - visually all Gold and all Females (only one copy of the Z chromosome, the other being the W)
Carrying out the cross in reverse (ie. Silver Male homozygous for the sex-linked dominant gene S, cross with a Gold Female hemizygous for the sex-linked recessive gene s+) will produce all offspring with the dominant gene S. Hence, all offspring regardless of sex, will be Silver. Try it with the Punnet square.
The above is but a tiny starting point towards understanding how genetics works. The important and very basic points to keep in mind are:
- Genes come in pairs (alleles) - although there can be exceptions to the rule;
- A pair of alleles are not different genes, but are variants of the same gene - with differences as minor as a single DNA base change;
- These variants often comprise a version that codes for a character / functions well / is switched on / etc; and a version that codes for a variant character / doesn't funtion well / is switched off /etc.
- Genes at different loci can interact with each other giving rise to other (perhaps unexpected) characteristics;
The above information has been adapted from the following website and courtesy of Karen Hall: Chicken Genetics