Inheritance

Single Gene Inheritance

To consider simple inheritance looking at the example of the Black Syrian Hamster: One of the genes mutated in the normal wild coloured Golden to give a self black hamster and the mutation is recessive - the symbol a was assigned to the Black muation. In other words the normal A gene mutated to give a and the result of this mutation is a self black hamster.

If a Golden female is mated with a Black male the chromosomes in the egg contain the normal unmutated gene A and the chromosomes in the sperm contain the Black gene a. These chromosomes (and genes) combine during fertilisation to form a cell containing Aa. Therefore all offspring from this mating have one normal gene A and one Black gene a.

The normal gene is dominant and so the characteristics of that gene are seen in the offspring but the Black gene although not seen in the offspring is part of their genetic make up. Therefore the phenotype (appearance) is Golden but the genotype (genetic make up) is Aa.

If two of these offspring are mated (Aa to Aa) the chromosomes containing these genes are passed from the parents in the egg and sperm. Therefore A and a are passed from the mother and an individual egg could equally contain A or a. In the same way A or a are passed from the father in each sperm. Therefore there are four possible outcomes during fertilisation (depending on the way in which the genes are passed to the germ cells from the parents) and these are:

Egg

Sperm

Offspring

A

A

AA

A

a

Aa

a

A

aA

a

a

aa

Another way to show this is in the form of a table:

Father

Mother

A

a

A

AA

Aa

a

aA

aa

The AA offspring contain only the dominant normal colour genes, the Aa and the aA offspring contain both the dominant normal gene and the recessive Black colour gene and the aa offspring contain only the recessive Black colour gene.

AA, Aa and aA all have the phenotype (appearance) of a Golden hamster due to the dominant normal A gene being present whilst aa are Black containing only the mutant Black gene.

Therefore the mating of Golden AA to Black aa results in 100% Aa (Golden carrying Black) offspring and mating two of these together (Aa x Aa) results in 25% Golden AA, 50% Golden carrying Black Aa and 25% Black aa offspring - ie 75% Golden, 25% Black.

If a Golden carrying the Black gene (Aa) is mated with a Black (aa) the germ cell from the Golden parent can equally contain A or a whilst the germ cell from the Black parent can only contain a.

Showing Golden carrying Black (Aa) mated to Black (aa) in the form of a table:

Golden carrying Black

Black

A

a

a

Aa

aa

Therefore offspring will be 50% Golden (carrying Black) Aa and 50% Black aa.

In the same way mating Golden AA to Golden carrying Black Aa the possible outcomes of fertilisation are:

Golden carrying Black

Golden

A

a

A

AA

Aa

Therefore offspring will be 100% Golden but 50% will be Golden carrying the Black gene (Aa) and 50% will be pure Golden (AA).

Where a hamster is said to carry another colour gene ie Golden carrying Black it is said to be split for that colour ie Golden split for Black.

Two Gene Inheritence

Where two mutant colour genes are involved things are a little more complicated but the basic principles of inheritance still apply.

Considering the Black Eyed Cream Syrian Hamster mutant gene e which removes black and brown pigment to produce a self cream hamster and the Cinnamon mutant gene p which changes the eye colour to red and reduces pigmentation it should be remembered that both genes are mutations from the 'normal'. Therefore the normal E gene mutated to e which gave a Black Eyed Cream Hamster and the normal P gene mutated to p to produce a Cinnamon Hamster.

The E and P genes are entirely different genes located at different loci on the chromosomes. Therefore the Black Eyed Cream has a mutant version of the E (e) gene but has the normal P gene. In the same way the Cinnamon has a mutant version of the P (p) gene but has the normal E gene.

Therefore when considering two gene inheritance both genes involved must be taken into account for both hamsters. Therefore the Black Eyed Cream should be considered as being eePP and the Cinnamon should be considered as being EEpp.

During reproduction the sister chromosomes with the genes will come together and then move to opposite ends before the production of the germ cell which will contain half the number of chromosomes and genes. The germ cell produced by the Black Eyed Cream will therefore contain eP. In the same way the germ cell from the Cinnamon hamster will contain Ep. On fertilisation the two germ cells will combine to produce a cell containing EePp.

Therefore all offspring will be of genotype EePp ie Golden. The reason the offspring are Golden is because there is a dominant normal gene E hiding the mutant Black Eyed Cream gene e and also a dominant normal gene P hiding the mutant Cinnamon gene p. However, both mutant genes are carried and therefore the offspring are Goldens which carry both Black Eyed Cream and Cinnamon.

If two of these offspring are mated (EePp to EePp) the chromosomes containing these genes are passed from the parents in the egg and sperm. Both parents will pass one E or e gene and one P or p gene. Therefore both parents can produce a germ cell that contains one of the following combinations: EP, Ep, eP or ep.

Showing this in the form of a table as before enables calculation of the genotypes of the offspring that will be produced from this mating.

Parent 1

Parent 2

EP

Ep

eP

ep

EP

EPEP
Golden

EPEp
Golden

EPeP
Golden

EPep
Golden

Ep

EpEP
Golden

EpEp
Cinnamon

EpeP
Golden

Epep
Cinnamon

eP

ePEP
Golden

ePEp
Golden

ePeP
Black Eyed Cream

ePep
Black Eyed Cream

ep

epEP
Golden

epEp
Cinnamon

epeP
Black Eyed Cream

epep
Red Eyed Cream

Wherever one of the offspring contains both a normal E and a normal E gene the offspring is Golden as it contains both dominant normal genes. Wherever one of the offspring contains a normal E gene but only contains the mutant p gene the hamster is a Cinnamon. In the same way wherever one of the offspring contains the normal P gene but contains only the mutant e gene the hamster is Black Eyed Cream. Where the hamster contains only both mutant e and p genes a new colour is created - the Red Eyed Cream with genotype eepp. Therefore the expectation is Golden:Black Eyed Cream:Cinnamon:Red Eyed Cream in the ratio 9:3:3:1.

Three Gene Inheritence

Where three or more mutant genes are involved the same principles apply. For example, mating a Red Eyed Cream eepp to a Dark Grey dgdg, the Red Eyed Cream contains the mutant genes e and p and the Dark Grey contains the mutant gene dg. Again it should be remembered that all genes are mutations of a normal gene and therefore the normal E mutated to give e, P to give p and Dg to give dg. Considering all the genes for both hamsters the Red Eyed Cream can be symbolised as DgDgeepp (with the normal Dg gene but mutant e and p genes)and the Dark Grey as dgdgEEPP (with the normal E and P genes and the mutant dg genes).

During reproduction the sister chromosomes with the genes will come together and then move to opposite ends before the production of the germ cell which will contain half the number of chromosomes and genes. The germ cell produced by the Red Eyed Cream will therefore contain Dgep. In the same way the germ cell from the Dark Grey hamster will contain dgEP. On fertilisation the two germ cells will combine to produce a cell containing DgdgEePp.

Therefore all offspring will be of genotype DgdgEePp ie Golden. The reason the offspring are Golden is because there is a dominant normal gene E hiding the mutant Black Eyed Cream gene e, a dominant normal gene P hiding the mutant Cinnamon gene p and a normal Dg gene hiding the mutant dg gene. However, all mutant genes are carried and therefore the offspring are Goldens which carry Black Eyed Cream, Cinnamon and Dark Grey.

If two of these offspring are mated (DgdgEePp to DgdgEePp) the chromosomes containing these genes are passed from the parents in the egg and sperm. Both parents will pass one E or e gene, one P or p gene and one Dg or dg gene. Therefore both parents can produce a germ cell that contains one of the following combinations: DgEP, DgEp, DgeP, dgEP, dgEp, dgeP, Dgep, dgep.

Showing this in the form of a table as before enables calculation of the genotypes of the offspring that will be produced from this mating.

Parent 1

Parent 2

DgEP

DgEp

DgeP

Dgep

dgEP

dgEp

dgeP

dgep

DgEP

DgEPDgEP

DgEpDgEP

DgePDgEP

DgePDgEp

dgEPDgEP

dgEpDgEP

dgePDgEP

dgepDgEP

DgEp

DgEPDgEp

DgEpDgEp

DgePDgEp

DgePDgEp

dgEPDgEp

dgEpDgEp

dgePDgEp

dgepDgEp

DgeP

DgEPDgeP

DgEpDgeP

DgePDgeP

DgePDgep

dgEPDgeP

dgEpDgeP

dgePDgeP

dgepDgeP

Dgep

DgEPDgep

DgEpDgep

DgePDgep

DgePDgep

dgEPDgep

dgEpDgep

dgePDgep

dgepDgep

dgEP

DgEPdgEP

DgEpdgEP

DgePdgEP

DgePdgEp

dgEPdgEP

dgEpdgEP

dgePdgEP

dgepdgEP

dgEp

DgEPdgEp

DgEpdgEp

DgePdgEp

DgePdgEp

dgEPdgEp

dgEpdgEp

dgePdgEp

dgepdgEp

dgeP

DgEPdgeP

DgEpdgeP

DgePdgeP

DgePdgep

dgEPdgeP

dgEpdgeP

dgePdgeP

dgepdgeP

dgep

DgEPdgep

DgEpdgep

DgePdgep

DgePdgep

dgEPdgep

dgEpdgep

dgePdgep

dgepdgep

As can be seen, of the 64 possible combinations of germ cells only one contains only the three mutant genes (dgdgeepp) and none of the normal genes, producing a new colour - Red Eyed Ivory.