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Linked Genes

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There are four pairs of linked genes in Syrians known at this time. The X-Linked Yellow; Banded & Hair Length; C-locus & Cinnamon; Umbrous & Satin. For information on Yellow inheritance go to X-Linked below. The other 3 linked sets can recombine but with varying degrees of success.

Let's begin with Banded and Long Hair, since those genes are usually available and easier to understand. Banding is a dominant gene that only requires one gene to express it's phenotype. So BaBa and Baba will both produce a banded hamster. BaBa is called homozygous and Baba is heterozygous. Long Hair is a recessive gene. This means it has to be homozygous ll to express it's phenotype (physical appearance).

We'll start with a Homozygous Banded & Short Haired BaBa LL hamster mated to a Solid Long Haired baba ll hamster. All offspring will be Banded carrying a gene for Long Hair Baba Ll. Now we'll mate one of the offspring to the Long Haired baba ll hamster. You would expect the Punnet square to look like this:

Grandparent/
Parent baba ll
Long Haired
Grandparents are BaBa LL (Banded) and baba ll (Long Hair)
Parent Baba Ll Banded carrying Long Hair
Ba L ba l Ba l ba L
ba l Baba Ll
Heterozygous Banded
carrying Long Hair
baba ll
Golden Long Hair
Baba ll
Heterozygous
Banded Long Hair
baba Ll
Golden carrying
Long Hair

You would get one of each genotype and phenotype. However these genes are so close to each other they usually get inherited as a pair instead of individually. The ratio is roughly 9:1 for Parental type to Recombined type. So the actual outcome of this mating would be more like this:

Grandparent/
Parent bal/bal
Long Haired
Grandparents are BaL/BaL (Banded) and bal/bal (Long Hair)
Parent BaL/bal Banded carrying Long Hair
9 BaL 9 bal Bal baL
bal 9 BaL/bal
Heterozygous Banded
carrying Long Hair
9 bal/bal
Solid Long Hair
Bal/bal
Heterozygous
Banded Long Hair
baL/bal
Solid carrying
Long Hair

As you can see it is possible to get a cross-over but more difficult then the normal genetic pairings. You would get 9 Banded Short HairedBaL/bal, 9 Solid Long Hairedbal/bal, 1 Banded Long HairedBal/bal, and 1 Solid Short HairedbaL/bal. You'll notice that the parental genes are listed as BaL/bal instead of Baba Ll. This is used with linked genes to signify the link between the two genes from the parents. I've used different colors for the original parental linked genes (BaL and bal) and the recombined parental linked genes (Bal and baL) to show where the new linked pairs are. The above chart is a short one for Banded and Long Hair. Below is a sibling pairing.

Grandparents are BaL/BaL (Banded) and bal/bal (Long Hair)
Parents are Siblings Parent BaL/bal Banded carrying Long Hair
9 BaL 9 bal Bal baL
Parent
BaL/bal
Heterozygous
Banded carrying
Long Hair
9 BaL 81 BaL/BaL
Homozygous Banded
81 BaL/bal
Heterozygous Banded
carrying Long Hair
9 BaL/Bal
Homozygous Banded
carrying Long Hair
9 BaL/baL
Heterozygous Banded
9 bal 81 BaL/bal
Heterozygous Banded
carrying Long Hair
81 bal/bal
Solid Long Hair
9 Bal/bal
Heterozygous Banded Long Hair
9 baL/bal
Solid
carrying Long Hair
Bal 9 BaL/Bal
Homozygous Banded
carrying Long Hair
9 Bal/bal
Heterozygous Banded Long Hair
Bal/Bal
Homozygous Banded Long Hair
Bal/baL
Heterozygous Banded
carrying Long Hair
baL 9 BaL/baL
Heterozygous Banded
9 baL/bal
Solid
carrying Long Hair
Bal/baL
Heterozygous Banded
carrying Long Hair
baL/baL
Solid Short Haired

As you can see it is possible to get some recombining but more difficult then the normal genetic pairings. You would get the following phenotypes and genotypes:

  • 281 Banded Short Haired with genotypes: 81 BaL/BaL; 162 BaL/bal; 18 BaL/Bal; 18 BaL/baL; 2 Bal/baL
  • 81 Solid Long Haired with genotypes: 81 bal/bal
  • 19 Banded Long Haired with genotypes: 18 Bal/bal; 1 Bal/Bal
  • 19 Solid Short Haired with genotypes: 18 baL/bal; 1 baL/baL

The next linkage is with two recessive genes - C-locus and Cinnamon. There are currently 3 alleles for the Syrian C-locus: C is full color; cd is Dark Eared White; ce is Extreme Dilute. The linkage works the same for any allele on the C-locus since it is the location not the gene that causes the linkage. For this example I'll use the Dark Eared White (DEW).

Grandparents are cep/cep (Pale Eared White or PEW) and CP/CP (Golden)
Parents are Siblings Parent CP/cep Golden carrying DEW & Cinnamon
2 CP 2 cep Cp ceP
Parent
CP/cep
Golden
carrying DEW
& Cinnamon
2 CP 4 CP/CP
Golden
4 CP/cep
Golden
carrying DEW & Cinnamon
2 CP/Cp
Golden
carrying Cinnamon
2 CP/ceP
Golden
carrying DEW
2 cep 4 CP/cep
Golden
carrying DEW & Cinnamon
4 cep/cep
FEW
2 Cp/cep
Cinnamon
carrying DEW
2 ceP/cep
DEW
carrying Cinnamon
Cp 2 CP/Cp
Golden
carrying Cinnamon
2 Cp/cep
Cinnamon
carrying DEW
Cp/Cp
Cinnamon
Cp/ceP
Golden
carrying DEW & Cinnamon
ceP 2 CP/ceP
Golden
carrying DEW
2 ceP/cep
DEW
carrying Cinnamon
Cp/ceP
Golden
carrying DEW & Cinnamon
ceP/ceP
DEW

Phenotypes and genotypes will be:

  • 22 Golden with genotypes: 4 CP/CP; 8 CP/cep; 4 CP/Cp; 4 CP/ceP; 2 Cp/ceP
  • 4 FEW with genotypes: 4 cep/cep
  • 5 Cinnamon with genotypes: 4 Cp/cep; 1 Cp/Cp
  • 5 DEW with genotypes: 4 ceP/cep; 1 ceP/ceP

The last known linkage for Syrians is Satin and Umbrous. Both are dominant genes so there only needs to be one gene to express that phenotype. The ratio is also higher then the other two. This means it's less likely the parental genes will recombine.

Grandparents are SaU/SaU (Homozygous for Satin & Umbrous) and sau/sau (Golden)
Parents are Siblings Parent SaU/sau Heterozygous Satin & Umbrous
16 SaU 16 sau Sau saU
Parent
SaU/sau
Heterozygous
Satin & Umbrous
16 SaU 256 SaU/SaU
Homozygous
Satin & Umbrous
256 SaU/sau
Heterozygous
Satin & Umbrous
16 SaU/Sau
Homozygous Satin & Heterozygous Umbrous
16 SaU/saU
Heterozygous Satin & Homozygous Umbrous
16 sau 256 SaU/sau
Heterozygous
Satin & Umbrous
256 sau/sau
Golden
16 Sau/sau
Heterozygous
Satin
16 saU/saU
Heterozygous
Umbrous
Sau 16 SaU/Sau
Homozygous Satin & Heterozygous Umbrous
16 Sau/sau
Heterozygous Satin
Sau/Sau
Homozygous Satin
Sau/saU
Heterozygous
Satin & Umbrous
saU 16 SaU/saU
Heterozygous Satin & Homozygous Umbrous
16 saU/sau
Heterozygous
Umbrous
Sau/saU
Heterozygous
Satin & Umbrous
saU/saU
Homozygous Umbrous

Phenotypes and genotypes will be:

  • 834 Satin Umbrous with genotypes: 256 SaU/SaU; 512 SaU/sau; 32 SaU/Sau; 32 SaU/saU; 2 Sau/saU
  • 256 Golden with genotypes: 256 sau/sau
  • 33 Satin with genotypes: 32 Sau/sau; 1 Sau/Sau
  • 33 Umbrous with genotypes: 32 saU/sau; 1 saU/saU

As you can see it is much more difficult to get recombining with this pair especially since 1156 animals would have to be produced to reach these ratios.

If you think this last square had high numbers there is a gene pair in rabbits that has a ratio of 600:1.

X-Linked Genes

Yellow, Tortoiseshell & Non-Yellow

Normal females are XX and normal males are XY. The X chromosome is thought to be the primary chromosome from when we were one-celled creatures and didn't have two sexes. The Y chromosome is a truncated X chromosome and carries little genetic information other then maleness. For this reason the recombination that occurs with other linked genes does not happen with X-linked genes. A female that is X0 (only one X chromosome) will have all of the required traits of a female the same as a normal female. So only one X is required in females. Because of this a phenomenon called X-inactivation occurs. One of the two Xsbasically shrivels up and becomes inactive. This happens early in embryonic development. The embryo grows and more cells are produced from the cells that now have only one active X. These are daughter cells. When there is an X-linked gene it's possible to inherit two different alleles, one from each parent. This is the basis for the Tortoiseshell pattern. The female hamster inherits a Yellow gene from one parent and a non-Yellow from the other parent on her two X chromosomes. The size of the patterns depends on how early the X-inactivation occurs. Rarely more then one ova and one sperm join together or the splitting for the ova and/or sperm doesn't happen like it should. Since a male normally has only one X they can only be Yellow ToY or non-Yellow toY. Females can be Yellow ToTo, non-Yellow toto, or Tortoiseshell Toto. Below are four Punnet squares showing only yellow. Of course if you breed two Goldens together you'll get all Goldens. And two Yellows will give you all Yellows. This is assuming there are no recessives lurking in the gene pool. For a more in depth series of Punnet squares on Tortoiseshells go here. Most people put the Y in for the males to designate there are two genes but only the X can be Yellow.

Yellow Father ToY
To Y
Golden
Mother
toto
to Toto
Golden Tort girl
toY
Golden boy
to Toto
Golden Tort girl
toY
Golden boy


Yellow Father ToY
To Y
Golden Tort
Mother
Toto
To ToTo
Yellow girl
ToY
Yellow boy
to Toto
Golden Tort girl
toY
Golden boy


Golden Father toY
to Y
Yellow
Mother
ToTo
To Toto
Golden Tort girl
ToY
Yellow boy
To Toto
Golden Tort girl
ToY
Yellow boy


Golden Father toY
to Y
Golden Tort
Mother
Toto
To Toto
Golden Tort girl
ToY
Yellow boy
to toto
Golden girl
toY
Golden boy
This page was last edited on October 12, 2011
 
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