0:00:00.000,0:00:08.632
Following the theoretical lecture on coat color genetics in farm animals, we will now show several examples.

0:00:08.632,0:00:18.630
The presentation is part of Module 2: Animal genetics, that is a part of the ISAGREED project.

0:00:18.630,0:00:26.795
This presentation was supported by Erasmus+ KA2 Cooperation Partnerships Grant

0:00:26.795,0:00:39.026
"Innovation of the content and structure of study programmes in the field of management of animal genetic and food resources using digitalization".

0:00:40.026,0:00:42.859
I chose a horse as a model mammal species.

0:00:42.859,0:00:50.857
In horses, the basic coat color is determined by the genotype at the two loci listed here.

0:00:50.857,0:00:58.856
This is the EXTENSION locus, where the gene for the melanocortin receptor (MC1R), is located,

0:00:58.856,0:01:07.887
and the AGOUTI locus, where the gene for the agouti signaling protein (ASIP), is located.

0:01:07.887,0:01:14.219
The resulting coat color is then determined by the interaction between these loci.

0:01:14.219,0:01:20.218
The recessive epistasis of the EXTENSION locus to the AGOUTI locus applies here.

0:01:20.218,0:01:27.217
This means that in the case of a homozygous recessive genotype at the EXTENSION locus,

0:01:27.217,0:01:35.882
the alleles of the hypostatic AGOUTI locus will not express in the phenotype;

0:01:35.882,0:01:43.880
in other words, "it doesn't matter at all" what genotype the individual has at the AGOUTI locus.

0:01:43.880,0:01:54.545
In an individual with a recessive genotype at the EXTENSION locus, only a pigment known as pheomelanin is produced

0:01:54.545,0:02:03.543
and the resulting coat color is chestnut, which possibly has fur on the body in different shades of red

0:02:03.543,0:02:11.974
and a mane and tail in the same color, possibly even lighter or darker, never but not black.

0:02:11.974,0:02:19.440
In the case of the presence of at least one completely dominant allele at the EXTENSION locus,

0:02:19.440,0:02:26.438
the resulting coat color is determined by the genotype at the AGOUTI locus.

0:02:26.438,0:02:34.437
A completely dominant allele of this locus restricts the distribution of the black pigment eumelanin

0:02:34.437,0:02:42.602
only to the distal parts of the limbs, mane and tail, and the resulting coat color is bay.

0:02:42.602,0:02:47.767
In the case of a recessive genotype at the AGOUTI locus,

0:02:47.767,0:02:55.932
eumelanin is deposited evenly throughout the body, resulting in a black coat color.

0:02:55.932,0:03:03.298
We can see the possible genotypes and connected phenotypes in this table,

0:03:03.298,0:03:13.629
if a dot is used in the notation, it means that any allele can be at this position.

0:03:13.629,0:03:23.627
In the last column, the basic colors are sorted from the most recessive chestnut to the dominant bay.

0:03:25.027,0:03:27.959
Now we will show a practical example.

0:03:27.959,0:03:38.091
Make a Punnett square for mating of two bay parents, both heterozygous at AGOUTI and EXTENSION loci.

0:03:38.091,0:03:43.923
Calculate phenotypic segregation ratio in their offsping.

0:03:45.923,0:03:50.688
First, we write down the genotypes of both parents - bays.

0:03:50.688,0:03:58.987
We already know that a bay horse must have at least one dominant allele from each locus in its genotype.

0:03:58.987,0:04:10.984
In the assignment, it is stated that both parents are heterozygous individuals, so we write down the genotype as follows.

0:04:18.316,0:04:22.982
To determine the possible offspring, it is first necessary to deduce

0:04:22.982,0:04:28.381
what all possible gametic combinations the parents will form.

0:04:28.381,0:04:34.680
As these are heterozygous individuals at the two loci considered,

0:04:34.680,0:04:41.911
this number will be 2 squared, that means 4 different types of gametes.

0:04:41.911,0:04:48.244
Since the EXTENSION and AGOUTI loci are located on different chromosomes,

0:04:48.244,0:04:56.242
there is no genetic linkage between them and all the mentioned combinations will arise with equal probability.

0:04:58.242,0:05:03.574
Now we write the gametes of both parents in the Punnett square

0:05:03.574,0:05:12.739
and we derive the corresponding phenotypes according to the genotype that results from their fusion.

0:05:12.739,0:05:18.671
To simplify the work, it is better to write down the gametes

0:05:18.671,0:05:27.969
of both parents in the same order - that means in the first row and in the first column.

0:05:27.969,0:05:36.634
From the completed square, we count the number of times individual phenotypes occur here

0:05:36.634,0:05:40.866
and write them into the segregation ratio.

0:05:40.866,0:05:52.331
We see here 9 bays, 3 blacks, and 4 chestnuts; the phenotypic segregation ratio is therefore 9 : 3 : 4,

0:05:52.331,0:05:59.063
which is the segregation ratio corresponding to recessive epistasis between two loci,

0:05:59.063,0:06:07.694
if in each of them, complete dominance between the alleles applies, which is exactly the case.

0:06:09.694,0:06:16.793
If you find this procedure time-consuming, it is possible to solve the case in a slightly different way,

0:06:16.793,0:06:23.925
which would be applicable, especially if we are not interested in the exact segregation ratio,

0:06:23.925,0:06:31.923
but, for example, only in what is the probability of a particular color in offspring.

0:06:31.923,0:06:41.588
For example, if we were wondering what is the probability that these two parents will produce a chestnut foal?

0:06:41.588,0:06:45.587
We can find the answer either in this square,

0:06:45.587,0:06:47.587
when we see that it will be 4 cases out of 16 possible, that means in 25 %.

0:07:05.583,0:07:12.782
However, we can achieve the same result without having to fill in the entire square.

0:07:12.782,0:07:23.479
It is enough to write down a common genotype for chestnut, in this case it is the recessive genotype at the EXTENSION locus

0:07:23.479,0:07:31.345
and since we know that this genotype is epistatic over the alleles of the AGOUTI locus,

0:07:31.345,0:07:36.344
we just need to solve the result of this monohybrid crossing,

0:07:36.344,0:07:43.742
which will result in a recessive combination with a probability of 25 %.

0:07:43.742,0:07:53.440
We would solve the other two colors in a similar way, but here we would have to consider the genotypes at both loci.

0:07:53.440,0:08:03.605
If we were interested in the probability of a black foal, we would write down its common genotype

0:08:03.605,0:08:12.003
we calculate the probability of the desired genotype occurring for each locus separately

0:08:12.003,0:08:22.335
and then multiply the two probabilities because we need both of these phenomena to occur simultaneously.

0:08:22.335,0:08:40.164
It will be then ¾ for E locus multiply ¼ for A locus. Therefore, the probability of a black foal being born is 18.75%.

0:08:40.164,0:08:48.396
Well, the last possible case is the bay with common genotype as follows

0:08:48.396,0:08:50.396
and then the probability will be 0.75 multiply 0.75, so overal 56.25 %.

0:08:50.396,0:09:26.389
To check our results, we can add all three values together 25 % + 18.75 % + 56.25 % which is 100 %

0:09:26.389,0:09:31.554
and we are therefore sure that we have calculated correctly

0:09:31.554,0:09:39.553
and at the same time that we included all possible phenotypes that can appear in foals.

0:09:41.552,0:09:46.118
GREY is another common coat color in horses.

0:09:46.118,0:09:53.816
This coat color occurs in many breeds, in some such as Old Kladruber Horse or Lipizzaner,

0:09:53.816,0:09:57.516
it is considered a breed characteristic.

0:09:57.516,0:10:06.980
A grey horse has a coat color characterized by progressive depigmentation of the colored hairs of the coat.

0:10:06.980,0:10:11.746
Most grey horses have black skin and dark eyes.

0:10:11.746,0:10:18.612
Grey horses may be born any base color, depending on other color genes present.

0:10:18.612,0:10:27.010
White hairs begin to appear at or shortly after birth and become progressively more prevalent

0:10:27.010,0:10:35.208
as the horse ages as white hairs become intermingled with hairs of other colors.

0:10:35.208,0:10:45.206
Greying can occur at different rates—very quickly on one horse and very slowly on another.

0:10:45.206,0:10:56.071
As adults, most grey horses eventually become completely white, though some retain intermixed light and dark hairs.

0:10:58.070,0:10:58.071
At the molecular level, the GREY gene was described relatively late (as late as 2008)

0:10:58.071,0:11:12.468
compared to other color genes. It is the syntaxin 17 (STX17) gene.

0:11:12.468,0:11:19.000
Within the mentioned locus, the complete dominance of the allele for greying applies,

0:11:19.000,0:11:24.199
that means the presence of one is enough to start greying,

0:11:24.199,0:11:29.198
BUT it has been shown that greying in heterozygous individuals is slower

0:11:29.198,0:11:33.997
and sometimes incomplete compared to dominant homozygotes.

0:11:33.997,0:11:40.029
The allele for greying is epistatic to alleles of other pigmentation genes,

0:11:40.029,0:11:44.861
that means it gradually overlaps their expression.

0:11:44.861,0:11:53.226
Therefore, the carrier of this allele will turn grey regardless of the base color.

0:11:53.226,0:12:03.391
In other words, we can say that the G locus is in dominant epistasis to other coat color loci.

0:12:06.390,0:12:12.656
This slide shows the possible result (phenotypic segregation ratio) of the mating of two greys,

0:12:12.656,0:12:23.054
both heterozygous at the G locus and at the same time heterozygous at the AGOUTI and EXTENSION loci.

0:12:23.054,0:12:30.852
Due to the fact that, from a genetic point of view, these are so-called trihybrids,

0:12:30.852,0:12:37.184
each of the parents will create 8 different types of gametes.

0:12:37.184,0:12:43.183
Since each of the considered loci is located on a different chromosome,

0:12:43.183,0:12:47.182
all the mentioned combinations will arise with equal probability.

0:12:47.182,0:12:58.180
You can see that in this case the Punnett square is already quite large and it is time-consuming to complete it all.

0:12:58.180,0:13:07.745
However, we can clearly see that with the greatest probability (75%), a grey foal will be born,

0:13:07.745,0:13:14.210
but with a probability of 25%, a non-greying foal can appear.

0:13:14.210,0:13:23.208
Well, due to the heterozygosity of the parents, this foal can theoretically have any of the basic color.

0:13:26.208,0:13:33.940
However, if we would like to find out the probability of grey foal, it is enough to consider G locus.

0:13:33.940,0:13:47.937
Since both parents are heterozygotes at this locus, the genotype segregation ratio in the offspring will be 1: 2 : 1,

0:13:47.937,0:13:54.536
the phenotype segregation ratio is then 3 : 1.

0:13:54.536,0:13:57.968
Given that the mare is uniparous,

0:13:57.968,0:14:03.967
it is probably more appropriate to express the possible result as a probability,

0:14:03.967,0:14:12.632
that means that we have a 75% probability that greying will occur in foals.

0:14:14.632,0:14:25.629
What genotype at the G locus must a stallion have to obtain as many greys as possible after mating to Old Kladruber black mares?

0:14:26.629,0:14:33.628
Due to the dominant character of grey phenotype, every grey horse must have at least one grey parent.

0:14:33.628,0:14:40.626
That is, we exclude a non-grey stallion (recessive homozygote).

0:14:40.626,0:14:48.791
Only 50 % of the foals will turn grey if a heterozygote stallion is used.

0:14:48.791,0:14:55.190
If a stallion is homozygous dominant at the G locus,

0:14:55.190,0:15:07.521
he will pass the allele for greying to all his offspring and all his foals will turn grey, regardless of the color of the mares.

0:15:07.521,0:15:13.853
The third option is, therefore, the best for our purpose.

0:15:15.853,0:15:24.851
At this moment I would like to thank you for your attention and if you have any questions, you can use the email listed here.