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In mink, as most other mammals, most genes affect the color of the coat.
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There are particular interest to breeders since new and different fur command premium prices.
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You have 13 genes known to determine coat color in mink.
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Now you have three different crosses shown on the table.
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So wild type crossed with platinum.
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F1 shows wild type.
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F2 shows 3 to 1 wild type and platinum.
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Wild type crossed with elution apparent f1 wild type.
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Now f2 is again 3 to 1 ratio.
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The platinum crossed with elution f1 show wild type and f2 show 133 wild type, 41 platinum, 46 elution, and 17 sapphire.
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Now these gene phenotype due to one gene or multiple with multiple alleles or two genes that affect the same phenotype explain logic and fill in the parental genotype in a table below.
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If you look at the third cross platinum and elution, while wild type f1 but 9 to 3 to 3 to 1 ratio for the f2, we decide that most likely platinum and elution are two different genes.
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So you can see 133, 41, 46, and 17 is roughly 9 to 3 to 3 to 1 ratio.
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So this is a very classic dihybrid cross where two genes are involved.
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So let's give each of these gene a name.
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Let's say p platinum.
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Let's call it gene p and elution gene a.
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So let's go ahead and look at each of the cross.
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So this is p, this is a.
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Now, so for wild type, it obviously is wild type dominant for every single gene.
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So for the number one cross, wild type cross with platinum.
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As we just mentioned that platinum, we use gene p as a key.
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So wild type is going to have, let's say, homozygous, capital p, and platinum is a recessive gene.
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Let's call it homozygous lower p.
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So that goes into the first two box of the table, genotype of parent number one and genotype of pair number two, number one and number two.
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So now for this cross, as you can see that all f1 is going to have one allele from each of the parents, heterozygous, capital p, lower p.
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Due to have a wild type allele, it's going to show all wild phenotype.
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Now for f2, if you have this cross, capital p, lower p, and f2 is going to show three to one ratio.
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So according to law of segregation, capital p and lower p separate from each other.
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And then same for the other parent.
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So capital p, homozygous, heterozygous, heterozygous, and homozygous lower p.
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So all three with dominant allele will show a wild type, wild type three.
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The last one, homozygous lower p, show platinum.
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Homozygous recessive show recessive phenotype.
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And platinum is one out of four.
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So it's a three to one ratio.
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This is consistent with our first cross.
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Now let's move on to the second cross.
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The second cross is wild type crossed with allusion.
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So basically it's the same cross, but we're going to use a different key a.
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So again, parent number one wild type is homozygous capital a and allusion is recessive homozygous lower a.
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So that goes into the box of the table.
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Now again, it's the same cross with f1, 100 % heterozygous capital a lower a.
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And this is why it shows wild type.
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Now for f2, and then you have two heterozygous parents.
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Again, it is going to show the same punnett square with three of them have at least one dominant a wild type and homozygous lower a, one out of four allusion.
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So this is basically the same punnett square as the first cross.
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I'm just replacing p with a.
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So it's again, wild type to allusion three to one.
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The last cross is interesting because now we have two different gene involved.
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So let's take a look at the genome type.
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So platinum has homozygous recessive for p, but it actually is homozygous dominant for a.
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So the platinum actually has a genotype of homozygous capital a, but with a homozygous lower p, because when it comes to gene a, it's actually wild type.
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Now, then the other parent's allusion is the opposite.
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It has homozygous lower a because you have homozygous recessive show recessive phenotype, but it comes to the p gene, it's homozygous dominant capital p.
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So you can see that now we can fill in the last two box of the table.
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Platinum has homozygous dominant a, homozygous recessive p, and allusion is the opposite, homozygous lower a and homozygous capital p.
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Now for f1, simply just one allele from each of the parent.
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So dominant a, recessive a, lower p, and capital p.
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So heterozygous is actually wild type because for both gene you have a dominant a and dominant p.
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Although it is a heterozygous for both gene, but it shows a wild type phenotype due to the dominant allele.
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Now, let's do this f2.
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So again, you have intercross between f1.
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So both parents are heterozygous.
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So this is another f1.
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So according to our very typical dihybrid cross from mendel's, the hybrid cross, you can see that i'm making a bigger...