Learn about dihybrid crosses, test crosses, epistasis, polygenic traits and how to calculate probabilities in this video!
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You can’t tell just by looking at the phenotype if the dominant trait is homozygous dominant or heterozygous. In order to find out what their genotypes are, we can do something called a test cross.
during a test cross you would be breeding the unknown individual with a homozygous recessive individual and analyzing the offspring to learn the genotype of the unknown individual.
Both the homozygous dominant and heterozygous genotypes will make a straight-winged fly. if we breed this unknown straight-winged fly with a curly winged fly, the recessive form, we could get two possibilities. If the fly is homozygous dominant, all of the offspring will be straight-winged. If the fly is heterozygous, then half of the offspring will have straight-wings and half will have curly wings.
it’s also possible to cross more than one trait at a time. A dihybrid cross will test two traits at the same time.
if you look at all the possible gamete combinations you can figure out what the dimensions of the cross should be. with all of these different combinations, remember that only half of the genes are used to make the gamete. We get to pull one A and one B. For the first letters, no matter which A or B we choose, we’ll still get the same combination, capital A and capital B. There’s only 1 possible combination. With this group of alleles, there must be a capital A, but there could be a capital B or a lowercase b, so there are two total gamete combinations. In this allele group, both genes are heterozygous which will give us the maximum number of combinations. There could be capital A with a capital or lowercase b. or lowercase a with a capital or lowercase b. That’s four possible combinations. In this last group of alleles there could be capital or lowercase a, and both will have a lowercase b. so there are 2 combinations for this group.
The largest possible cross is made with two completely heterozygous parents. Each of the four gamete possibilities ends up on the sides of the Punnett square.There are 9 different genotypes here, but if we look at the phenotypes, there are only 4 phenotypes. Those show up in a 9:3:3:1 ratio, with 9 showing the dominant traits and the one representing both recessive traits.
This is the easiest possible cross we could make. Each group of alleles has only one possible gamete it could make. big B big H and little b little h. So we make a square, and cross it. So for the genotypic ratio, all of the offspring will be big B little b big H little h. and for the phenotypic ratio they will all be black and short-haired.
Using the same rabbits, we’ll list the genotypic and phenotypic ratios for the F2 offspring produced between crosses of two animals of the F1 generation. Across the top and the side are the four possible gamete combinations. two big, big B little h, little b big H and little little. Then fill in the punnett square. There are nine boxes that are short black hair, three that are long black hair, 3 that are short brown hair and one that is long brown hair. That’s the 9:3:3:1 ratio that we saw before!
The first cross is homozygous dominant and homozygous recessive which would explain all the cream colored animals. Then two of those heterozygous cream guinea pigs are crossed which would give a 1:2 : 1 ratio of yellow, cream and white guinea pigs. 16 to 33 to 15 is very close to 1:2:1 which confirms this explanation.
Sometimes in a dihybrid cross you don’t see the 9:3:3:1 ratio, and it could be cause by epistasis. that’s the effect of one gene depending on the presence of another gene. We’ll use the example of the fur on labradors. if they have a dominant color gene they get black fur, if recessive, it’s brown, but they also need the full melanin to show the color. if there is no melanin, then the fur is completely white. The ratio becomes 9:3:4 in this kind of epistasis.
polygenic traits have more than one gene that controls the phenotype. There are over nine genes that control eye color. other polygenic traits include hair color, skin color, and height.
There is a short cut to finding the probabilities of certain combinations of traits: The probability product rule. just take the product of the probabilities of the individual events!
All we need to do is turn the percentages into decimals by moving the decimal two places to the left, then multiply them together. This gives us .5625 which is 56.25%
we’ll multiply 1/20 times 1/20 and we get 1/400, the incidence of both parents being carriers. If they are both carriers, when you solve a punnet square you’ll see that there is a 1/4 chance of having a child with the recessive disease. Multiply 1/400 and 1/4 and get 1/1600. This answer represents the risk of two people from the general population having a child with cystic fibrosis.
