Solving Genetics Problems I: Monohybrid Crosses Classical genetics is a science of logic and statistics. While many find the latter intimidating, the mathematical side of most classical genetics puzzles is relatively simple — and there are actually ways to get around most of the math. The logic part is inescapable. All genetics problems are solved using the same basic logic structure. If you learn the sense of the approach, you can solve virtually any genetics problem, provided you are given enough basic information. Lets first consider the little puzzle you were given in the Five Rules essay. The problem involved two gerbils named Honey and Ritz. The gene in question was a fur color gene which has two alleles — dominant brown (B) and recessive black (b). Its a very good idea to write down the information you are given in a problem so that it will be easy for you to refer to it when necessary. So begin by writing something like this at the top of your work page: Using the Five Rules for Pedigree Analysis, we figured out that each of our parent gerbils was heterozygous for this gene. So here is our mating: Step One: Figure out the genotypes of the parents. Once youve got that settled, you need to address the question of all of the possible kinds of babies they could produce. Before any parent makes babies, of course, that parent makes gametes. So in order to find what kinds of babies they can have, you must first determine what kinds of gametes they can produce. Since Honey is a heterozygote (and paying attention to Rule #1), she can produce two kinds of eggs: B eggs and b eggs. Ritz is also a heterozygote, so he can produce two kinds of sperm: B sperm and b sperm. Something like this: Step Two: Figure out what kinds of gametes the parents can produce. Now you need to determine all the possible ways that his sperm can combine with her eggs. There are several different techniques used for this operation. The most popular among students is the Punnett Square. Punnett Squares are probability tables — a way to do statistics while avoiding as much math as possible. Step Three: Set up a Punnett Square for your mating. Setting up a Punnett Square is easy. You need to create a chart with one column for each of the females egg types, and one row for each of the males sperm types. For Honey and Ritz, your table would look like this: The little boxes on the inside of the table are the kinds of offspring this pair can produce. Note that, since Honeys two kinds of eggs will be produced in equal proportions, and Ritzs two sperm types will also be produced in equal proportions, each of the boxes inside should represent an equal proportion of potential offspring. A statistician would say that each of the offspring types has an equal probability. Now we fill in the inside of the table. The upper left square represents those babies that get a B from both egg and sperm; the upper right box represents babies that get B from the sperm and b from the egg. You can figure out the rest. Step Four: Fill in the babies inside the table by matching the egg allele at the top of the column with the sperm allele at the head of the row. When you are finished, your completed square should look like this: So we have now figured out that, if Honey and Ritz have a lot of babies, we can predict that 1/4 of them should be BB, 1/2 of them (2/4) should be Bb, and 1/4 should be bb. This conclusion is often expressed as a genotypic ratio: 1 BB:2Bb:1bb. This means that we are predicting that, for every BB baby, they should have 2 Bb babies (twice as many), and one bb baby. Step Five: Figure out the genotypic ratio for your predicted babies. Notice that dominance hasnt entered into the problem yet (other than in the application of the Five Rules to figure out the genotypes of the parents). Now its time to consider dominance. If you look back at the Five Rules, youll note that the question you were asked wasnt about genotypic ratios. You were asked to figure out what fraction of their babies would be expected to be brown. This means that you have to figure out the phenotypic ratio. Step Six: Figure out the phenotypic ratio for your predicted babies. To do this, you need to ask yourself one question: do any of these different genotypes produce the same phenotype? In other words, do any of these babies look alike? This is where dominance enters the picture. If B is completely dominant to b, all gerbils with at least one B will look pretty much alike, no matter whether their second allele is B or b. So BB and Bb have the same phenotype, and we can add them together. Thus, our phenotypic ratio is 3 Brown:1 Black. Or, there should be three times as many brown babies as black babies. So the answer to our question is, 3/4 of the babies should be brown. Step Seven: Answer the question youve been asked. The mating scheme weve just worked through is called a monohybrid cross. This means that we were paying attention to only one gene (mono=1), and both of our parents were heterozygous for that gene (hybrid=heterozygous). Here again are the steps we followed: Step 1: Figure out the genotypes of the parents. Step 2: Figure out what kinds of gametes the parents can produce. Step 3: Set up a Punnett Square for your mating. Step 4: Fill in the babies inside the table by matching the egg allele at the top of the column with the sperm allele at the head of the row. Step 5: Figure out the genotypic ratio for your predicted babies. Step 6: Figure out the phenotypic ratio for your predicted babies. Step 7: Answer the question youve been asked. Of course, the alleles you are using dont always have a complete dominance relationship. Other dominance relationships dont affect any of the early parts of the problem — they only enter the calculations when it comes time to look for the phenotypic ratio. Heres how you handle the most common situations: Incomplete Dominance: If two alleles have incomplete dominance, the heterozygote has a phenotype which is intermediate between the phenotypes of the two parents. For example, in many flowering plants, alleles for red and white flowers have an incomplete dominance relationship. If CR is the allele for red flowers and CW is the allele for white flowers, a CRCR plant will have red flowers, a CWCW plant will have white flowers, and a CRCW plant will have pink flowers. If our gerbil problem involved two pink flowering snapdragons, instead of two brown gerbils, we would have exactly the same kind of mating as weve just been through. Youd use different symbols for the alleles — capital and lower case letters usually predispose you to think in terms of complete dominance — but the procedure for working the problem would be precisely the same as the gerbil problem, right up to that moment when you ask yourself that phenotype question, Do any of these genotypes produce the same phenotypes? In this case, your answer would be, No. This heterozygote has its own phenotype. So your phenotypic ratio would be 1 Red:2 Pink:1 White. There is another dominance relationship called co-dominance which works pretty much the same as incomplete dominance does. In co-dominance, the heterozygote expresses both of the alleles, essentially independently of each other. Thats where blood type AB comes from — a heterozygote between the A-type allele and the B-type allele, which are co-dominant to each other. Again, these problems are solved exactly like our gerbils, except when it comes to figuring out the phenotypic ratio. Lethal Genotypes (Pseudodominance): In some cases, one of the three possible genotypes creates a condition which is lethal. In other words, the embryo which is conceived with that genotype doesnt survive to birth. Once again, problems involving lethal genotypes are worked exactly like our brown and black gerbils above until we get to the phenotypes. In fact, we can illustrate this situation with a different gerbil gene. Besides the fur color gene we dealt with earlier, gerbils also have a gene which can cause the gerbil to have white markings on his or her fur. Gerbils who have no white markings are homozygous for the wild type solid color allele; gerbils which are marked with white patterns are heterozygous. The homozygous white-marked genotype is lethal; these gerbils dont survive the development period, and are not among the babies delivered when a litter is born. If two white spotted gerbils mate, this is obviously a monohybrid cross, since both are automatically heterozygous. To figure out the prospects for their offspring, you solve the problem precisely the same way you solved the brown/black problem for Honey and Ritz. Again, the difference comes when you figure out the phenotypic ratio. Since homozygous white-marked gerbils are simply missing from the babies who are actually born, you have to subtract them from the final phenotypic ratio. Thus, that ratio becomes: 2 white-marked gerbils:1 solid gerbil. Or 2/3 white marked, 1/3 solid.
Posted on: Sun, 01 Dec 2013 06:27:41 +0000
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