usually when we think of math on the MCAT we think chemistry and physics but in this video we're going to talk about the math that you're going to see in the biology and biochemistry section of the MCAT we'll focus on punet squares and mandelian Basics to answer even the most complex probability questions that will show up in genetics inheritance problems first let's go through a couple definitions that we'll be using in this video autosomal this just means chromosomes and genes that are not related in determining sex for mammals that means everything except for the X and the Y chromosomes on the other hand Sex Link genes refer to genes that are on the sex determining chromosomes like X and Y these will have different inheritance patterns so it's important to recognize if you're working with autosomal genes or sexlink genes next up dominant traits dominant traits are where if you even have just one copy of that Gene it's going to be displayed in a phenotype that's in contrast to recessive traits where we need both copies of the gene in order for it to be expressed as a phenotype finally homozygous versus heterozygous for a homozygous inheritance that means you've gotten two copies of the same Al either dominant or recessive for heterozygous you got one of each one dominant and one recessive Alo all right now for some inheritance rules remember that for mammals including humans which is what we're going to focus on in this video we have two copies of each gene in our DNA when a Maman zygote is formed we get one Al of a gene from the male parent and one from the female parent but the male and the female parents each have two alals to choose from so we can end up with four different combinations that's why we use pet squares to figure out and view the probabilities of each of those combinations good so far all right let's now work through some melan math problems starting with some probability and melan Basics let's do this one first go ahead try this on your own by pausing the video and then we'll come back and work through the probabilities together okay so in this situation they tell us that hemophilia is determined by XL or sexlink Gene inheritance they tell us that a man with normal blood clotting and remember males have X and Y so they only have one copy of the X Gene and he has normal blood clotting so he does not have hemophilia and he marries a woman with normal blood clotting whose father was a hemophiliac so let's do a mini punet Square for the woman woman's genotype so her father had hemophilia we'll go ahead and write little H for that one I am inferring that hemophilia is a recessive gene because if her dad had the gene then it's a guarantee that the daughter the woman here had at least one copy of the hemophilia Gene and if it was dominant that would mean she would also have hemophilia and she would not be normal I'll show you what I mean here so let's suppose that her mother would was normal so we'll say big H Big H and we can see here that if we do the inheritance pattern for the xlink gene any daughters that this grandfather would have are going to be heterozygotes they're going to be carriers where they'll have one copy of the recessive gene but won't display the phenotype and the Suns would be normal right but we're not worried about that because this is a woman so we're focused on her pattern so she's going to be heterozygote so let's go ahead and draw her genotype here again also known as a carrier now the uh husband right who's marrying this woman is does not have the gene which we can affirm means that he has the homozygous dominant Alo so so far so good let's go ahead and fill out this punet Square for daughters there's a chance that one of the daughters options would be to be homozygous dominant no chance of having the hemophilia and then the other daughter is hetero zygote for the sons one of the sun probability is that they would be normal because they would get the normal alil from the mother and the other probability is that they would have hemophilia because they get the hemophilia carrier Al from their mother now in order to do the probability we have to determine are we looking at all of the children or just the sons or just the daughters they're asked if the couple has three sons so right away we can eliminate the daughters from our probability calculations if they let us know gender and we can eliminate the gender from our punet Square we'll do so to simplify our probability so then they're asking us what is the probability that hemophilia will be transmitted to all three of them in other words what is the chance that all three sons ended up getting this genotype so what is the chance that the oldest son oldest son would get this genotype 50/50 right so we'll say one since that is what our uh probabilities look like on our answer choice so our oldest son has a 1/ half probability but we weren't just asked about the oldest we were asked about all three so what's the chance that the middle Sun would have it again it's the same probability yep one2 and then the youngest son again one half right we have the same probabilities for all three it does not change between um multiple children right multiple progyny so how do we do prob ility when we have multiple different probabilities between the Suns and we want to calculate the overall probability well when the oldest and the middle and the youngest all need to have uh the gene in order for us to determine this probability what we do is we multiply so when it's an and question the oldest son and the middle and the youngest you multiply the individual probabilities to get the overall probability when you multiply fractions you just multiply the the numerator times and multiply the denominators and we have 1/8 so that is our answer here all right so the key thing with this question is a getting rid of the probabilities for the daughters since we know that there are only Sons recognizing that each son has a 50/50 chance of getting the hemophilia Al and then multiplying those probabilities because we're looking for the chances of all of the Suns having hemophilia not just one or the other all right if you have any questions on that practice problem go ahead and pop them in the comments below let's move on to practice question number two okay you know the drill go ahead pause this video give this question a shot on your own first and then we'll walk through it together okay so we're going to have to take this one step by step this is a multigeneration probability inheritance pattern we're going to need to go and figure out all the information they provide us in the question stem first and then figure out how to do our probabilities so let's get started here in a species of fly green body color is dominant to Brown that means if we have big G Big G they'll be green if we have big G little G they'll be green but if we have little G little G then the Flies will be brown two green flies are crossed that means that they're just giving us the phenotype it could be either homozygous dominant or heterozygous and we're crossed and we produce 29 green and eight Brown Offspring so by having Brown Offspring we actually know what the genotype needed to be for these flies and the reason why I'll sketch it for you here is because if we had even one homozygous dominant right Big G Big G B G little G you can do this punet Square quickly and see that we would only get green phenotypes so if even one of the parents is homozygous dominant we're going to have only green Offspring in the F1 generation but we didn't we had some brown so if we had two green flies but we ended up with brown that means that we had to have had a heterozygote cross all right so if you cross two uh organisms with a dominant phenotype and you get the recessive phenotype that means we had a heterozygote cross so let me go ahead and draw that heterozygote cross here all right we're going to have big G little G right and so these guys are going to be homozygous dominant we'll have two heterozygotes and then we'll have one homozygous recessive so this is always the pattern with a heterozygote cross is a 1 to 2: one ratio homozygous dominant to heterozygote to homozygous recessive so we know right away that this is the progyny but that's the F1 generation right these are the parent Generations so now they're saying okay now we take two more green F1 flies all right so we're going to take our options here are these three guys right we're going to take from only the green flies and cross and then they're ask what is the probability that both green and red flies appear in that next Generation so to reward that what they're really asking is what is the probability that we'll have another heterozygote cross right that we pick both the male and the female being heterozygotes so here's how we do the probability it's going to take a couple different probabilties to make this happen the first is that we know we're only taking green flies so we do not count the brown fly in our probability so what is the probability that the male fly is hetero zgo well out of our green fly options that's two out of three now what's the probability that the female is heterozygote well that probability is two out of three we need both the male and the female to be heterozygote in order to get brown flies right the same rule as before so therefore it needs to be an and which means we need to multiply this probability again it's two out of three not two out of four because we're not including the brown flies we took only green flies so we can ignore the brown flies because they're not part of our probability so multiplying across 2 * 2 is 4 3 * 3 is 9 we have a 4 9th probability all right so again again I hope you're seeing the theme first off see if you can eliminate any genotypes from the probability because of the phenotype we're choosing from and then making sure that you're multiplying if we need both situations or all situations to be true in order to get our final probability all right if you have any questions on the Fly problem go ahead and put them in the comments below all right our final practice problem is going to be on die hybrid crosses where we have multiple genes going on at once before we get into that I'm Amanda Bram and I've been coaching students on their MCAT Journeys since 2019 please make sure to subscribe to this channel for more videos on MCAT content test taking strategies and mental fitness tips to help you perform your best on test day and if you'd like more in-depth interactive lessons on topics like this one that also include active practice and test day strategies please check out the link in the caption below which will take you to register for my next available MCAT course all right go ahead and give this challenging question a try first and then we'll walk through it together so we are told that in cats suppose that in cats short hair is dominant to long hair and Amber eyes are dominant to green eyes right and so we have two different genes going on a female cat that is heterozygous for both traits is made up with a male cat that is also heterozygous so we have a double heterozygote cross both of the H&H traits sort independently for each other this is just a phrase that they have to say that they sort assort independently so that we can do this punet square if they did not assort dependently we would not be able to do this probability then they ask what is the probability that they will have an offspring that has long hair and Amber eyes so here's the deal in order to do a DI hypd cross we can't do the punet square separately because the probabilities may not work out separately we actually have to combine the different alals in the way that they would be combined in a zygo so we have a female cat that's heterozygous so she's going to be h h a a and crossed with a male cat that's heterozygote HH AA and what we're looking for you always want to figure out what you're looking for first is long hair so that means they need to have little H little H and Amber eyes so they just need to have at least one dominant a Al all right so this is what we're looking for in our progyny now again we have to do what's called a DI hybrid cross which is going to be a 16 cell punet Square so I'm going to draw it out for you here just as if I was doing this on test day 1 2 3 4 1 2 3 4 okay so we'll put the female options across the top they're going to be the same for both female and male in this cross so what do we do we have to show all the possible combinations of alal in each gam so remember we have one of each so option one would be dominant H dominant a right for the female option two would be dominant AG lowercase a option three would be lowercase H dominant a and the final option would be lowercase H lowercase a little H little a all right so these are the four possible combinations that we'll get from the female gametes and we're going to have the same possible combinations from the male gametes because these are autosomal traits so we'll go ahead and just draw those same traits down the side here I want you to do this on test a where you draw out the axes and the combinations but I don't want you to fill in all these cells all right it's going to take too much time on teste instead now we want to look for the genotype that we need we need little H little H so that's going to knock out any dominant H's right so we can cross out all of these cells because they're going to include at least one dominant H so we're only left with four out of 16 so right away we can eliminate B which says nine out of 16 then we need a dominant a so we do have a dominant a here we do have a dominant a here we do have a dominant a here oh but we have little a little a here so we do not have a dominant a here so one two three are left so our answer here is three out of 16 all right so you don't need to fill in all the cells but you do need to set up the possible combinations and put the axes together and then you can quickly go through and eliminate any genotypes that don't fall into our probability that we're looking for this one was a little tricky so if you have any questions on this final practice problem please go ahead and put them in the comments below all right I hope you enjoyed those practice problems on melan math if this was helpful for you please share it with your Premed Community you know that the MCAT is hard and stressful and sometimes we could all use a little help thanks so much and happy [Music] studying