let us practice working with multiple traits computing genotype phenotypes genotypic proportions phenotypic proportions and so on in this first problem we are asked to compute the fraction of the progeny that will have the genotype literally literally little b little b big c big c big d little d little e little e from across a meeting between these two individuals as always the main thing to remember when working with two or more traits is that because of mendel's law of independent assortment we can work with each trait independently therefore we can write down big a little a crossed with big a little a separately and this is a dihybrid cross which yields a quarter big a big a individuals half heads and a quarter individuals homozygous for little a the next one is big b little b crossed with little b little b and this is a test cross and will yield half big b little b or heads and half little b little b the third cross is also a dihybrid cross big c little c crossed with big c little c and will yield just like in the case of a a quarter big c big c half hex and a quarter little c little c the fourth cross is a test cross for d and gives big d little d crossed with little d little d and we get a half heterozygotes and the other half are homozygous for little d and finally the fifth cross is a dihybrid cross giving a quarter homozygotes for big e half heads and a quarter little e over little e now in order to answer the question we must ask what is the probability of obtaining little a little a in the dihybrid cross for a and therefore the probability of little a little a is one-fourth similarly the probability of having little b little b is a half the probability of having a big c big c is a quarter the probability of having a big d little d is a half and finally the probability of having a little e little e in the progeny is a quarter finally we can compute the probability of having this combination of genotypes using the product rule since the traits are independent and we are asking what is the probability of having little a little a and little b little b and big c big c and big d little d and little e little e so we simply multiply these probabilities together applying the product rule and we obtain an answer of 1 divided by 2 2 to the power 8 or 1 in 256. now one thing to keep um in mind while doing such problems is that there is really no formula that you can use to compute the answer and the answer depends on the particular genotypes that that are being presented to you and it's worthwhile paying attention to the genotypes for example i could change the genotype of one of the parents to big b over big b homozygous for big b instead of big b over little b and if i do that then the calculations and the answer will change for now we no longer have a test cross instead we are crossing two homozygous individuals and all the progeny will be hets and therefore the probability of getting little b over little b is zero you cannot get uh an individual that is homozygous for little b and therefore the final answer is also zero next let us do a variation of these type of questions where instead of being asked to compute the fraction or the frequency of progeny of different genotypes we are asked to compute how many different genotypes the progeny can have and since the cross is the same the genotypes of the parents are the same we can utilize the crosses that we had worked out before but instead of computing the fraction of progeny having a particular genotype now we will compute the number of different combinations that are possible when we combine these five different traits and the best way to do that is to make branching diagrams so you can be big a big a or a head or little a little a and then you could be a head for b or homozygous for little b same here and also for the third possibility for a which is homozygous for little a now for each leaf in this tree you will have three more branches and i'm not going to draw all of them but you could be big a little a big b little b and then big c big c or big c little c or little c little c and then for each leaf at this level you can have two more branches one for big d little d and another for little d little d again i'm not drawing all the branches here and finally for each leaf in the tree at this point you can have three further branches big e big e big e little e little e literally and if we compute how many leafs or endpoints we have at the end we will know how many combinations are possible and so here we had three combinations then each leaf developed two branches so we multiply three with two the third gene multiplies the number of branches by three and the fourth gene multiplies the number of branches by by two and the fifth gene multiplies the number of branches by three because there are three different genotypes possible and that tells us that the total number of leaves at the end or the total number of possible combinations are one hundred and eight next let's do a similar type of problem but instead of computing genotypic proportions let's compute phenotypic proportions and so the cross is given and we have to determine what fraction of progeny will have the dominant phenotype for all of the traits as before let's work with individual crates and then combine them using the product rule to determine the probability of of obtaining of an individual who has the dominant phenotype for all the traits and so here are my crosses and now we know that the first cross for a is a dihybrid cross and there will be a quarter big a big a half heads and a quarter little a little a individuals but in this problem we are not interested in their genotypes but in their phenotypes and there will be three quarters of individual individuals who will have the dominant phenotype and we can write their genotype as a over dash because they will have the dominant phenotype and one quarter of the individuals will have the recessive phenotype and their genotype will be little a over little a and therefore we obtain the phenotypic proportions where this is dominant and this is recessive for the second trait we are crossing two homozygous individuals and therefore all of the progeny are going to be heterozygotes and since big b is dominant they will all be dominant in the third trait we have again a dihybrid cross and so a three quarters of the individual are going to be big c over dash or dominant and a quarter of the individuals are going to be little c over little c or recessive the fourth trait is a test cross and we will have a half heterozygotes are dominant and a half homozygous for the recessive trait and showing the recessive phenotype and the fifth cross is again a dihybrid cross and we will have a three-quarters dominant and a quarter recessive we want to know the probability of having an individual who has the dominant trait dominant phenotype for all the traits and therefore we would like an individual who is a over dash and who is also b over dash and c over dash big c over dash big d over dash and big e over dash and using the product rule we can simply multiply these probabilities to obtain the probability of determining the probability of an individual having this combination of phenotypes and the probability of getting an a dash a over dash individual or having the dominant phenotype is three quarters times the probability of obtaining a dominant individual in this cross for the second rate is one and the probability of obtaining a dominant phenotype for the third trait is three quarters for the fourth trait it is half and for the last trait it is three quarters and doing the arithmetic this gives us the probability of obtaining an individual who has a dominant phenotype for all the traits is 27 over 128