all right so welcome to the chapter three of our lecture in genetics So today we're going to discussed uh a three unit or a three-part lecture on this topic unit one involved the basic melan genetics where in we're going to discuss the experimentation of Gregor Mandel that led him to develop or to uh establish the transmission genetics we're also going to discuss some genetic process hybrid to trihybrid or polyhybrid crosses alongside with back and test crosses as well as the three law of inheritance and the loss of probabilities on unit two we're going to discuss the autosomes and sex chromosomes and lastly for the unit three we're going to discuss the non- mallan inheritance so these days scientists know how you inherit characteristics from your parents they are also able to calculate probabilities of having a specific Tre or getting a genetic disease according to the information they have from your parents or even from your family history this was of course made possible by Gregor Mandel Gregor Mandel was the one who proposed the what they call transmission genetics and he was able to uh perform this by breeding the P plants or pum Saum he was growing in the monasteries Garden after all he is an augustinian Monk he discovered the principles that rule heredity in which we know now as what we called mandelian genetics so mandelian genetics is also known as transmission genetics in which its principle it is the principle of describing how traits are passed from parent to offspring uh of course it was proposed by Gregor Mel as stated earlier and melan inheritance technically or it refers to the certain patterns that are passed down from generations to Generations so who is exactly uh Gregor Mandel so he is an augustinian Monk and a biologist and because of his contributions on Mel and genetics he became the father of genetics but a quick trivia after publishing his work it was not recognized until the turn of the 20th century when a group of scientists um re opened his proposed Theory or his published work and they began to realize that what was proposed by Mandel is the most plausible and the most um acceptable in the scientific community so Mel's experimentation on on transmission genetics began between 1856 and 1863 and between this span he cultivated and tested around 28,000 Garden pee plants he also found out that the plants Offspring retained the traits of the parents in which um such traits of the parents are passed down from different Generations at the time of his experimentation he was not aware of these particles and he just called it particles that are inherited from one parent to another and now given the advancement in the scientific research we know now that these particles he was referring to were actually chromosomes and your DNA Mel was able to establish these through different types of genetic process and and he was able to establish this by using a model organism known as pum sativum in scientific name or Peace by by common name so peace uh he he used this type of plants because they can be grown in a in a small area he cultivated these piece in a very small space in his monasteries and in in a garden in his Monas and these already produce a lots of Offspring in a given period of time so uh through the speci was also able to produce pure plants which he allowed to self-pollinate through several generations and these model organisms can be artificially crossbreed just a quick uh botany uh in this one uh po the plant in this one as you can see in this illustration they contain pollin no pollen contains the sperm that is produced by the stam as you can see in this one and of course in a given plant they also contain the ovary which contains the egg if the pollen pollinated technically the ovary they can undergo what we call Self fertilization and if the plant undergo self fertilization they will be producing Pure or a plant or an organism containing pure strains so I'm discussing a little bit of this background because Mendel uh involve these experimental methods wherein he hand-pollinated the flowers using a paintbrush as depicted in this image particularly he snipped or he cut the stamens to prevent self-pollination of these plant for example and covered each flower with a cloth bag in order to prevent contamination from the other pollin coming from the other plants and to ensure that he will be pollinating from his his plants from his observed plants through this he was able to trace the traits through the several generations and these TRS that we're actually talking about are are technically uh summarized in some of the genetics book I will be showing that to you uh later on so how Mel began Mendel produced a pure strains by allowing the plants to self-pollinate for several Generations so technically uh these plants if they self-pollinate they will be producing uh a pure strain for several Generations until such time that Mel will be able to crossbreed other F2 to F3 Generations but anyways Mel began observing different pant traits to determine which characteristics were inherited and which characteristics were not and these traits that I was referring to are actually in this table these are some of the few examples we have from the seed shape all the way from the stem height this column of F1 and F2 results are technically The Offspring from the first and the second generations and some of these uh detail are presented in a ratio we call this a genotype ratio um you will be conducting uh some of this sooner or later Mel particularly use the term parent or P1 to F2 Generations um when we say parent or P1 generation this is actually the original plant that he bred and uh do take note that parent generation contains pure strains and then uh the resulting offspring of our F1 generation is none other than the F1 Generations as you can see uh most likely when we cross two pure uh parent plants one that is pure dominant or one that is containing a homozygous dominant and one containing a heterozygous a homo zos recessive rather when we crossbreed these two parents most likely the F1 Generations are all heterozygous uh contains heteros alil so uh with that being said if they are heteros alil they will be containing one dominant alil so that being said what was reflected in all of these uh F1 Generations are all dominant alil as well so F2 Generations actually are um organisms or technically the generations resulting from breeding from two parents from F1 generation and as we can see in F2 Generations there are there's a recessive treat that that reflected or expressed in one of The Offspring so we've been talking about genetic process and um and there are many different types of genetic process we have from monoh hybrid to poly hybrid crosses and this varies depending on the number of traits that we will be Crossing mono we will be Crossing only one single trade from both of the parents di hybrid cross two traits and poly hybrid cross three or more traits so genetic Rosses are usually carried out through a what we call Planet Square which is just a diagram that is used to predict specific combinations that we can actually arrive to from given the genotype of a given parent so in making AET Square let's consider this problem determine the genotype and phenotype ratio of the resulting Offspring if both of the parents contain heterozygous yellow P alil before we start of course it's very important for you to familiarize yourself with some terminologies when we say traits technically these are characteristics that of an organism so these are this may refer to hair color seed color or shape flower color flower shape all characteristics that are technically expressed in a given organism are known as traits and in each given trats do take note that that they contain two alyss so for this one for this example is a trat for the color of the piece let's say um this is a p color trait and then in each trait it contains two different alyss Each of which are represented by an uppercase letter or a lowercase letter or two uppercase letters or two lowercase letters it depends on the organisms one of the traits may contain dominant alil and when we say dominant Al it is usually represented by an uppercase letters these will carry the trait that is usually expressed in an organism if there are two two types of alil in a given trait let's say like this one the dominant alil will be expressed in an organism for example that this uppercase letter is a dominant alil which carries the yellow p alil and this lower case letter technically is the recessive alil which carries the Green P Al if there is a dominant if uh if it if there's a dominant alil in a given trait it is the dominant alil that will be expressed in an organism so with that being said this plant will have a yellow P since the yellow PE is the dominant alil over the recessive one which is the green alil the recessive alil on the other hand is represented by lowercase letters and this carries the trait that does not usually appear in an organism um it does not produce a trait at all when only one copy in a given trait is present especially when it is with a dominant alil for a recessive tree to be expressed in an organism two copies of this recessive alil must be expressed in a given trait in this example the recessive alil carries the Green P trait but since as I said earlier we have here a dominant P alil which carries the yellow P alil the plant will Express yellow color in uh in its p lastly uh some of the last reminders here you might encounter terminologies such as heterozygous homozygous dominant and homozygous recessive when we say heterozygous um the plant or the parent carries two different types of alil one that is dominant and one that is recessive when we say homozygous dominant the plant may carry two dominant or two uppercase uh letters as you can see in this one and when we say homozygous recessive the plant carries two recessive alil as you can see in this one so this is the only time that the plant will Express the recessive traits so going back to our step-by-step process in making a planet Square considering this um considering this problem as we can see step one is to make the Grid or make a 2X two pet Square just like this one so place the alyss of the Gam of one parent along the top portion of the greed and those of the other parent along the left hand side as you can see in this one for your Vis uh visualization I particularly created this one wherein the parent one carries heterozygous yellow PE alil since both of the parent carries uh actually both of the parent carries heteros yellow PE alil so meaning to say one of which is dominant which is the yellow p and one of which is recessive which which is the Green P alil so in making your Panet Square just input both of these alil into our table as you can see in this one one is in the column and one is a row step two is basically to fill in the greed just combine the parent alals from both of these geds and um you will be able to come up with a genotype and phenotype ratio but for the purpose of this lecture I particularly assign numbers and letters in each stable as you can see um by means of combining the alals uh the first step is technically is is technically to take the first alil of the first parent or the uh y1 and multiply it to the first alil of the second parent which is technically the Y3 and our answers are reflected in box a in this one which is technically y why this one is homozygous dominant the second step is just to carry on with the remaining boxes so we have here y1 multiplied with y4 now we have y uh dominant Y and recessive y for our box C and just carry on with the rest of the boxes for step number three it's time for you to fill in The Offspring by using the law of dominance as stated earlier you have to uh uh determine the the genotype and phenotype ratio this particular Offspring will have a combination of a homozygous dominant why why with that being said um we are expecting that this organism will have the yellow P Al because it contains both of the dominant trats on the other hand two of The Offspring might actually display a heterozy alil so with that being said they contain one dominant and one recessive and according to to our discussion earlier by using the law of dominance if there is one dominant alil in our threats the dominant treat will be expressed hence uh the the color of the PE will still be color yellow lastly this organism or this Offspring inherited both of the recessive alals of both of the parents so with that being said what will be expressed in the organism is the recessive threat which is technically the Green P the next step is to determine the genotype ratio this is just a very simple test because you technically have to segregate this different output the genotype ratio is 1 is 2 is is to one the one here represents this one technically we have one homozygous dominant the two here represents two heterozygous and the one here represents this one which is technically which leaves us with only one homozygous recessive the phenotype ratio is much easier than the genotype ratio because technically you will have to uh determine their appearances how many plants will have the yellow py and how many plants will have the Green P Aly since three of them contains the dominant treats as um stated in our genotype ratio 1 two three three of them contains the dominant alil three of them will have yellow p alil and um only one contains the recessive traits or the green P Al I particularly attach here a YouTube um practice you might want to click on this icon if you need more practice on setting up your planet Square these are some of the summary of terminologies that I used in the previous discussion but I already explained them so I just want to skip this part but you might want to uh review this uh terminologies as it will help you to further understand our discussion in this particular part of this recorded lecture please uh stop this a lecture and try to answer this by yourself all right so the answer for this uh short exercise is technically 2 is to two for our genotype ratio and for our phenotype um technically we don't we don't have a phenotype ratio because all of The Offspring will have the same characteristics our problem here is that we need to predict our our Offspring if our parent one contains the homozygous dominant yellow p alil and the other contains the heterozygous yellow p alil so parent one contains y y or contains both dominant alil and parent two contains one dominant and one recessive so we we just have to in uh put that in our planet Square this one technically represents parent two as we can see one dominant and one recessive this one technically represents our parent one which contains both dominant alil so conducting or multiplying this one or conducting cross genetic crosses in between of our traits we arrive to a genotype ratio of 2 is to two so two here represents that there will be two homozygous dominant and two here represents that there will be two heterozygous alil in our Offspring if we're going to um predict it's uh which traits will be expressed in these organisms since all of them technically contains one at least one dominant alil the dominant alil is technically the yellow piece it means that our penot typ Rao all of our Offspring as expected will contain the yellow seeds or yellow PE rather going back to the experiments of Mandel he particularly um crossed uh or he particularly observe the uh shape or the texture rather of the bees as you can in this one um Mel wanted to know wanted to determine the genotype and phenotype ratio of the F2 Generation which is technically the third generation from this cross if both of the parents have homozygous dominant round calil meaning to say parent one contains both dominant alil having the dominant tra is the round seed and parent two contains homozygous recessive which is technically a ring called seed alil so uh what Mandel conducted was to cross these two parents and if you're going to predict The Offspring or the F1 generation from this through a planet Square uh the resulting Offspring will be like this one so if we're going to cross uh two pure parents or two homozygous dominant and homozygous recessive um all of The Offspring will contain heterozygous alyss which is already expected because we have here dominant and we have here um recessive trats and such 100% of our um Offspring will contain heterozygous alil and if I'm going to ask you what do you think is the characteristics of the resulting or what do you think is the phenotype ratio of or phenotype rather of the resulting Offspring from this cross or from this breed you might want to pause the video if you want to answer but the answer in this one is that technically all plants will contain the round seeds since all of them round here or the round um piece technically is the dominant trats and since all of The Offspring contains at least one dominant alil they will be um they will be expressing round piece than the wrinkled seed again uh recessive traits could only be expressed if a specific um traits contain two recessive alyss on the other hand Mandel is not yet finished on this one as he wanted to know what will be the result a resulting Offspring if he will be um cross breathing two Offspring or two parents from the F1 generation so RR technically times RR and if we want to predict that one we have to put that in our planet Square just like this one and this will be our resulting Offspring which is technically 1 is to two is to one or one homozygous dominant two heterozygous and one homozygous recessive so our expected offspring would definitely look like this so three of them which is technically this one this one and this one would contain round seeds round seeds because each of them contains at least one of the dominant alyss and only one of The Offspring exhibited The wrinkled seeds so some of these piece have a smooth texture while others are wrinkled because technically um in the F2 Generations the recessive treats are already reflected now let's talk about the test cross so what is a test cross so test cross is technically a continuation of what Mel conducted so last time in the in the previous slide we uh we stopped from this generation but in testc Cross um it a further uh genetic Cress or further breeding of these of the plants from the F2 generation will be conducted so there are two possible test crosses here one that is between between a homozygous dominant and a a hybrid or a plant containing the heterozygous alil and one between a heterozygous and a heterozygous recessive and a hybrid the result of course um if we want if we want to predict that one we will be using another planet squares in this one and our genotype ratio is technically one is to one in which we will be having um the ratio of the homozygous dominant and the ratio of the heterozygous or hybrid is just the same so this is 50 50% the same can be observed in heterozygous recessive and uh I mean to say homozygous recessive I'm sorry this one is a mistake I would be editing this one the same can be observed in homozygous recessive and across between a heterozygous as you can see in this one so most geneticist nowadays they utilize this test cross in order for them to determine a specific phenotype or genotype ratio of a given parent um especially if that parent has a known genotype ratio so for example I have here a a problem set determine the genotype of this plant with dominant yellow P so what is known technically is that this plant contains a dominant yellow P Al but we are not sure if uh that if this plant carries a homozygous dominant or a heterozygous alil so if you want to identify this one geneticist usually use testcross because the resulting Offspring will tell you everything you need to know about the genotype of that parent so the first step in test Crossing is that you breed the plant or technically you breed the plant the the plant with a known genotype ratio um with another plant however um uh to an organism you you breed this plant to an organism with a recessive phenotype so technically that is a small y y as you can see in this one and then if the resulting offspring are all all exhibited um all exhibited the dominant alil this means that our parent with a known genotype contains or contains a homozygous dominant alil this one why why on the other hand if the offspring are technically um 5050 meaning to say 50% of them exhibited yellow P seeds or yellow P Al and 50% % of them exhibited Green Seed alil this means that our unnown that a parent with a known genotype contains a heterozygous alil so this could be predicted through the planet squares as well now we're done with the test cross in this part we're going to discuss the back cross Naman back cross technically is a cross between the F1 generation and the parent in the in the test cross this is a cross between two parents from the F3 Generations now if the test cross is used technically to determine the aown genotype ratio of a given organism back Ross is used to uh uh usually back Ross is being utilized by Farmers as it can be helpful to them for the or the plant breeders to design the breeding as per the demand or technically if they want want to produce a specific strain of the plant according to the demand of the economy or according to the to the demand of the market so how is back cross being conducted so number one is to take one parent from uh one one parent which is a pure uh which contains a pure strain and take one organism from the F1 generation and conduct genetic process and the resulting Offspring it would give you 50% Pure or homozygous it's either homozygous dominant if you take this parent or homozygous recessive if you take this parent and the other half would give you 50% heterozygous so in summary of the results of an hybrid chosis inheritable factors or genes are technically responsible for all heritable characteristics phenotype is based on genotype but the difference say is that in phenotype um technically it's the appearance of our organisms each trait is based on two genes one from the mother and other from the father and true beaing individuals are homozygous in which um both alals are either dominant or recessive and are pure now what are the three laws of inheritance these three laws of inheritance are very important to consider especially here in melan genetics because this one uh technically holds the keys for um some explanations regarding uh inheriting characteristics from parent Offspring the law law number one is the law of dominance in this one in a cross apparance that are pure for contrasting traits only one form of the trait will appear in the Next Generation and all of The Offspring will be heterozygous and express only the dominanta in summary law of dominant uh states that um the dominant rates will always be expressed in organism as you can see in this one all of our Offspring from the F1 Generations exhibited the dominant trait because technically a cross between a homozygous dominant and homozygous recessive will give one dominant alil to each of our to each of our Offspring so with that being said all of the dominant Al here will be expressed the law of segregation on the other hand states that during the formation of GTs the two Al responsible for a treat separate from each other so for example this is from your mother and this one is from your father technically this one uh separates the two traits technically separates and they are the ones that are being randomized during genetic process this one is uh basically your planet Square so Al for a treat are then recombined at fertilization producing the genotype for the trats of The Offspring so that's a very random um possible combinations in law of independent assorment to summarize it states that the traits are inherited independent L and these traits are never in group so for example if your mom has a curly hair and your dad has freckles and of course um the resulting Offspring could be a boy with curly hair and freckles so technically these um treats are inherited independently and they do not come in group so the Law of Independent Assortment can be Illustrated further using what we called dihybrid crosses in dihybrid Cross this is actually a breeding experiment that tracks the inheritance of two traits in the past uh in the previous example we have the monohybrid cross wherein we are trying to predict the possible gamut combinations or gamut combinations uh the possible traits of a given Offspring considering only one traits in this one in the hybrid cross we are now considering already two traits so it could be either wrinkled seeds and yellow color or yellow PE rather or it could be color of of flower color or flower of the seeds uh it depends on the traits but in the hybrid again we're talking about two traits so this could be uh the DI hybrid cross is a very perfect uh it's a perfect example rather of uh law of Independence assortment wherein each pair of alil segregates independently during gamut formation so uh in in conducting at the Hybrid cross we usually conduct a calculation wherein this is the actually formula to rais to n where n is the number of heterozygotes in order for us to predict the possible number of Gamin combination in a given traits so let's have a an example so how many gamuts will be produced for the following alil Arrangement so do remember our Formula 2 rais to n where n is the number of heterozygotes you might want to stop this video and try to answer this on your own all right so the answer here is for the first one it it would be four gamuts because there are two heterozygotes here substituting that to our Formula 2 ra to two is technically four in this one I have here 1 2 three we have three heterozygotes that two double C is not heterozygote that's actually a homozygote so because it contains two dominant uh alyss remember that heterozy are uh contains two alyss that are opposite to one another so that's 2 ra to three so that's why the answer is eight gamuts so 2 * 2 * 2 and in in the last one we have a total of 64 gamuts okay now in the hybrid cause we also create a P Square in that one in order for us to predict the possible genotype and phenotype ratio of our parent let's consider this problem determine the genotype and phenotype ratio of the resulting Offspring if both parents contain heterozygous yellow round p alil so as we can see in our plant organism the yellow and round p are reflected or expressed in our plants this means that the yellow and round traits are both dominant atates or dominant alyss all right in this case we are already dealing with a hybrid crosses because we are now dealing with one color of the p and second shape of the P so this is already two traits now the question is asking us what could be the possible um uh characteristics of of these two combinations so before we start uh again in the hybrid Cress each parent now car carries two traits and each trait now has four alyss a while ago in the monohybrid Cress each each traits carries two alyss one could be dominant one could be recessive in the hybrid process since we are now considering two traits we will be also we will also be considering four different types of alyss so in this example the two traits contains heterozygous alyss as you can see in this one with each trait having one dominant and one and recessive alyss all right next slide so the trait RR which is technically the trait one carries the shape of the seed having round seed as a dominant trait while wrinkled seeds as a recessive trait these characteristics are just brief examples trade two YY on the other hand carries the color of the seeds having yellow seeds as a dominant tra while green seeds as the recessive tra now we're um to just to visualize our problem I specifically created this um representation as you can see right now uh again step one is to predict the possible gamut combination on each parent so we're going to use our example 2 rais to n so in this one the number of heterozygotes are two because there are two um uh there are two heterozygotes in our parent one also two heterozygotes in our parent two calculating these possibilities we have four possible gamut combinations in each parent so remember there are two traits in in this example trait one carries the shape of the P while trait Two Carries the color of the p now we have to predict the possible gamid combination on each parent the first one is you have to employ the foil method step a is to multiply is to multiply the first Al of trade one to the first Al of trade two thus our first possible gamut combination is r y this one step two on the other hand technically is um multiply the first alil of uh um uh as you can see here multiply the first alil of trait one to the second alil of trait two thus our possible our second possible Gamin combination is R dominant R and recessive Y in Step C you'll have to multiply the second alil of trait one the recessive R to the first Al of tra two thus our third possible Gamin combination is R recessive r and dominant Y in Step D you have to multiply the second alil which is technically the recessive R of trait one to the second alil of trait two so this is recessive to recessive thus our fourth possible gam combination are recessive R and Y so in this case as you can see just like when you were in Elementary uh this is technically a a foil method no so you will have to uh multiply um each each of these traits together in order for you just to predict possible gamut combinations so with that being said since both of these parents are just the same they just they contain the same genetic traits uh both of these parent uh has this possible gamut com com combinations this is based from our foil method now you'll have to take these possible gamet combinations and you will be uh putting them in our Planet squares just like the monohybrid cross you will have to uh input both of these possible comment combinations on the top and left corner for the top corner it's for either of the parent parent one parent one or two and for the left corner here as you can see you'll also have to input these gamut combinations of the other parent as you can see in that one okay so just like the monohybrid cross you can interchange but in this point it's pointless because technically they have the same gamut combinations all right so fill in the grid and combine the parent alyss inside the boxes so just like the monohybrid cross we're going to take this one and then multiply it with this one the resulting answer is this one dominant R and dominant y likewise uh continue while you fill in the boxes we're going to take this one and multiply this one to this one and the answer will be this one we're going to take this one and multiply it to the third column and then the answer is this one we're going to take this this one and multiply to the fourth column and the answer is this one so technically you you'll just have to follow the steps in each of the row of our pet squares and if you notice our pet square is now a 4x4 pet Square now you'll have to identify the genotype and phenotype ratio such that you will have to quantify each of these traits that are unique to one another like for example this one represents the r r y y because there are only one R dominant RR and dominant y y across our um results in this one and continue your way up to the second column up to the third column and Ford column until such time that you able to account every single possible genotypic ratio on the other hand we also have here's uh the phenotype ratio wherein um you will have to predict the color and the shape of the seeds or the PE rather again if there is a presence of um of a dominant alil it is the dominant alil that will be reflected like for example in this one this is dominant R and dominant y with that being said the this Offspring or this possible gam combination will have a phenotype of round and yellow as represented by this one on the other hand in this example we have recessive R and dominant y so that being said this one will have a phenotype of wrinkled and yellow so once that you are able to identify the phenotype of every single genotypic ratio of every Offspring you will have all you will have to uh quantify them as well and present them in a ratio like for example in this planet Square there are nine round yellow in this example there are three round greens there is there are three wrinkled yellow and only one wrinkled green thus our phenotype ratio would be 9 is three is to three is to one if you find this a bit confusing I particularly link a YouTube link on uh I particular I particularly link a YouTube um uh a short video on setting up a DI hybrid cross in a planet Square now let's move on to the try hybrid and poly hybrid cross considering this problem determine the genotype and phenotype ratio of the F1 generation if both of the parents contain heterozygous yellow round seed and white flour alil and parent two is homozygous to the recessive traits meaning to say that the parent one carries all of the dominant traits the dominant traits with regards to color of the seed are yellow round with regards to the shaped and white with regards to the color of the flower so these are the dominant traits because they were expressed in the organisms on the other hand the phenotype of the parent two is the exact opposite of the phenotype of parent one because according to our problem they carry all the recessive traits now before we start in TR in Tri hybrid and polyhybrid Crosses each parent carries multiple traits and multiple alals since this is a trihybrid this parent now carries three traits trait one which is represented by the letter A carries the traits for the seed shape trait two as represented by letter B carries the traits for seed color well trait three as represented by letter C carries the traits for flower color the more traits that we consider in a given cross the more possible number of gamut combinations and the more possible number of gam combinations the larger our planet Square will be for the monohybrid cross we only have a a 2x2 Planet Square in the hybrid it's 4x4 in this one it's 8x8 Planet square but before we jump in into creating our planet Square it's important for us to predict the possible gamut combinations applying our Formula 2 rais to n where in we uh n here technically is the number of cyot heterozygotes and since there are three heterozygotes in this example 2 ra to 3 is equal to 8 so meaning to say we need to predict at least eight possible gam combinations for this problem in order for us to set up our planet Square so in in predicting possible gin combinations like this one you'll have to take letter per letter in each trits step a take the first letter of the first trit and multiply it to the first letter of the second and third trat in this example we got all the dominant traits of letter A B and C step two is take the first letter of the first trate and multiply it to the first letter of the second trate as well as to the second letter of the of the third trate and as such we got dominant trats A and B but the recessive letter c as stated in this one step C take the first letter of the first trate and multiply it to the second letter of the second trait as well as to the first letter of the third trait now we got in this example we got the dominant A and C but the recessive B which is which brings us to our third possible gamut combinations in Step D you will have to still take the first letter of the First Rate and multiply it to the second letter of the second as well as to the third tra in this one we got only one dominant traits which is letter a and the rest of the traits contains recessive traits B and C in Step letter e since we already have four possible gam combinations but according to our calculations we need eight so we still continue in Step e you take the second letter of the first trate and multiply to the first letter of the second tra as well as to the first letter of the third trait and here we got one recessive letter a and the rest of our traits are dominant step f take the second letter of the first grate and multiply it to the first letter of the second grate as well as to the second letter of the third tra our six gamit combination contains two recessive letter A and C but a dominant letter B as we can see in this one for step G take the second letter of the First Rate and multiply to the second letter of the second rate as well as to the first letter of the third trate in this possible gam combination we have two recessive traits recessive A and B but dominant letter c as seen in this one our last possible gam combination take the second letter of the first trate and multiply it to the second letter of the second trate as well as to the second letter of the third tra in this case our possible gam combinations are all recessive letter A B and C once that we are able to uh to predict the possible gam combinations of each parent we can now we can now be able to create our planet squares so these technically are the the eight possible gaming combinations that we calculated earlier from parent one and since parent two only contains a recessive traits as you can see if you apply the same method you will be arriving in the same answers it's all recessive traits and as such just continue creating your pet Square for for example if if I'm going to take this one uh the the cell number one of this column I'm going to multiply it to cell number one of this Row the answer between these two is this one and continue moving through each boxes until such time that you will be able to finish and fill in the possible gamut combination probability and genetics so Mendel was the first person to realize that probability can be used to predict the results of genetic process in other words he could use probability to know that all of The Offspring of the first generation would be tall without even seeing them and just mere calculations now we have two um we have two portions or technically two rules in probabilities it's either rule of multiplication or rule of addition in rule of multiplication uh technically we use this for independent events in quins so let's have an example in a cross between PE plants that are heterozygous for purple flower color what is the probability that The Offspring will be homozygous recessive so we are being asked to calculate the probability given these the traits of these parents so let's try to understand these the following traits so what was stated in this one is that if we have this parent what is the probability that their offspring will be will have the homo will be homozygous recessive so to be honest a planet Square can actually tell you the answer so let's try to build a planet Square here so I have here four boxes this one is blank and then our planet Square P letter P P recessive and then begin constructing your pet Square all right the question here is we are only interested with the homozygous recessive but if we're going to analyze our planet Square here to be honest there are 25% of The Offspring is homozygous dominant on the other hand um 50% could be heterozygous and lastly there is only 25% as well by percentage that our Offspring will be homozygous recessive so the answer in this one is 25% 25% or technically 1 four or 1/4 either way is okay now this one may take some time especially if you're answering an examination um there's another uh method to follow which I call this chit where I follow technically this derived answers from um every possible uh parent cross breeds so for example if you cross a heterozygous and a heterozygous parent um um 25% or 1/4 or there is a 1/4 or 25% that The Offspring will be homozygous recessive and 25% would be homozygous dominant and 50% would be heterozygous so technically this one the ones that we created are the same with this one however if you cross two homozygous parents let's say pure dominant multiplied with pure recessive you will be arriving to 100% heterozygous offspring on the other hand um we have pure dominant and pure dominant of course our Offspring will be 100% homozygous dominant but if pure recessive and pure recessive our upspring will also be 100% homozygous recessive however in the cross between heterozygous and homozygous in this one um 50% or 1/ half of the time or 50% there is a 50% chances or probability that The Offspring would be homozygous dominant and the other 50% are technically heterozygous right so if we're going to apply the cheat sheet this one in our question so so it's PP times dominant P recessive P what are the odds that The Offspring will be homos hyos recessive you just have to look in our cheat sheet sheet you can actually memorize this one so since they are both heterozygous both heterozygous parents we're going these are both heteros parents we're going to look at this portion right because it's heteros times heteros and according to our cheat sheets there will be 25% that The Offspring would be homos recessive and that's your answer right it's easier this way I think you just have to memorize this let's have another example especially if it's already a DI hybrid cross right so this one is an example of a dihybrid cross uh in a cross between P plants that are heterozygous so both of our parents are heterozygous as well for two traits purple spotted flower what is the probability that The Offspring will be homozygous recessive so uh technically if I'm going to write that down that's y y r r again as discussed earlier this could be um sorry let me just complete the equation let's say as discussed earlier maybe this uh specific trait is trait one this hold uh the trait for the color of the flower and this Tre technically is stet two which holds the trait for the spots or of texture of the flower so the question is asking what are the odds that these two parents will be producing an offspring containing homozygous recessive right now I'm not going to conduct you can actually create a planet square and you will be arriving in the same answer but supposedly that this is an examination you no longer have time to create a planet square and that is the time where our cheat sheet comes into play first determine the parents parents are technically homozygous right so this is a crossbreed between homozygous and homozygous the technique in di hybrid process is that first you'll have to determine or calculate them trait by trait and it asks what are the odds that it will be producing recessive traits let's first determine trait one so y y or the heterozygous y multiplied with a heterozygous Y what are the odds that we will be arriving to our recessive traits let's look at our chit sheet so we we should be looking on this um on this data because both of our parents are heterozygous and then we are looking for a homozy for the probability of homozygous recessive which is Tech techically this one right so with that being said for this alone for this alone we have one4 or 25% chances now we still have tra two so trade two here is technically this one multiplied with this one what is the chance that they will be having this look at our uh chat sheet again so let me just delete this one cheat sheet again so both of our parents in R are technically heterozygous so we're going to look at this data and we are looking for a homozygous recessive which is also this one so it's also 1/4 so 1/4 now it we have two data already one from trit one and the other from tra two now what what what you can do here is technically to apply the rule of multiplication so 1/4 * 1/4 is 1 over 16 now in most genetic books um um um it is actually preferred to have this or Express this in percentage so how do you express this in percentage simply divide 1 by 16 and then multiply it by 100 the answer is 6.25% or 1 over 16 all right let's have another example for TR hybrid crosses so in a cross between a a BBC and the other parent what is the probability that The Offspring will be this which is technically an offspring which contains um heteros alyss so again I'm not going to conduct my Planet squares here but you can even you can actually try to construct a planet square and you will arrive in the same answer but in this one in the for tri hybrid cross I'm not going to construct any pet squares since it's too time consuming especially if this is an examination so uh according to this one the answer is 1 over8 or 12 12.5% chances but let's try to calculate that using our chit sheit so I have here AA let me just copy my equation a a BB do I got that correct pure heteros zycus multiply a a BP and CC what are the odds that they will be producing the same trait as the parent one which is technically heterozygous alil to all traits all right let's apply ours technique so uh let's first uh calculate them tradeit per trait so we have three traits here here tra trait number one multiply with this one trait number one what are the odds that they will be producing this one which is technically heterozygous as well let's look at our CH shet heterozygous time heteros which is this one and we what are the odds that they will be producing another heterozygous which is technically this one so the answer is 1/2 so I'm just going to input here one half plus let's move on to the second trait we have here trait number two multiply to this one now this one is a different combination because we have here um we have here one heterozygous and one homozygous dominant so let's try to look at on our cheat sheet which is this one since one het and one homus dominant so the an so we are currently looking for the probability of of this one if we cross this one to this one what is the probability that they will be producing this which is a heterozygous and since we are currently looking at this one and we want heterozygous the answer is 1/2 so plus 1/2 now we are done with our last tra which is this one and this one this is a cross between a heterozygous and homozygous dominant the same with the previous one if we're going to cross this one and this one what are the odds that they will be producing this right a heterozygous as well so we're going to look at this um cheat sheet because it's heterozygous times homozygous and we are currently looking for this one for the heterozygous so the answer is 1 12 plus 12 oh I mean multiply because this is Rule of multiplication so 2 * 2 is 4 * 2 is 8 the answer is 1/ 8 the same answer with this one right so 1/ 8 you can actually convert that to percentage 1 uh divided by 8 uh the answer is is uh the answer should be uh multiplied by 100 the answer is 0.125 we multiply it by 100 the answer is 12.5% so meaning to say there are 12.5% chances that The Offspring will be the same with the parent one okay in rule of addition essentially they are just similar with the rule of multiplication but from the name itself of course we use addition so we also we will also be using our cheat sheets here but just another definition uh rule of addition is usually applicable for mutually exclusive events or meaning to say two or more events that cannot happen simultaneously let's have an example in a cross between P plants that are heterozygous for purple purple flower color what is the probability of The Offspring being heterozygote converting that into an equation form we have PE uh heterozygous alil uh both of them are are heterozygous alil both parents are and then it is asking what is the probability what is the probability that our hopes there that our Offspring will be hetero as well right so you can again you can create a planet square and to be honest I already know the answer from our past calculation and I I do hope you also know the answer with this one so I have here P homozygous dominant heterozygous heterozygous homozygous recessive so the question is asking about this heterozygous recessive which is technically this by percentage it is 50% so the answer here is 50% or 1/2 next in a cross between these stats what is the probability that The Offspring will be AA will be the following and don't want to read that anymore so I'm going to try and um copy this to another page which is this one so we have so we will have enough space to calculate okay so the offspring that we wanted to um find is either the other one or the other one which is this one or this one okay so I have no I have to write it Pala talaga a a uh BB c c but technically heterozygous times a a b b dominant dominan C what are the odds that um The Offspring would be this or this okay now in rule of addition when you see the word or normally you will be using the rule of addition than the rule of multiplication but if you see the word end you will be using the rule of uh multiplication so as you can see based from our previous definition of rule of addition mutually exclusive events meaning to say uh events that can happen at the same time so technically uh they are actually describing this one these are events that cannot happen um at the same time because it's asking you uh what is the probability that you will have either this one or this one but anyways going back let's calculate the probability of these mutually exclusive events using our cheat sheets once again so let me just transfer this one so I could access my cheat sheets easier just like the rule of multiplication we're going to um we're going to calculate this by uh traits per trats so starting with letter A a and then cross that one with the other parent which is technically this one this is a cross between heterozygous parents and we wanted to know what is the probability if we cross two heterozygous parents what is the probability of The Offspring to have a homozygous dominant treat so going back to this one both of our parents is heterozygous and we wanted to know if what's the probability of our Offspring having homozygous dominant tra technically the answer is 1/4 this one is oh wait let me oh no let me just delete that okay so technically the answer in this one is 14 now we're not yet done because we have we still have two remaining traits left I'm going to calculate this one as well as with this one you can see so this is heteros this is a cross between heterozygous parent and a homozygous parent as you can see here and what are the odds that they will be producing this which is heterozygous now let's try to look that let's try to look that up in our cheat sheet so this is a cross between heterozygous and homozygous so hetero time homo which is technically this one and what are the odds that the offspring would be a heteros the answer is 50% or 1/2 so I'm just going to write it down in this one 1/2 right last rate I have here um this one which is heterozygous multiply with a homozygous technically the same but let's just let's not skip the process so hetero time homo technically this one and you wanted to know what are the odds between this two parents that will be producing a heterozygous Offspring so technically the same answer so that's 1/2 as well 1/2 Now set that aside now we need to calculate as well this one the other possibilities so but um let's set that aside our information so far oh see let me delete Nang so that 1 12 okay 1 12 and then I still have this other side that needs to be calculated so I have here a a b b CC multiply with a a BB and C C working our way towards these streets first starting with letter A multiply with this one so take what are the odds that they will be producing this trits hetero time hetero and we're looking for homozygous dominant going back to our cheat sheets technically hetero times hetero and we're looking for dominant the answer is one4 one uh 1/4 and then moving on to the next trate we have letter B here which is hetero times dominant what are the odds that they will also be producing a dominant rate so hetero times um hetero times homozygous dominant so we are looking here on our cheat sheet what are the odds that they will be producing a homozygous dominant um the answer is 1 half one half lastly I would be taking out the last rates multiply this one what are the odds that they will also have this homozygous dominant rate so hetero time homo hetero time homo in our cheat sheets that is this one hetero times homozygous what are the odds that they will be producing using this a homozygous dominant which is this one technically the answer is also 1/2 1/2 oh now before we add you add these two uh possibilities but before you add them you have to multiply each number in the parenthesis so let me just erase this one for more space and rewrite them oh no so anyways I'm just going to calculate so anyways 1/4 * 12 * 12 + 1/4 * 1/2 * 1/2 the answer is this one as you can see so uh 4 * 2 that would be 8 * 2 that would be 16 so this one is 1 over 16 we have the same number numers here so this one is 1 over 16 as well so you'll have to add this one all right so going back so for example 1 over 16 + 1 / 16 = 2 over 16 and we're going to simplify that uh the uh the simplified form of 2 over 16 is 1/8 the answer is of course 1/8 so meaning to say um in a cross between these two parents the probability of them that will be that the oping will be this one or this one would be 1/ 8 you can actually um uh you can actually convert them into a percentage so 1 / 8 * 100 the answer is 12.5% right okay so when do we use the rule of multiplication so when we have independent events in sequence you multiply the probabilities of the independent sequential events so for example in here in heterozygous we plans what is the probability that The Offspring will be hom Moos recessive as we can see in this one we only have independent events in this one that are separated to one another but um my clue as I said before if you see in the question the word end most likely you will have to use the rule of multiplication but for example what is the probability that The Offspring form across between AB C while parent two is wyz will be ABC end ABC so for example uh just watch out for the word end on the other hand for the rule of addition you will be using them for two or more mutually exclusive events the clue word here is the word or so if you see the word or in your question most likely you will have to use the rule of addition right oh sorry okay I have some another examples here this one is uh much more difficult not really that difficult the question is suppose that parent one carries heterozygous green stem yellow seed and white flow treat so this is Tri hybrid cross while parent Two Carries the homozygous recessive traits what is the probability that The Offspring will also contain the homozygous recessive traits now again as I said before we can definitely carry that out through a planet square but you can definitely uh but you don't you no longer have the time to create this very complicated Planet Square especially if you are in in the middle of an examination you can just carry out on our cheat sheets which is my derived formula that I created for you so through cheat sheets we are able to calculate it that the answer would be 1/ 8 and it actually similar with our planet Square here as you can see uh the planet Square tells us that 8 out of 64 passible G it combination will have the homozygous recessive trat so 8 over 64 is technically 1/ 8 essentially if you're going to use the cheit it will also give you 1 over8 answer essentially the same answer with our planet Square so um 1/ 8 is converted to a percentage so that's equivalent is 12.5% okay moving on from unit one now we're going to discuss about non- Mandela inheritance so not all um not all uh genetic Offspring or genetic variations rather could be explained by Mendel's Law And there are some few exceptions that happens during the fertilization process going to inheritance of traits that quite devat from the rule of Mandel and this will be discussed under non-mendelian inheritance hence the term non-mendelian now uh While most bearings led to offspring with genotypic and phenotypic ratios that that matches those expected from Mendel's loss some deviations were also found these natural deviations are examples of non-mendelian inheritance which could be two types mendal law do not apply at all and second genotypic ratio is that expected from mendal law but the phenotype ratio is not exactly the same or does not match with our genotypic ratio so let's have an example a type of non-m inheritance is incomplete dominance as we can see in in complete dominance our Offspring here have an appearance that is somewhat in between the phenotypes of our two parents so we have our parent here is that we have uh Two Roses uh one parent the red one carries h a homozygous dominant red uh flower a little while the white roses carries a homozygous recessive white flower alil supposedly our expected output or expected offspring for F1 generation should also have the red coloration because um a cross between these two would be producing pure heterozygous Offspring however in incomplete dominance um the dominant alil was not able to be fully expressed in The Offspring and as such it displayed a color a different coloration so in this case it's color pink we also have here codominance in Co dominance uh two Al are expressed in heterozygous individuals um Co means uh two and dominance which refers to dominant alyss so this means that two traits are expressed in a given organism so for example we have here um two different animals or two parent animals which uh the other one contains white fur and the other ones has a red fur or brown fur and their resulting Offspring technically is spotted which contains both of their colors another example of codominance is the flower R rododendron wherein the flower here appears to be spotted one that is red and the other is white neither of these neither of the dominant or recessive treats are um was able to um to be fully expressed and a such uh two traits were expressed in this organism so this is also another example of codominance wherein type A and type B are technically codominant however there is a recessive alil for type O but if type A type A and B show Regular dominance over this recessive alil so that being said there is a very low chances for one to have type O thus the only way to be type O is to be homozygous recessive now I would like you to ponder about this question and pause the video for a while is viigo which is the skin discoloration as you can see in this image a case of codominance you might want to post the video all right so actually this is not a case of codominance because Vio is a case of autoimmune disease wherein the skin fails to produce pigmentation all right so again this is not a case of codominance our next part is multiple alyss of polygenic inheritance wherein a specific trat is controlled by more than one gene or by more than one uh yeah Gene instead there are several genes that may contribute to the final phenotype of a given trat so the um the that can be passed down from the parent to offspring there are many different variations because of that also polygenic inheritance could be influenced by environmental factors a good example of that is the hydrogena flower wherein um normally in a normal pH it's around purple but under acidic soil it becomes Bluer or bluish but under a basic soil it becomes more pink to Red in color okay now we're done with unit two and down to our last unit on autosomes and sex chromosomes so autosomes are the first 22 homologous pair of human chromosomes that do not influence the sex of an individual so I already showed you a complete cotype of this one so the autosomes here technically is from number one all the way to number two so they autosomes are us are numbered and sex chromosomes are not since sex chromosomes should be represented by either XX or XY so an autosome is one of the numbered chromosomes as I stated before and in humans they have 22 pairs of this now autosomes play a role in uh in um um passing down traits from one parent to another from one parent to its Offspring because as you can tell these autosomes are are uh may actually contain dominant or recessive traits so if dominant alil is present on the autosomes then the individual will Express the trait this is already discussed in the previous slide however sometimes the dominant alil in your autosomal in on in your autosomes may carry some um variations that may lead to a disorder or to a syndrome perhaps like this one we have here Progeria is an example of autosomal dominant inheritance wherein the parents was able to pass this down from from them to their children no so uh a patient with progeria have a shorter lifespan in fact um the oldest uh Progeria uh the oldest patient with progeria uh was 20 years old Huntington Huntington this is is already is is also an example of autosomal dominant inheritance now if you have a dominant treats we also have an autosomal recessive treats in which um the parents may actually trans uh may may actually um what you call this pass the recessive alil into their offspring as we can see in this one so this one is a very good example of you of the autosomal recessive inheritance pattern for example both of the parents are carriers but they have no conditions they have no conditions because uh both of them uh uh carries the the trait or carries the recessive trait so that being said it will not be reflected or it will not be expressed uh to these to the parents so for example they will have their own children someday uh they will be producing um Offspring and there is a slight chances that the child May inherit both the recessive traits and at this point uh it will be expressed in the organism so like all this one albis albinism is an example of an autosomal recessive inheritance albino CLE cell anemia is already is also an example of autosomal in recessive inheritance but back then in our first lecture this can also happen because of genetic mutations not not only with uh inheritance remember that some of the amino acids during our first lecture was um modified and it also led to Cle cell anemia um CLE cell anemia may also be a result of an autosomal recessive inheritance now down to the sex chromosomes so the sex chromosomes as discussed earlier is actually our last pair in our human cotype and they are not numbered they specifically have the X or Y chromosomes X or Y for M for male and xx for female now in every animals um it's actually different no uh in humans it's xx and XY but in other animals um it varies so the structure of the X chromosome of most organisms are straight rodlike and comparatively longer than than our Y chromosome as you can see in this one this one is the Y chromosome and this one is the X chromosome so if you're going to compare them the X chromosome is much larger or longer than our Y chromosomes [Applause] no so so the X chromosomes have a very large UK chromatine and a small amount of heterochromatin and UK chromatine have a large amount of DNA and with that being said much genetic information are in the X chromosomes in comparison to your Y chromosomes but the Y chromosome is particularly very very important because this is where um sex determination or or this is where the uh sex of a uh h a human will be determined no so in the Y chromosomes they actually contain the par1 and the pa2 or the pseudo autosomal regions as you can see this takes up about 5% uh space or region of the said chromosome in in our Y chromosome they contain genes shared with the X chromosomes the white chromosome also contains the male specific regions such as uh actually 95% of the chromosome contains uh the M specific regions and 5% of which contains the par one and the pa2 uh and this one it was stated as the male specific regions because majority of the genes here includes Sr Y and the a a f genes or the AO azospermia genes so sry and the ASF genes are responsible for the development of testicles and um necessary Ducks uh for a male um human this one is a very good uh chart of how uh sex is determined in the case of humans so for example uh we begin with an egg and a sperm cell uniting together forming a zygote as you can see in that one uh if if if the zygote inherited the Y chromosome from its parent you can actually track it through the planet Square uh there will be a presence of the SR Gene and because there's a presence of s Sr y Gene it will uh technically um influence the development ment of the testes and if the testes is developed meaning to say there would be a presence of the testosterone wolfian ducks and as such male reproductive structures will continue to develop right as we can see in this one on the other hand the presence of the sry gene also um influences the presence of mi mi Mis genes in which its function is technically for the regression of the Marian ducts that uh soon enough that will also halt the development of female reproductive structures okay if we're going to compare this one for example this is for female uh the zygote contains the XX chromosomes and since it it does not contain the Y chromosome it will not produce the it it does not have the SR y Gene so there's an absence of Sr y Gene so no testes will be uh produced or will be developed rather in the process but ovaries will develop so if ovaries will be will develop in our growing uh embryo for example or growing human uh there would also be absence of testosterone there would be absence as well of wolfian ducks and as such the male reproductive structures would regress on the other hand since there's also absence of the Mis Gene the Maran Ducks will develop but at the same time uh because of this because of the absence of both of these hormones uh especially this one the female reproductive structures will continue to develop now let's move on to the sex link trates as your X and Y chromosomes are also carriers of different traits but ah I'm sorry the your X chromosomes are technically carriers but the Y chromosome does not carry any traits but the Y chromosomes is very important for sex determination as discussed in the previous slides so many sex L many Sex Link trats are carried on the X chromosomes a good example of this one is the red eye is the um if we're going to cross a Redeye male and a white eyed uh female and apply it in a pet Square for example the resulting structure would be 50% red eye female and 50% are white eyed female so uh technically since the white chromosome does not carry any traits at all what um whatever the uh the alyss that are carried on the X chromosomes will be expressed in the organism so even though our X chromosome here carries the recessive gene since the Y chromosome does not carry any any traits at all the organism will have um red eye or white eye rather I'm sorry white eye hemophilia is actually another example of our sex link trate problems wherein majority of the carriers are female and major and this is usually expressed among male patients no as we can see in this planet Square Mal I'm so sorry um in this patient although the X chromosome only carries recessive hemophilia are related treats since the Y chromosome does not carry any uh any treats at all um this patient would Express the hemophilia no in comparison to this one for example we have two actually this one would also Express the um the hemophilia or the or the condition rather but let's say that there is another X chromosome that carries a dominant trait and the dominant rate carries a normal Gene and uh most likely uh among among female they are less uh there is a a lesser chances for them to develop or to have this condition of Hemophilia because they have twox chromosomes in comparison to male color blindness is another example of our sex link recessive traits now lastly all let's discuss about pedigree and genetic disorders in order for some geneticist to track um if genetic diseases are inherited from one family member to another we they actually employ the pedigree charts so this is a diagram that shows the occurrence and appearance of a phenotypes and this diagram usually uses shapes and colors to represent if a given individual inherited a given genetic disease so um we do have a rule in our pedigree chart as you can see here square if it's male Circle if it's female a horizontal line would equate that both of them are meeting or both of them are parents no and then as we can see here uh a line that is connecting our parents a vertical line connecting the parents all the way to another shape uh represents Offspring the F1 generation or uh children no so as we can see here we also have um we also have symbols for our twins either they zygotic or monozygotic twins we also have here um a diamond sheap if the sex is unspecified and um most the most important um symbols or Legends here are technically the male female and the lines as well as if if the uh shapes are already colored because in pedigree charts when the shapes are colored it means that they they inherited the genetic disease so for example let's make a pedigree for the following couple Dana is color blind her husband Jeff is not they have two boys and two girls let's try to draw a pedigree chart on this one so again going back if it's male it's square if it's female it's Circle so Dan is represented by Circle and Jeff is represented by square or box no and then since um it's explicitly stated that Jeff is Dana's husband I'm just going to draw a horizontal line in between and they also have four op Springs two boys so I'm going to illustrate two squares and two girls so dalawang circle now the next question is um s Dana she's color blind so I'm going to highlight Dana with black color because according to our pedigree charts May coloration you affected individuals now the real question now is um how many or how many of their offspring or children will inherit color blindness as well so actually you can actually I'm sorry you can actually predict this using a planet Square for a while oh come on so let's say uh Dana is color blind XX color blind so she's a carrier so uh so to speak so for example color blind here and then Jeff is not he's normal let's try to have a planet square of that one so I have here X chromosome containing the color blindness Sex Link traes though by the way and then I have here y uh X Y okay so let's try to derive on a given answer I have here XC X and then uh XC y x x XY meaning to say uh 50% anak no 50% of their children will inherit or might inherit especially if this one is dominant alila might inherit the color blindness and uh 50% Naman on the other hand is normal no so going back to our pedigree charts in order for us to uh represent this um our result uh one boy will have color blindness and one girl so K one boy and one girl so that's actually how you make pedigree charts right so actually that's the last of my discussion you have assignment two and three so for these assignments for the submission dates of these assignments please do refer to our calendar I know that there's too much on this lecture so if you have any questions please do not hesitate to ask I would be opening my calendly for your consultation in the past few weeks it's not yet open so I'm not receiving any request from any students but um I will be opening them for if you have any questions or do approach me right after classes thank you everyone