Bio 101 Ms. Bledsoe Making Sense of Geneics roblems Monohbrid crosses Le s sar wih somehing simle: crossing wo organisms and waching how one single rai comes ou in he offsring. Le s use eas, as Mendel did. Suose ou are working wih he heigh gene in eas. he heigh gene has wo alleles: all and shor. We ll use o reresen he all allele, and for he shor allele. Each individual ea lan is diloid, so i mus have wo coies of he gene (one from each aren) in ever bod cell. However, gamees are haloid. When a ea lan reroduces, i can onl donae one of is heigh genes o each offsring eiher he one i go from he female aren, or he one i go from he male aren. Wha heigh alleles can an one ea lan u in is gamees? Deends uon he genoe of he lan. Here is how he alleles will be disribued o he gamees in he following individuals: condiion of aren genoe gamees Homozgous dominan ------> onl Heerozgous ------> and Homozgous recessive ------> onl A unne square is a wa of finding ou wha he raios of henoes and genoes of he offsring will be. When we make a unne square, we u he gamees of he arens ouside he square and combine hem in all ossible was o ge he offsring. In monohbrid crosses, he gamees will all be reresened b a single leer, since we re following onl one rai and each gamee is haloid. he resuling offsring should each be reresened b a air of leers, since he will all be diloid individuals. Here are some examles, crossing various arens: x x x x (noe: his could be a single-square unne square) (noe: hese wo could be wo-square unne squares) However, hese squares do no describe acual oucomes. he redic he chances of he offsring having a aricular rai. For insance, he second square shows us ha each offsring has a 3 in 4 chance (75% or 0.75) of being all.
Dihbrid Crosses Here s where man eole ge ino rouble when he do geneics roblems. he do fine uing single leers on o of boxes, even if he don undersand wha he leers mean. Bu give hem wo rais a once, and he sar uing single leers on o of he unne square boxes and don undersand how o inerre he resuls. Le s have a look a how o handle crosses where we follow wo rais a once. Again, we ll work wih ea lans. Suose we wan o follow he rais of heigh a ea color. he heigh gene has wo alleles: all and shor. We ll use o reresen he all allele, and for he shor allele. he color gene also has wo alleles: ellow and green, where ellow is dominan and green is recessive. We ll use o reresen he ellow allele, and for he green allele. Remember, ea lan bod cells are diloid, so each cell will have a air of he heigh alleles and a air of he color alleles. Gamees are haloid, so each gamee will have one of he heigh alleles and one of he color alleles. Wha heigh and color alleles can an one ea lan u in is gamees? Deends uon he genoe of he lan. Here is how he alleles will be disribued o he gamees in he following individuals: genoe gamees onl and and and and and Noe ha each individual us one allele of each gene in each gamee. Since we re following wo rais a once, each gamee mus have one allele of each of he wo genes ha s wh here are wo leers in each gamee. Wha gamees would he following individuals be able o make? genoe gamees Now how do we cross wo individuals and rack he rais as he sor ou in he offsring? Remember, he leers ha go across he o and he sides of he unne square reresen he ossible gamees each aren roduces. Use he able above o hel ou figure ou wha ossible gamees each aren in a cross can donae.
For examle, wha if we cross wo individuals who are boh heerozgous for each rai? We wrie he genoes of hese ea lans as. Using he able above, ou can see ha hese ea lans can roduce four kinds of gamees:,,, and. We re going o have o use a unne square ha is four squares b four squares: he henoic raio of his cross (beween wo individuals who are heerozgous for boh rais is he classic 9:3:3:1 raio. Bu nine wha? hree wha? One wha? If ou coun u he individuals wih each of he following henoe, we ll see: henoe frequenc all, ellow 9/16 all, green 3/16 shor, ellow 3/16 shor, green 1/16 So, on our exams and quizzes, be sure o sae wha he numbers of our raios refer o. Bu is his he onl raio we can ge wih a dihbrid cross? Le s find ou. Suose we sar wih an individual who is all and ellow, and one ha is all and green. Suose also we know he genoe of each individual:,. Le s do his cross as follows: 1) Figure ou wha alleles each aren can u in is gamees. aren ossible gamees and and and and 2) Se u he square for his cross: henoe frequenc all, ellow 5/8 all, green 1/8 shor, ellow 1/8 shor, green 1/8
rihbrid crosses and beond Oka, unne squares for monohbrid crosses are fairl simle. Dihbrid crosses are harder, bu undersandable. Bu wha if we wan o wach hree, four, or more rais all a once? Well, we could sar wih he rais of he arens, find heir genoes, and hen figure ou heir gamees, and make a unne square. Le s see how we could do ha wih hree rais in eas. Suose ou have a lan ha is all and has green eas. Suose i also has urle flowers, which are dominan over whie. And suose ou know he ea lan is heerozgous for all hree rais (). Wha alleles will his individual u in is gamees? ou can se u a forked diagram o figure his ou, like so: gamees Le s see ha s eigh differen ossible gamees. If ou cross wo ea lans ha are heerozgous for all hree rais, ou ll have o make a unne square ha is eigh b eigh, or sixfour squares oal. Do ou reall wan o do ha? I didn hink so. Would ou like an easier wa? (his will work for dihbrid crosses as well.) Le s consider each of he rais searael, hen use he rules of robabili o figure ou he robabili of several rais occurring a once. he basic rules of robabili are as follows: robabili of wo evens haening a once (even 1 AND even 2) = (robabili of even 1) (robabili of even 2) robabili of eiher even 1 OR even 2 occurring = robabili of even 1 + robabili of even 2 So, le s consider he above ea lan,. Le s r crossing his lan wih anoher lan of he same genoe, ha is, anoher. Firs, use unne squares o figure ou he robabiliies for each individual rai in he offsring:
heigh ea color flower color all: 3/4 or 0.75 ellow: 3/4 or 0.75 urle: 3/4 or 0.75 shor: 1/4 or 0.25 green: 1/4 or 0.25 whie: 1/4 or 0.25 Suose we cross hose wo eas ha are heerozgous for all hree rais and wan o know wha he robabili is of geing offsring ha are all, wih ellow eas and urle flowers. Here s how i works: robabili of an offsring being all = 0.75 robabili of an offsring having ellow eas = 0.75 robabili of an offsring having urle flowers = 0.75 robabili of an offsring being all AND having ellow eas AND having urle flowers = (0.75)(0.75)(0.75) = 0.42 (ha is, 42%) We can also find ou he robabili of an oher combinaion of rais in he offsring of hese wo arens: robabili of an offsring being shor AND having green eas AND having whie flowers = (0.25)(0.25)(0.25) = 0.015 (or 1.5%) robabili of an offsring being all AND having green eas AND having whie flowers = (0.75)(0.25)(0.25) = 0.047 (or 4.7%) Wha if he arens aren heerozgous for all hree rais? Bu suose hose eas we alread discussed aren he wo arens we sar wih? Le s see wha haens when we cross he following wo eas: x Now if ou work ou he unne squares for each searae rai, he robabiliies are: all = 0.5 ellow = 0.75 urle = 0.5 shor = 0.5 green = 0.25 whie = 0.5 Now consider he offsring: robabili of an offsring being all AND having ellow eas AND having urle flowers = (0.5)(0.75)(0.5) = 0.19 (or 19%) robabili of an offsring being shor AND having green eas AND having whie flowers = (0.5)(.25)(0.5) = 0.06 (or 6%) robabili of an offsring being all AND having green eas AND having whie flowers = (0.5)(0.25)(0.5) = 0.06 (or 6%)
Le s hrow in anoher comlicaion. Wha if he rais include one ha shows incomlee dominance, codominance, or sex linkage? We can sill solve he roblem using he robabili mehod. Suose a married coule knows ha he are boh carriers for Csic Fibrosis. In addiion, he woman has a faher who was color blind (a sex-linked rai), which means she has o carr one recessive color blind allele on he X chromosome she go from her faher. Wha are he odds ha a son of heirs will have Csic Fibrosis and be color blind? Csic fibrosis gene Color blind gene normal child = 0.75 normal girl =.5 CF child = 0.25 normal bo =.25 color blind bo =.25 o find ou he robabili of geing a color blind son wih CF, we muli he robabili of an one child having CF b he robabili of geing a color blind bo: (0.25)(0.25) = 0.062 (or 6.2%) chance of geing a color blind CF bo. Consider incomlee dominance, also. Le s r crossing wo of he common garden flowers, he Four O clocks. hese flowers come in all or dwarf (shor) forms, and heir flowers are red, ink, or whie. Use for all and for shor as in eas. Use R for red and R for whie, where RR resuls in red flowers, RR will roduce ink flowers, and R R will resul in whie flowers. Cross a lan ha is heerozgous for heigh () and is red (RR) wih one ha is heerozgous for heigh () and whie (R R ). he robabiliies for hese rais showing u in he offsring are: Heigh gene Flower color gene all = 0.75 Red = 0.25 shor = 0.25 ink = 0.5 Whie = 0.25 Wha is he robabili of geing lans ha are: all and red flowered? (0.75)(0.25) = 0.187 all and ink flowered? (0.75)(0.5) = 0.375 shor and whie flowered? (0.25)(0.25) = 0.0625 And so on wih an oher combinaion of rais.