Bio Exam Final Nov 14 Chromosomal Basis For Mendel s Independent Assortment If the individual is heterozygous for both genes gene A and gene B according to Mendel s principle of independent assortment when the gametes are formed you can get 4 types f gametes AB Ab aB ab The idea is that the homologous chromosome pairs line up on the metaphase plate independently in metaphase I and this arrangement is just as likely as the other Genes on the same chromosome are called Linked Genes o Linked genes don t show independent assortment dependent assortment Dependent vs Independent Assortment Parents AABB x aabb F1 AaBb x aabb o o o Gametes of F1 Independent Assortment AaBb parental Aabb non parental aaBb non parental aabb parental o Gametes of F1 Dependent Assortment AaBb parental aabb parental Two of the four gametes have the parental phenotype 1 1 1 1 ratio Both of the gametes have parental phenotypes 1 1 ratio Non Parentals recombination new combination of traits that was not in the parents Parental same phenotype as the parent Sweet Peas P purple flower p red flower L long pollen l round pollen Find out if the genes are linked o Homozygous for both parents PPLL x ppll o Do a test cross with a recessive for both genes F1 PpLl x ppll o If it was independently assorted we would expect Purple and Long as one parental 900 Red and long as a non parental 10 Purple and round as a non parental 15 Red and round as another parental 912 o Not a 1 1 1 1 ratio but its not linked either because there are non parentals Crossing over generated gametes that were part paternal and part maternal o Gamete LP and lp When combine them with ones from test cross get parentals only In prophase I of meiosis if the maternal and paternal lined up maternal lp and paternal LP Produce chromosomes part paternal and part maternal Crossing over accounted for the non parental summary if the genes show IA 4 phenotypes 1 1 1 1 ratio If the genes are linked 2 phenotypes 1 1 ratio all parental If neither its call partial linkage Generate some recombinant gametes by crossing over which creates chromosomes with both paternal and maternal parts and that can result in a recombinant gamete which will result in a recombinant individual when you do a test cross Morgan scientist the studied fruit flies Had an undergraduate student working for him named Sturtevant Came up with a way to map genes on a chromosome by determining the frequency of recombinants If you have two chromosomes one pat and one mat they have genes A B and D on them when crossing over occurs the place where it occurs is random The cross over can be anywhere on the chromosomes o He realized that two genes will be more likely switched exchanged during a cross over if they are far apart versus close together The cross over can only be in one place on a chromosome Said you can map you can map genes on a chromosome by determining the frequency of the recombinants To map genes on a chromosome start with a cross between a double heterozygote o o o o o The results of the test cross are AaBb 20 45 Aabb 18 5 aaBb 22 4 aabb 25 46 Black numbers are close to a 1 1 1 1 ratio Independent Assortment Green numbers are not independent assortment so the frequency of recombinance is the total number of non parentals recombinants the total number of individuals 9 100 9 9 map unit apart 1 map unit 1 recombinants o Summary to map genes on a chromosome first you if start with pure strain parents create a double heterozygote then you test cross the double heterozygote If you get something close to a 1 1 1 1 ratio IA not linked genes If you get an excess of parentals and not many recombinants then that means the genes are linked The map units are the total recombinants divided by the total number of individuals Aneuploidy A chromosomal aberration When there is one too many or one too few chromosomes Trisomy 2N 1 one too many chromosomes 47 chromosomes o o Monosomy 2N 1 one too few chromosomes 45 Arises from chromosomal non disjunction in anaphase I the chromosomes are supposed to separate disjoin if they don t separate its called non disjunction
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