Objectives Unit Three:  Mendelian Genetics

1.  Define the following terms and utilize these terms in an appropriate context:
allele                      chromosome theory                      dihybrid cross                      dominant
F1 generation         F2 generation                               genotype                              heterozygote
homozygote           independent assortment                 monohybrid cross                 multiple allele
P generation           phenotype                                    recessive                              reciprocal cross
segregation             test cross                                     trihybrid cross                      Punnett Square

2.  State Mendel's four postulates of inheritance.

3.  Correlate Mendel's four postulates with what is now known about homologous chromosomes, genes, alleles and the process of meiosis.

4.  Distinguish between homozygosity and heterozygosity.

5.  Demonstrate the ability to determine possible gametes to be passed given a parents genotype in monohybrid and dihybrid circumstances.
6.  Given the designations of + for wild type and bw for brown eyes in Drosophila melanogaster, outline a monohybrid cross between normal flies and brown eyed flies to produce the F1, and an F1 cross to produce the F2.  Give the possible genotypes and phenotypes in each generation, and the expected ratios.

7.  Explain what is done in a test cross, and how the possible results are interpreted.

8.  Given select non-linked traits from the Drosophila genome, outline a dihybrid cross to produce an F1, and an F1 cross to produce the F2 generation.  Include the expected genotypes, phenotypes, and ratios.

9.  For any total of F2 individuals, calculate the number expected to be of each phenotype.

10.  Determine the relationship between the number of gene pairs for which an individual is heterozygous and the number of possible gamete types which can be produced by that individual with respect to those genes.

11.  Determine the relationship between the number of phenotypes which can be produced by a cross between individuals which are heterozygous for any number of genes.

12.  Discuss the considerations of possibility, probability and reality when dealing with genetic crosses.

Resources:  Text Chapter 3, Cartoon Guide pgs. 36-55
 

 Mendelian Genetics/ Classical Transmission Genetics

 Possibility vs. Probability vs. Reality

        -Inherited characteristics are the result of particulate factors called genes composed of DNA that are transmitted from
           generation to generation on vehicles called chromosomes.

        -1866 Gregor Mendel parented formal discipline of genetics
            22 different varieties of peas, objective, simple, mathematical

Mendels Postulates (Laws):
    1.  Genetic characters are controlled by unit factors that exist in pairs in   individual organisms.

    2.  when two unlike unit factors responsible for a single character are present in   a single individual, one unit factor is
           dominant to the other, which is said   to be recessive.

    3.  During the formation of gametes, the paired unit factors separate or   segregate randomly so that each gamete receives
            one or the other.

    4.  During gamete formation, segregating pairs of unit factors assort   independently of each other.

        **Important to consider that Mendel had no knowledge of chromosomes or meiosis

Modern modifications to Mendel:
    Allele- alternate forms of a gene (a gene is a location on a chromosome, an   allele is the info that fits in that place,  like a
        cassette tape)
        -Dominant allele
        -Recessive allele
        Gene locus
        Homozygous
        Heterozygous
        Genotype
        Phenotype
        Punnett Square

        Monohybrid Inheritance

        A.  Crosses involving only one trait
            -if blending were true, offspring should all be intermediates of parents

        B.  P- parental generation;  F1 first filial generation;  F2  second filial generation

            F = filial -pertaining to or approptiate to a son or daughter

        Mendel-  Experiment with Tall X Short produced all tall

        then tall plants self pollinated and produced 3/4 tall, 1/4 short, 0 intermediate

        Conclusions:  3:1 ratio possible if:
            1. F1 contained two seperate copies of each hereditary factor, one dom one rec
            2.  Factors seperated when gametes were formed
            3.  random joining of gametes upon fertilization

            Example  Normal vs. nerd (please correct me if I seem to be wandering from political      correctness!!)
                Alleles =  N - normal    n - nerd

            Genotypes vs. phenotypes

            Cross a homozygous normal individual with a nerd

            Monohybrid Reciprocal cross - reverse roles of male and female

            Monohybrid Test cross-  used to identify pure lines of descent, identify carriers of recessives
                -an individual with a dominant phenotype is crossed with one having a  recessive phenotype.  Allows one to
                determine whether an individual is  homozygous dominant or heterozygous (determines genotype).

DIHYBRID INHERITANCE- cross of true breeding lines that differ in two traits

    Assumptions:
        1.  if dominants segregated together and recessives segregated together, we would only see two phenotypes in F2

        2.  if dominants and recessives segregated independently then we would see 4 phenotypes in F2

        Mendel's law of Independent Assortment:
                Members of one pair of factors segregate (assort) independently of members of another pair of factors.  Therefore,
                all possible combinations of factors can occur in the gametes.  (***we now know these "factors" to be
                chromosomes and genes, however, chromosomes are the predominant mode of organization)

            For each genotype, give all possible gametes:

                Remember, mother nature says that sexually reproducing organisms must pass  on one set of information to the
                offspring, and there are two contributors to the  genetic component of the offspring

            Find all the ways to pass on one piece of each set of information:

                A.  ttGG  B. TtGG  C. TtGg  D. TTGg

                Dihybrid example:
                    T= tall     t= dwarf     R= rough      r= smooth

                Mate a homozygous tall/rough with an individual that is dwarf/smooth

                    Find F1 and then mate two F1 to find the F2

            Dihybrid Testcross:
                as in mono cross questionable dominant genotype with individual that is homozygous recessive for both traits

                From our example above, are our tall/rough individuals  TTRR, TtRR,  TTRr or    TtRr?

                Cross questionable with ttrr (can check for both traits or singles)

                    T_R_   X   ttrr

Trihybrid  inheritance and beyond

Probabilities for each seperately

AABBCC   X    aabbcc

ABC    AbC  ABc  aBC  Abc  aBc  abC  abc

This is an 8 X 8 Punnett Square with 64 possible outcomes.

Multiple gene inheritance produces tremendous variability

with 2 or more genes calculations become complex

        Simple rules for determining gametes (and from there probability)
            1.  determine number of heterozygous gene pairs and let this = n

            2.  2n is the number of different gametes that can fe formed by each parent

            3.  3n is the number of different genotypes that result following fertilization.

            4.  2n is the number of different phenotypes that are produced by these    genotypes

            # of het gene pairs     #of diff gametes      #of diff genotypes     #of diff phenos
                        n                            2n                            3n                            2n

                        1                            2                              3                              2

                        2                            4                              9                              4

                        3                            8                              27                            8

                        4                            16                            81                            16
 

            Ex.  AaBbCc n= 3
                   AABbCc n= 2
                   AAbbCc n= 1
                   AABBCC n= 0

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