Lesson 15

Fundamentals of Biology
Lesson 15

Mendelian Genetics

I. Gregor Mendel.

    A. “Father of Genetics.”
        1. Austrian monk, 1822-1884.
        2. Studied the genetics of pea plants.
        3. After 8 years of research published his results in 1865.

    B. Experiments.
        1. Began with true-breeding plants; e.g. a tall plant always produced tall offspring.
        2. Tore off the anthers so they would not self-pollinate.
        3. Crossed a tall plant with a short one.
        4. Common sense would lead us to expect a medium-size plant.
        5. But all the offspring were tall.
        6. Then he allowed the offspring to self-pollinate.
        7. In the second generation, 1/4 were short and 3/4 were tall.

    C. Conclusions.
        1. Characteristics must be determined by factors which occur in pairs.
        2. When two different factors are present, one will express itself and the other will be masked. We call these the dominant and the recessive traits.
        3. When gametes are formed, these paired factors separate and each gamete receives only one factor.

P1 Generation T T x t t

Gametes formed T T | t t

F1 Generation T t x T t

Gametes formed T t | T t

F2 Generation T T T t T t t t

II. Genetic Terms.

    A. Phenotype: The expression of the genes. What the organism looks like.
    B. Genotype: Which genes the organism has.
    C. Allele: One of a pair of genes that code for the same sort of trait (e.g. height). They have equivalent positions on homologous chromosomes.
    D. Homozygous: Both alleles are the same; e.g. both height genes are for shortness.
    E. Heterozygous: The alleles are different; a hybrid; e.g. one height gene for tallness and one for shortness.
    F. Monohybrid Cross: Only crossing one trait.
    G. Dihybrid Cross: Crossing two traits at once.

III. Monohybrid Crosses.

A. Homozygous dominant with homozygous recessive.

T T

t T t T t

t T t T t

1. Phenotype: 4 tall : 0 short.
2. Genotype: 0 homozygous tall : 4 heterozygous : 0 homozygous short.

B. Heterozygous with heterozygous.

T t

T T T T t

t T t t t

1. Phenotype: 3 tall : 1 short.
2. Genotype: 1 homozygous tall : 2 heterozygous : 1 homozygous short.

C. Homozygous dominant with heterozygous.

T T

T T T T T

t T t T t

1. Phenotype: 4 tall : 0 short.
2. Genotype: 2 homozygous tall : 2 heterozygous : 0 homozygous short.

D. Homozygous recessive with heterozygous.

T t

t T t t t

t T t t t

1. Phenotype: 2 tall : 2 short.
2. Genotype: 0 homozygous tall : 2 heterozygous : 2 homozygous short.

    E. A test cross can reveal whether a tall pea plant is heterozygous or homozygous.
        1. Need to cross in such a way that the phenotype clearly shows this.
        2. If we cross with a known homozygous dominant, all the plants will be tall in either case. This tells us nothing.
        3. If we cross with a known homozygous recessive, all of the offspring will be tall if the parent was homozygous,
            but half of the offspring will be short if the parent was heterozygous.

IV. Incomplete Dominance.

A. Cross a red snapdragon with a white one.

r r

w r w r w

w r w r w

        1. Phenotype: 0 red : 4 pink : 0 white.
        2. Genotype: 0 homozygous red : 4 heterozygous : 0 homozygous white.
        3. Neither red nor white is dominant.

B. Cross two pink snapdragons.

r w

r r r r w

w r w w w

1. Phenotype: 1 red : 2 pink : 1 white.
2. Genotype: 1 homozygous red : 2 heterozygous : 1 homozygous white.

C. The phenotype and the genotype are identical where there is incomplete dominance.

V. Dihybrid Crosses.

    A. Considering two traits at once.
        1. Mendel performed dihybrid crosses.
        2. He found that the inheritance of one trait is not affected by the inheritance of the other.
        3. This is his Law of Independent Assortment.
        4. The alleles for each trait segregate without interference from the alleles for other traits during gamete formation.

B. Crossing a homozygous green inflated-pod pea with a homozygous yellow constricted-pod pea.

P1 Generation G G I I x g g i i

Gametes formed G I G I | g i g i

F1 Generation G g I i x G g I i

Gametes formed G I G i g I g i | G I G i gI gi

F2 Generation GGII GGIi GGii GgII GgIi Ggii ggII ggIi ggii
1 2 1 2 4 2 1 2 1

G I G i g I g i

G I G G I I G G I i G g I I G g I i

G i G G I i G G i i GgIi G g i i

g I G g I I G g I i g g I I g g I i

g i G g I i G g i i g g I i g g i i

      1. Phenotype: 9 green inflated : 3 green constricted : 3 yellow inflated : 1 yellow constricted
      2. Genotype: 1 hom. green, hom. inflated : 2 hom. green, het. inflated : 1 hom. green, hom. con.:
                           2 het. green, hom. inflated : 4 het. green, het. inflated : 2 het. green, hom. con.:
                           1 hom. yellow, hom. inflated : 2 hom. yellow, het. inflated : 1 hom. yellow, hom. con.

C. The Law of Independent Assortment only holds true when the genes are on different chromosomes.

VI. Multiple Alleles.

    A. Most traits are determined by one of two possible alleles; e.g. tall peas or short.
        1. There are cases where there are more than two possible alleles.
        2. Of course, only two are present at any time in an organism, one on each homolog.

    B. Human blood type.
        1. There are three alleles: A, B, i.
        2. “A” produces the A glycoprotein.
        3. “B” produces the B glycoprotein.
        4. “i” produces no glycoprotein.
        5. A and B are both dominant.
        6. Thus blood type A could be homozygous (IA IA) or heterogygous (IA i).
        7. So could type B: homozygous (IB IB) or heterogygous (IB i).
        8. If some one has both A and B alleles (IA IB), he is heterozygous for both, and both A and B glycoproteins are produced.
        9. If he is homozygous recessive (i i), he produces no glycoproteins and has blood type O.
        10. Heterozygous type A crossed with heterozygous type B.

I^A i

I^B I^AI^B I^Bi

i I^Ai i i

1 type AB : 1 type A : 1 type B : 1 type O

VII. Sex-linked Traits.

    A. Sex chromosomes.
        1. Human beings have 22 pairs of autosomes and 1 pair of sex chromosomes.
        2. Females have 2 X chromosomes; males have an X and a Y.
        3. Thus the father determines the sex of his children.

X X

X XX XX

Y XY XY

4. Half of the children are male and half are female.

    B. Sex-linkage.
        1. The X chromosome is much larger than the Y chromosome, and can accommodate more genes.
        2. A trait that is determined by a gene on the X chromosome will be expressed differently in men than in women, since men only receive one copy of it.

    C. Colorblindness.
        1. A gene on the X chromosome codes for a chemical that enables proper color apprehension.
        2. An alternate form of the gene produces less of the chemical, resulting in colorblindness.
        3. This alternate form of the gene is recessive.
        4. Consider a cross of a normal woman who possesses one copy of the colorblind gene with a normal man.

X^G X^g

X^G X^GX^G X^GX^g

Y X^GY X^gY

        5. All the girls will have normal vision, but half will be “carriers” for colorblindness - they can pass on the condition even though they do not have it themselves.
        6. Half of the boys will have normal vision, but half will be colorblind.
        7. Note that boys cannot be carriers because they cannot be heterozygous - they only get one copy of the gene since they only have one X chromosome.
        8. Now consider a cross of a colorblind man and a normal woman who is not a carrier.

X^G X^G

X^g X^GX^g X^GX^g

Y X^GY X^GY

9. All the girls will be carriers, but all the boys will be normal.

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