The Chromosomal basis of inheritance provides an understanding of the pattern of passage (transmission) of genes form parent to offspring Heredity Part 1

Ideas about inheritance Blending Inheritance Particulate Inheritance The “blending” hypothesis is the idea that genetic material from the two parents blends together (like blue and yellow paint blend to make green) The “particulate” hypothesis is the idea that parents pass on discrete heritable units (genes) Mendel documented a particulate mechanism through his experiments with garden peas

Gregor Mendel Mendel discovered the basic principles of heredity by breeding garden peas in carefully planned experiments z7SUbnc z7SUbnc

Why study pea plants? ◦There are many varieties with distinct heritable features, or characters (such as flower color); character variants (such as purple or white flowers) are called traits ◦Mating of plants can be controlled ◦Each pea plant has sperm-producing organs (stamens) and egg-producing organs (carpels) ◦Cross-pollination (fertilization between different plants) can be achieved by dusting one plant with pollen from another

TECHNIQUE RESULTS Parental generation (P) Stamens Carpel First filial gener- ation offspring (F 1 ) 5 Mendel chose to track only those characters that varied in an either-or manner He also used varieties that were true-breeding (plants that produce offspring of the same variety when they self- pollinate)

Table 14-1

Fig EXPERIMENT P Generation (true-breeding parents) Purple flowers White flowers  F 1 Generation (hybrids) All plants had purple flowers F 2 Generation 705 purple-flowered plants 224 white-flowered plants

Fig Allele for purple flowers Homologous pair of chromosomes Locus for flower-color gene Allele for white flowers

Fig EXPERIMENT RESULTS P Generation F 1 Generation Predictions Gametes Hypothesis of dependent assortment YYRRyyrr YR yr YyRr  Hypothesis of independent assortment or Predicted offspring of F 2 generation Sperm YR yr Yr YR yR Yr yR yr YR YYRR YyRr YYRr YyRR YYrr Yyrr yyRR yyRr yyrr Phenotypic ratio 3:1 Eggs Phenotypic ratio 9:3:3:1 1/21/2 1/21/2 1/21/2 1/21/2 1/41/4 yr 1/41/4 1/41/4 1/41/4 1/41/4 1/41/4 1/41/4 1/41/4 1/41/4 3/43/4 9 / 16 3 / 16 1 / 16 Phenotypic ratio approximately 9:3:3:

Segregation and Independent Assortment of chromosomes result in genetic variation

Vocab Dominant Recessive Homozygous Heterozygous Genotype Phenotype Monohybrid Cross Test Cross Dihybrid Cross

Fig TECHNIQUE RESULTS Dominant phenotype, unknown genotype: PP or Pp ? Predictions Recessive phenotype, known genotype: pp  If PP If Pp or Sperm ppp p P P P p Eggs Pp pp or All offspring purple 1 / 2 offspring purple and 1 / 2 offspring white Test Cross: used to determine genotype of individuals with dominant phenotype

Fig Rr  Segregation of alleles into eggs Sperm R R R R R R r r r r r r 1/21/2 1/21/2 1/21/2 1/21/2 Segregation of alleles into sperm Eggs 1/41/4 1/41/4 1/41/4 1/41/4

Rules of Probability can be applied Rule of multiplication ◦The probability of a compound event (like flipping 2 coins and both coins landing heads up) is the product of the separate probabilities of the independent events ½ x ½ = ¼ Rule of addition ◦The probability that an event can occur in two or more alternative ways (like the chance of rolling a 1 or a 2 when rolling a die) is the sum of the events 1/6 + 1/6 = 2/6 = 1/3

We can apply the multiplication and addition rules to predict the outcome of crosses involving multiple characters A dihybrid or other multicharacter cross is equivalent to two or more independent monohybrid crosses occurring simultaneously In calculating the chances for various genotypes, each character is considered separately, and then the individual probabilities are multiplied together Copyright © 2008 Pearson Education Inc., publishing as Pearson Benjamin Cummings

Example Dihybrid Cross: TtGg X ttGG What is the probability of producing offspring that is heterozygous for both traits. You could do a dihybrid Punnett square. You could do 2 monohybrid Punnett squares then use the rule of multiplication.

Example 2 monohybrids… ◦Tt X ttprobability of Tt offspring = ½ ◦Gg X GG probability of Gg offspring = ½ ◦Probability of both TtGg = ½ X ½ = ¼ Or do dihybrid – 16 box Punnett Square