Chapter 3 – Basic Principles of Heredity

Johann Gregor Mendel (1822 – 1884) Pisum sativum Rapid growth; lots of offspring Self fertilize with a single plant; cross fertilize between two plants

Pisum sativum 7 characteristics –Each had only 2 forms –True-breeding varieties When allowed to self-fertilize, all offspring had same parental trait

Modern Genetic Terminology Gene – inherited factor that codes for a specific characteristic Locus – physical location of a gene on a chromosome Allele – alternate forms of a gene –what specifically the gene codes for (black hair, blond hair)

Modern Genetic Terminology Genotype – set al individual’s alleles; its genetic makeup –Homozygous – 2 of the same allele for a gene –Heterozygous – 2 different alleles for a gene Phenotype – outward expression of a gene –An allele may be present but not expressed in the phenotype

Monohybrid cross Cross between plants that differ in a single characteristic P (paternal) generation –True-breeding for trait

Monohybrid cross F 1 (filial) generation –All have trait of one parent –Reciprocal cross – sex of parent with trait made no difference

Monohybrid cross F 2 generation –Phenotype ratio 3:1 3 = trait in F1 1 = trait not seen in F 1 ; seen in P generation –“lost” phenotype reappeared

Monohybrid cross conclusions Each plant has two “factors” (modern terms - genes) In heterozygotes, one allele will be expressed; other will be masked, but can be passed on and expressed in offspring –Dominant allele – expressed Capital letter –Recessive allele – masked Lowercase letter 2 alleles separate with equal probability

Principle of Segregation Each diploid organism has 2 alleles for each gene Alleles segregate from each other randomly in gamete formation

Punnett square Illustrates possible gametes formed and possible fertilization combinations

Probability Likelihood of the occurrence of a particular event expressed as a fraction or a decimal Multiplication rule “AND” –Probability of two or more independent events occurring together

Multiplication rule Probability of rolling a four – 1/6 Probability of rolling a four AND then a 3 –1/6 x 1/6 = 1/36

Multiplication rule Cross between two heterozygous purple flowered plants (Pp x Pp) ? Probability of having a purple offspring, AND then a white ? Probability of having two white offspring

Addition rule “either/or” Probability of having 2 or more mutually exclusive events occur together Probability of rolling a three OR a four –1/6 + 1/6 = 2/6 (1/3)

Albinism – autosomal recessive disorder 2 carriers mate (Aa x Aa) ? Probability of having three children with albinism –¼ x ¼ x ¼ = 1/64 ? Probability of having 2 “normal” and 1 albino (order not important) –1 st affected = ¼ x ¾ x ¾ = 9/64 –2 nd affected = ¾ x ¼ x ¾ = 9/64 –3 rd affected = ¾ x ¾ x ¼ = 9/64 –Add all possible combinations = 27/64

Binomial expansion a = probability of albinism (1/4) b = probability of “normal” pigmentation (3/4) 5 children (a + b) 5 a 5 + 5a 4 b + 10a 3 b a 2 b 3 + 5ab 4 + b 5

Binomial expansion a 5 + 5a 4 b + 10a 3 b a 2 b 3 + 5ab 4 + b 5 Term number = n +1 First term has a n, second term a n-1 b, etc –a always loses 1; b gains 1 For coefficient (number in the front) –1 st term is always 1 –2 nd term – same power as binomial (5) –3 rd term – multiply preceding coefficient (5 from 2 nd term) by exponent of a in the 2 nd term (4), then divide by term # (2) = (5x4)/2 = 10 Coefficient for 3 rd term is 10

Binomial expansion ? probability of 2 carriers of albinism having 5 albino children and 1 “normally” pigmented child (a + b) 6

Test Cross Determination of genotype of a dominant phenotype individual Cross with homozygous recessive individual –If any offspring demonstrate recessive phenotype, unknown must be heterozygous

Types of genetic crosses Reciprocal –Sex of parents with a specific trait is switched Test –Cross of unknown dominant with recessive Back –Cross of individual with a parent

Dihybrid cross 2 different traits are examined at the same time P generation – true breeding for both traits

Dihybrid cross F 1 exhibits both dominant forms of the traits Heterozygous for both – can form 4 different types of gametes

Dihybrid cross F 2 generation 9:3:3:1 ratio 9 – both dominant traits 3 – dominant for color; recessive for shape 3 – recessive for color; dominant for shape 1 – both recessive traits Principle of Independent Assortment –Alleles at different loci segregate independent from one another

Branch diagram Uses probability rules 1 st column lists proportions of phenotypes of 1 st trait 2 nd column lists proportions of phenotypes of 2 nd trait, etc Faster than a Punnett square when dealing with multiple loci –Specifically when you need one particular phenotype

Ratios Punnett squares and Branch diagrams deal with probability Observed ratio is rarely EXACTLY the expected ratio Goodness of fit Chi-Square test –Indicates probability that deviation between observed and expected ratio is due to chance alone

Chi-Square example

Chi-Square example cont Number is squared, so it’s always a positive number X 2 = 2.0 Need Table Degrees of freedom = n – 1, where n = possible phenotypes

Chi-Square example cont Df = 1 X 2 = 2.1< p <.5 10% < p < 50% that variability is due to chance – hypothesis is accepted Cut off is usually p = 0.05 (5% variation due to chance)

Cat example – Chi-square Assuming black is dominant to gray, a cross between Bb x Bb yields an expected ratio is 3:1 Offspring = 30 black cats and 20 gray Accept or reject hypothesis?

Cat example calculations