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

• F1 (filial) generation– All have trait of one

parent

– Reciprocal cross – sex of parent with trait made no difference

Monohybrid cross

• F2 generation– Phenotype ratio 3:1

• 3 = trait in F1• 1 = trait not seen in

F1; 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)– 1st affected = ¼ x ¾ x ¾ = 9/64– 2nd affected = ¾ x ¼ x ¾ = 9/64– 3rd 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

• a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5

Binomial expansion

• a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5

• Term number = n +1• First term has an, second term an-1b, etc

– a always loses 1; b gains 1• For coefficient (number in the front)

– 1st term is always 1– 2nd term – same power as binomial (5)– 3rd term – multiply preceding coefficient (5 from 2nd

term) by exponent of a in the 2nd term (4), then divide by term # (2) = (5x4)/2 = 10

• Coefficient for 3rd 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

• F1 exhibits both dominant forms of the traits

• Heterozygous for both – can form 4 different types of gametes

Dihybrid cross

• F2 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• 1st column lists

proportions of phenotypes of 1st trait

• 2nd column lists proportions of phenotypes of 2nd 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

• X2 = 2.0

• Need Table• Degrees of freedom =

n – 1, where n = possible phenotypes

Chi-Square example cont

• Df = 1• X2 = 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