Mendelian Genetics: Genetic Segregation Part 1 (Lecture 1)

Front 1

heterozygous vs homozygous

Back 1

1. heterozygous: inherited different alleles at one or more loci, different alleles between homologous chromosomes

-e.g. diploid Aa, tetraploid AAaa or Aaaa or AAAa

2. homozygous: inherited identical alleles at one or more loci, identical alleles in homologous chromosomes

-e.g. diploid AA, tetraploid AAAA or aaaa

Front 2

heterogenous vs homogenous

Back 2

1. heterogenous: population of individuals with different genotypes

-e.g. population of AA/Aa/aa

2. homogenous: population of individuals with the same genotype

-e.g. population of AA/AA/AA or population of Aa/Aa/Aa

Front 3

seed shape of pea example of phenotype expression

Back 3

-W = round, w = wrinked

-WW and Ww produce SBE1: converts unbranched starch molecules into branched starch molecules and causes peas to be round

-ww has abnormal alleles that don't function properly

-SBE1 not synthesized: reduced starch conversion (starch remains unbranched), peas wrinkle as the seed lose water when maturing, seed coat collapsing

Front 4

Gregor Mendel

Back 4

-1866 published work on inheritance of "elements or units" in pea plants

-examined clear-cut alternative traits: flower color, seed coat color, plant height

-figured out why traits can disappear and reappear in different generations

Front 5

why pea plants were a good choice for Mendel

Back 5

-self-fertilizing

-"true-breeding" for the traits he studied: AA x AA, no genetic segregation, identical to parent, after many generations eventually become homozygous

-many seeds per pod: large populations = big sample size = more accurate statistics

-traits were qualitative: controlled by a single gene, no environmental effects

-flower color, seed coat color, plant height

Front 6

Mendel's genetic principals

Back 6

1. theory of particulate inheritance

2. dominance/recessiveness

3. law of segregation

4. law of independent assortment

Front 7

theory of particulate inheritance

Back 7

-hypothesized that observable traits were determined by discrete units of inheritance and different units make up a trait

-each plant carries two "particles of heredity" for each trait

-heredity particles, elements, units = all mean alleles

Front 8

dominance/recessiveness

Back 8

-when 2 different but homozygous individuals are crossed --> heterozygous F1 progeny uniform and genetically identical

-F1 expressed trait: dominant allele

-trait not expressed: recessive allele

Front 9

law of segregation

Back 9

-during gamete formation paired alleles of a gene segregate randomly so that each gamete receives one allele or another with equal probability

-anaphase I *

-if individual contains a pair of identical alleles: all gamete receive same allele

-if an individual contains a pair of unlike alleles: each gamete has 50% chance of receiving one allele vs the other

Front 10

F1 selfing

Back 10

-(Dd x Dd)

-progeny: DD, Dd, dD, dd

-genotypic ratio 1:2:1

-phenotypic ratio 3:1 ( 3 D_ : 1 dd)

Front 11

where we see segregation in population

Back 11

F2

Front 12

test-cross

Back 12

-cross an individual with a dominant phenotype (A_) but unknown genotype with an individual with a recessive phenotype/genotype (aa) to determine genotype of the first individual

Front 13

multiplication rule

Back 13

-probability of 2+ indepenent events occurring simultaneously = product of individual probabilities

-events are mutually exclusive if the outcome of one event doesn't affect the outcome of other events

-e.g. tossing a penny and a nickel, what is the probability you will get heads for both? (1/2) x (1/2) = (1/4)

Front 14

sum rule

Back 14

-probability of independent events being accomplished in 1+ ways

-one event OR another can happen to prove question true

-e.g. tossing a penny and a nickel, what is the probability you will get one head and one tail?

-two ways: head/tail or tail/head

-each has a 1/4 chance of happening so (1/4) + (1/4) = (1/2)

Front 15

1. what is the probability of getting a 5 with a single die?

2. what is the probability of getting a 5 on each of two dice thrown simultaneously?

3. what is the probability of getting either a 3 or a 4?

Back 15

1. (1/6)

2. (1/6) x (1/6) = (1/36)

3. (1/6) + (1/6) = (1/3)

Front 16

In humans, disease occurs in homozygous recessive individuals for the allele a. If two normal parents have a daughter and a normal son, what is the probability the son is a carrier of the recessive allele? *

Back 16

(2/3)

Front 17

binomial distribution

Back 17

-mutually exclusive events: both events can't occur together

-mutual independence: occurrence of the first event

-"binomial distribution": only two outcomes are possible (2 alleles for a trait)

Front 18

binomial distribution

Back 18

-mutually exclusive events: both events can't occur together

-mutual independence: occurrence of the first event

-"binomial distribution": only two outcomes are possible (2 alleles for a trait)

Front 19

binomial distribution math

Back 19

-n = # of progeny

1. Assign initial probabilities to each outcome

2. Locate correct equation in the binomial based on n

3. Pick the correct term by the distribution of n

4. Plug in outcomes from step #1

Front 20

binomial distribution math example: probability that a couple will have two children with albinism and two with normal pigmentation

Back 20

1. Assign initial probabilities to each outcome: p = 3/4, q = 1/4

2. Locate correct equation in the binomial based on n: n = 4 so p4 + 4p3q + 6p2q2 + 4pq3 + q4

3. Pick the correct term by the distribution of n: 6p2q2 (2 and 2 are the exponents)

4. Plug in outcomes from step #1: 6 * (3/4)2 * (1/4)2 = 27/128

-27/128 probability the couple will have two with albinism and 2 with normal pigmentation

-aka 27 out of 128 families with 4 children will have two albino and two normal