LAB #6: Mendelian Genetics Lab Manual Exercise 13

G Genotype & Phenotype Ph • Genotype refers to particular genes an individual carries • Phenotype yp refers to an individual’s observable traits • Cannot always determine genotype by g phenotype p yp observing

Alleles • Different forms of a gene – distinct in DNA sequences

• Originally arose by mutation – which changes DNA sequence

F

f

• Alleles come in dominant and recessive genotypes – Dominant allele masks a recessive allele of a gene We denote dominant m alleles with CAPITAL L letters – W

Homologous H l Chromosomes

When talking about genotypes: • Homozygous

– two identical alleles at a locus - AA: homozygous dominant - aa: homozygous recessive

• Heterozygous yg

– two different alleles at a locus - Aa

Mendel’ss Theory of Segregation Mendel • An individual inherits a particulate unit of i f information ti (allele) ( ll l ) about b t a trait t it f from each h parent formation the two alleles of • During gamete formation, a gene in an organism segregate from each other. (MEIOSIS!) • Genes are inherited as discrete “particles.” • Blending does not occur, so that an allele th t is that i presentt iin th the parental t l generation ti and is not evident in the F1 generation can reappear in the F2 generation.

Law of Independent Assortment • Mendel concluded that the two “units” (alleles) for the first trait were to be assorted d into gametes independently d d l of the two “units” (alleles) for the other trait • The two chromosome in each homologous pair are sorted into gametes during meiosis

Nucleus of a diploid (2n) Reproductive cell with two pairs of homologous chromosomes

OR

Possible alignments of the two homologous chromosomes during metaphase I of meiosis

The resulting alignments at metaphase II:

allelic combinations possible in gametes:

A

A a

a

A

A a

a

B

B b

b

b

b B

B

A

A

a

a

A

A

a

a

B

B

b

b

b

b

B

B

B A

B A

1/4 AB

b a

b a

1/4 ab

b A

b A

1/4 Ab

B a

B a

1/4 aB

Fig. 10.8, p. 158

Th Garden The G d Pea P Plant Pl • • • •

Used to study genetics Self pollinating Self-pollinating True breeding C be Can b experimentally i t ll crosspollinated

Mendel’s Experiments p Monohybrid Cross: A cross between two individuals in the same species in which ONE genetic trait is documented.

purple flowers AA

white flowers aa

TRUEBREEDING PARENTS:

purple flowers

white Flowers

x

aa

AA

GAMETES:

A

A

a

a

Aa

F1 HYBRID

OFFSPRING:

In-text, p. 158

Monohybrid Punnett Square True-breeding homozygous recessive parent plant

F1 PHENOTYPE S aa

True-breeding homozygous dominant a parent plant

Aa

Aa

Aa

Aa

a

A

Aa Aa

A

Aa Aa

AA

F1 PHENOTYPES

Aa An F1 plant self-fertilizes and produces gametes:

F2 PHENOTYPES Aa

A

AA

A Aa

Aa

aa

a

A AA Aa a

Aa aa

Dihybrid y Cross Dihybrid Cross: A cross between two indi id als in the same species in which individuals hich TWO genetic traits are documented.

purple flowers AA t ll tall

BB

white flowers aa dwarf

What are parental genotypes?

bb

TRUEBREEDING PARENTS:

purple Flowers, Tall

x

white Flowers, dwarf

aabb

AABB

GAMETES:

AB

AB

ab

ab

AaBb

F1 HYBRID

OFFSPRING:

What is the F1 generation phenotype? In-text, p. 158

F1 Generation: flowered, tall AaBb

AaBb

PurplePurple

Trait Studied SEED SHAPE H PE SEED COLOR POD SHAPE POD COLOR FLOWER COLOR

Dominant Form

Recessive Form

F2 Dominant-toRecessive Ratio

5,474 round

1,850 wrinkled

6,022 yellow

2,001 green

882 inflated 428 green 705 purple

FLOWER POSITION

651 long stem

STEM LENGTH

787 tall

299 wrinkled

2.96: 1 3.01: 1 2.95: 2 95: 1

152 yellow

2.82: 1

224 white

3.15: 1

207 at tip

3.14: 1

277 dwarf

2.84: 1 Fig. 10.5, p. 156

Monohybrid Cross Exercise 13.1 B pg. 183

Monohybrid Cross Lab handout exercise •

•

•

– – –

Tobacco Seedlings •

Make individual counts of white and green phenotypes DO NOT PULL PLANTS OUT OF PETRI DISH

Pool class data Data analysis y

NOTE: Difference between corn and seedling counts = one piece of corn is THE population, l ti while hil each hP Petri t i dish di h of f seedlings is a population. TAKE HOME: each seed is an individual (whether its part of an ear of corn or laying in a Petri dish).

Dihybrid Cross Exercise 13.2 B pg. 189

Observable human traits Exercise 13.3, pg. 190 E

• There are many traits that follow the M d li pattern Mendelian tt of f inheritance. i h it • Ex: Earlobe Anatomy depends on a single gene n with ith ttwo alleles ll l s – Dominant allele specifies detached earlobes – E – Recessive allele specifies attached lobes - e

In today today’ss lab, lab we will examine the genotypes for several human traits.

We’ll Use Statistics Today y –

Scientists use statistical tests to examine the likelihood that their results are due to random events (vs. “real” results)

–

For example, if you flip a coin 100 times, you probably won’t get a clean 50 heads, 50 tails. The same problem arises for genetic ratios (e.g., (e.g. you might get 2.81:1.03 ratio, ratio instead of 3:1 ratio)

–

Chi-square (X2) is one such statistical procedure that m the difference ff between y your actual data versus examines some prediction

–

If you calculate a probability less than or equal to 5% (p ≤ ) then y your results significantly g y differ from the 0.05), predicted d d genetic ratio

–

Don’t get bogged down in the math behind chi-squares (e.g., there are many electrical tools we use in our day-to-day lif without life ith t knowing k i the th actual t l electrical l t i l circuitry…but i it b t they still get the job done for us!)

Chi Squared Analysis Lab and handout exercise • Scientists use statistical tests to examine the likelihood that their results are due to random events (vs. “real” “ results) • Otherwise stated: Chi-square (χ2) tests if an observed b d frequency f distribution di ib i fits fi an expectedd distribution

(O − E ) Χ =∑ E 2

2

Monohybrid Chi-square Chi square (χ2) Phenotype

Geno- Observed type (O)

Expected (O-E)2 (E)

(O-E)2 E

Purple

P_

64

60.75

10.56

0.174

White

pp

17

20.25

10.56

0.522

81

81

Total

Expected p ratio in a monohybrid y cross? How do we calculate E? ((3/4)) x 81 = 60.75 (1/4) x 81 = 20.25

0 696 0.696 3:1

χ2

Chi-square

2 (χ )

• Once you have calculated your chi-squared compare p it to the table on page p g 186. • If your chi-squared is GREATER than the .05 05 cutoff = chance alone cannot explain the deviation • Your number is LESS than the .05 05 cutoff = variation is due to chance.

Monohybrid Chi-square

2 (χ )

• What is the expected ratio? – 9:3:3:1

• How do we calculate expected values? – – – –

(9/16) x Total = E (3/16) x Total = E (3/16) x Total T t l=E (1/16) x Total = E

TODAY’S TODAY S PLAN • Work in pairs, but complete the post-lab BY YOURSELF! • Follow the lab book and do all of the exercises in order, except SKIP section 13.1.C1, 2, & 3 (all Fern exercises). • Look at the tobacco seedlings & count number of green vs. albino. As soon as you have your tobacco data, give it to your TA TA. Do the X2 test for these data… data Show all work! • Answer all the questions in your lab book. • Post-lab P t l b

– Have TA check off your lab book work – Turn in post-lab sheet with completed tables and problems

• When you’re all finished turn in the post-lab assignment.

FOR POST-LAB QUESTIONS, RECALL:

Inheritance of X-Linked Genes • X and Y chromosomes function as a homologus pair •X X is i chromosome h on which diseasecausing g allele is found. •No comparable allele is found on the Y chromosome.