heredity

40
Heredity Mendel’s Laws S-B-8-1_Heredity PowerPoint

Upload: aimon

Post on 19-Mar-2016

31 views

Category:

Documents


1 download

DESCRIPTION

Heredity. Mendel’s Laws. Heredity. Source: http://en.wikipedia.org/wiki/Habsburg. How are traits inherited?. Selective Breeding Domestication of plants Higher yielding crops Mature faster, set fruit earlier Attractive fruit Domestication of animals Desired behaviors Desired qualities. - PowerPoint PPT Presentation

TRANSCRIPT

Page 1: Heredity

Heredity

Mendel’s Laws

S-B-8-1_Heredity PowerPoint

Page 2: Heredity

Heredity

Source: http://en.wikipedia.org/wiki/Habsburg

S-B-8-1_Heredity PowerPoint

Page 3: Heredity

How are traits inherited? • Selective Breeding

– Domestication of plants• Higher yielding crops• Mature faster, set fruit earlier• Attractive fruit

– Domestication of animals• Desired behaviors• Desired qualities

Gray Wolf

http://animal.discovery.com/mammals/wolf/ http://www.akc.org/breeds/

S-B-8-1_Heredity PowerPoint

Page 4: Heredity

Gregor Mendel• Born in 1822, Brno, Moravia (now

the Czech Republic), the only son of a farming family

• Showed promise in school and entered the monastery to continue his education

• Went to the University of Vienna, studying physics and math

• Came back to the Abbey in Brno to teach and experiment in the garden

S-B-8-1_Heredity PowerPoint

Page 5: Heredity

How can we calculate the probability of traits being

inherited?

• Sum Law• Product Law

http://farm1.static.flickr.com/231/464511629_d09fb664af.jpg

S-B-8-1_Heredity PowerPoint

Page 6: Heredity

Product Law• States that the probability of two

independent events occurring simultaneously is equal to the product of the probabilities of each event occurring alone.

Example:• What is the probability of a coin toss

resulting in 2 consecutive heads?

S-B-8-1_Heredity PowerPoint

Page 7: Heredity

S-B-8-1_Heredity PowerPoint

Product Law Example Answer:

The probability for each toss individually resulting in heads is 1/2.

To obtain the probability of two consecutive heads, multiply their individual probabilities.

(1/2) × (1/2) = 1/4

Page 8: Heredity

Sum Law:• States that the probability of two

independent events occurring simultaneously is equal to the product of the probabilities of each event occurring alone.

Example: • What is the probability if we flip a coin twice,

that it will result in heads one time, and tails the next, if we do not specify the order in which they come?

S-B-8-1_Heredity PowerPoint

Page 9: Heredity

S-B-8-1_Heredity PowerPoint

Sum Law Example Answer:There are two mutually exclusive ways this can occur:•Heads the first time, tails the second.•Tails the first time heads the second.

The probability for each event is ½.The probability of each situation can be derived using the product law: (1/2) × (1/2) = 1/4

Since the individual events are mutually exclusive, the total probability of either occurring can be derived using the sum law: (1/4) + (1/4) = 1/2

Page 10: Heredity

Mendel’s Experiment

• Mendel investigates inheritance in pea plants for eight years.

• He chose peas because: – They are easy to cultivate.– They reproduced sexually.– They have easy to

distinguish traits.

http://www.scientificamerican.com/article.cfm?id=gene-behind-mendels-green&ref=rss

S-B-8-1_Heredity PowerPoint

Page 11: Heredity

Pisum sativum-The Pea

http://www.tutorvista.com/content/biology/biology-iii/heredity-and-variation/mendels-experiments.php

S-B-8-1_Heredity PowerPoint

Page 12: Heredity

THINK-PAIR-SHARE1. What made Gregor

Mendel uniquely qualified for his experiments in heredity?

2. Pea plants have more than 7 observable traits. What condition did the trait need to meet in order for Mendel to study it?

http://commons.wikimedia.org/wiki/

File:Gregor_Johann_Mendel_bust.jpg

S-B-8-1_Heredity PowerPoint

Page 13: Heredity

Artificial Pollination

Removing Stamen from flower

Exposing Stigma and dusting with pollen

Covering the flower with bag and labeling

stem

S-B-8-1_Heredity PowerPoint

Page 14: Heredity

Monohybrid Cross• Crossed Tall plants with

Short plants, the parents or P

• Called the hybrid offspring, the First Filial, F1, generation

What would you expect the F1 Generation to look like?

X

F1

P

All offspring are tall.

S-B-8-1_Heredity PowerPoint

Page 15: Heredity

Rule of Dominance• recessive traits

– The trait that is not expressed, or hidden, in the F1 generation

• DOMINANT traits – The trait expressed in

hybrid offspring, the F1 generation

S-B-8-1_Heredity PowerPoint

Page 16: Heredity

Dominant/Recessive People Search

Widow’s Peak

No Widow’s Peak

Straight thumb Hitchhiker’s thumb

http://www.ncrtec.org/tl/camp/gene/thumbs.htm https://mrsmaine.wikispaces.com/Jade+D

Mid-digit hair No mid-digit hair

http://www.nydailynews.com/lifestyle/2010/06/09/2010-06-09_new_cosmetic_surgery_dimpleplasty_gives_you_dimples_like_mario_lopezs_in_ten_min.html

Dimples No Dimples

Smile

Rolling tongue Non-rolling tongue

http://udel.edu/~mcdonald/mythtongueroll.html

Kinky/naturally curly hair Straight or wavy hair

http://www.naturallycurly.com/curlreading/curl-products/curl-expert-creates-hair-rules-line-for-curls-and-kinks

http://udel.edu/~mcdonald/mythearlobe.html

Free Earlobe

Attached Earlobe

https://mrsmaine.wikispaces.com/Jade+D

Longer second toeBiggest toe longest

S-B-8-1_Heredity PowerPoint

Page 17: Heredity

The F2 Generation

Mendel allowed the F1 plants to self pollinate.

What would you expect the offspring to look like?

F1

P

F2

X

787 tall plants and 277 dwarf plants

S-B-8-1_Heredity PowerPoint

Page 18: Heredity

Rule of Unit FactorsMendel’s Explanation

• Individuals have two “heritable” factors for a trait .

• The two factors may or may not contain the same information:– Homozygous individuals have

identical factors .• TT or tt

– Heterozygous individuals have 2 different factors.

• Tt

F1

P

F2

X

TT

Tt

tt

TT ttTt Tt

S-B-8-1_Heredity PowerPoint

Page 19: Heredity

Genetics

Modern Genetics• Individuals have two copies of a

gene (one one each homologous chromosome).

• The various forms of a gene are called alleles:– Dominant alleles are expressed

when they are present.– Recessive alleles are expressed

only when the individual is homozygous for the gene form.

Dwarf allele

Tall allele

Genes at the same location

HOMOLOGOUS CHROMOSOMES

S-B-8-1_Heredity PowerPoint

Page 20: Heredity

Phenotypes vs. Genotypes

• Phenotypes refer to what is expressed or the organism’s outward appearance.– TT and Tt pea plant

will be over 6 feet tall.– tt pea plant will be

under 2 feet tall.

• Genotypes refers to the specific alleles that make up the organism.– A TT pea plant is

homozygous Tall while a tt pea plant is homozygous dwarf.

– A Tt pea plant is heterozygous Tall.

S-B-8-1_Heredity PowerPoint

Page 21: Heredity

THINK-PAIR-SHARE1. If the black rabbit and the

white rabbit were bred and all the offspring were black, what conclusion can you make?

2. If the hybrid offspring were allowed to mate, would you expect some of the litter to be white? Why?

Xhttp://www.countryfarm-lifestyles.com/raising-rabbits.html

http://athomepets.blogspot.com/2008/03/bunnies-born-for-easter-crowd.html

S-B-8-1_Heredity PowerPoint

Page 22: Heredity

THINK-PAIR-SHARE1. Are the genotypes homozygous dominant,

homozygous recessive, or heterozygous?• DdFF• RrYy• LLww• ggSs

2. The letters used in genotypes usually represent the dominant trait, e.g., Y-yellow seed and y-green seed. Decide on traits to represent the genotypes above, and describe the phenotypes.

S-B-8-1_Heredity PowerPoint

Page 23: Heredity

Law of SegregationMendel’s Explanation

• Alleles separate when forming gametes.

• Parents pass only one allele for each trait.

Modern Genetics• What process ensures

that alleles separate during gamete formation?

http://www-bcf.usc.edu/~forsburg/meiosis.html

S-B-8-1_Heredity PowerPoint

Page 24: Heredity

Dihybrid Cross

• Mendel repeated his experiment with plants that are bred true for two traits.

• He again allowed the F1 generation to self pollinate.

• Phenotypic ratio: 9:3:3:1

http://image.wistatutor.com/content/feed/u2044/Punnet%20square.gif

XP

F1

F2

32101108315

Y-Yellow

y-green

R-Round

r-wrinkled

S-B-8-1_Heredity PowerPoint

Page 25: Heredity

Law of Independent AssortmentMendel’s Explanation

• Traits are inherited independently from one another.

Modern Genetics• What process ensures

that genes are randomly assorted among the gametes?

http://www.palaeos.com/Fungi/Lists/Glossary/GlossaryM.html

S-B-8-1_Heredity PowerPoint

Page 26: Heredity

Punnett Squares

• British biologist Reginald Punnett used a matrix to show the Laws of Segregation and Independent Assortment.

• The Punnett Square shows the probability of an offspring having a particular genotype.

http://image.wistatutor.com/content/feed/u2044/Punnet%20square.gif

Punnett Square of Dihybrid CrossParent Genotype RrYy

Gametes’ Genotype

Gam

etes

’ Gen

otyp

e

S-B-8-1_Heredity PowerPoint

Page 27: Heredity

Using Punnett Squares

Both parents are hybrids or heterozygous for

black fur.

♂Bb and ♀Bb

Step 1

Write the genotypes of the parents.

Black colored fur (B) is dominant to tan colored fur (b) in dogs. Suppose you cross two hybrid dogs with black fur. What are the possible phenotypes and genotypes of the puppies? What are the expected ratios for each genotype and phenotype?

S-B-8-1_Heredity PowerPoint

Page 28: Heredity

Using Punnett Squares

STEP 2

List the different gametes that each parent can

make.

To find the number of different gametes is 2n,

where n is the number of heterozygous gene pairs

Black colored fur (B) is dominant to tan colored fur (b) in dogs. Suppose you cross two hybrid dogs with black fur. What are the possible phenotypes and genotypes of the puppies? What are the expected ratios for each genotype and phenotype?

♂ gametes ♀ gametes B and b B and b

S-B-8-1_Heredity PowerPoint

Page 29: Heredity

Using Punnett Squares

STEP 3

Make a grid of horizontal and vertical lines.

Black colored fur (B) is dominant to tan colored fur (b) in dogs. Suppose you cross two hybrid dogs with black fur. What are the possible phenotypes and genotypes of the puppies? What are the expected ratios for each genotype and phenotype? Since the male and female can

make 2 different gametes the grid will be 2 x 2.

S-B-8-1_Heredity PowerPoint

Page 30: Heredity

Using Punnett SquaresBlack colored fur (B) is dominant to tan colored fur (b) in dogs. Suppose you cross two hybrid dogs with black fur. What are the possible phenotypes and genotypes of the puppies? What are the expected ratios for each genotype and phenotype?

STEP 4

Put the gametes from one parent on the top and the other parent’s gametes on the side.

B b

B

b

♂ gametes

♀ ga

met

es

S-B-8-1_Heredity PowerPoint

Page 31: Heredity

Using Punnett SquaresBlack colored fur (B) is dominant to tan colored fur (b) in dogs. Suppose you cross two hybrid dogs with black fur. What are the possible phenotypes and genotypes of the puppies? What are the expected ratios for each genotype and phenotype?

STEP 5

Fill in the boxes by copying the row and

column headings.

B b

B

b

BB Bb

Bb bb

S-B-8-1_Heredity PowerPoint

Page 32: Heredity

Using Punnett SquaresBlack colored fur (B) is dominant to tan colored fur (b) in dogs. Suppose you cross two hybrid dogs with black fur. What are the possible phenotypes and genotypes of the puppies? What are the expected ratios for each genotype and phenotype?

STEP 6

Determine the phenotypes of the possible offspring.

B b

B

b

BB BbBB Bb

Bb bb1-BB

2-Bb

1-bb

3-Black colored fur

1-Tan colored fur}

S-B-8-1_Heredity PowerPoint

Page 33: Heredity

THINK-PAIR-SHARE1. What would a Punnett

square look like for a monohybrid cross of a purple flowered pea plant allowed to self pollinate?

2. What are the genotypic ratios of the monohybrid cross?

http://en.wikipedia.org/wiki/File:Punnett_square_mendel_flowers.svg

S-B-8-1_Heredity PowerPoint

Page 34: Heredity

Problems with Two TraitsBlack colored fur (B) is dominant to tan colored fur (b) in dogs while smooth hair (h) is recessive to wire hair texture (H). Suppose you cross two hybrid dogs with black wiry coats. What are the possible phenotypes and genotypes of the puppies? What are the expected ratios for each genotype and phenotype?

Step 1

Write the genotypes of the parents.

Both parents are hybrids or

heterozygous for black and wiry coats.

♂BbHh and ♀BbHh

S-B-8-1_Heredity PowerPoint

Page 35: Heredity

Problems with Two Traits

STEP 2

List the different gametes that each parent can

make.

BH,

Step 1

Write the genotypes of the parents.

♂BbHh and ♀BbHh

♂ gametesBh bH bh

BH, Bh, bH, bh ♂ gametes

BH, Bh, bH, bh ♀ gametes

S-B-8-1_Heredity PowerPoint

Page 36: Heredity

Problems with Two TraitsSTEP 3

Make a grid of horizontal and vertical lines.

STEP 4Put the gametes from one parent on the top and the other parent’s gametes on the side.

BH

Bh

bH

bh

♀ g

amet

es

♂ gametes

S-B-8-1_Heredity PowerPoint

Page 37: Heredity

Problems with Two Traits

BH

Bh

bH

bh

♂ gametes

BH Bh bH bh

♀ g

amet

es

STEP 5Fill in the boxes by

copying the row and column headings.

BBHH BBHh BbHH BbHh

BBHh BBhh BbHh Bbhh

BbHH BbHh bbHH bbHh

BbHh Bbhh bbHh bbhh

S-B-8-1_Heredity PowerPoint

Page 38: Heredity

Problems with Two TraitsSTEP 6

Determine the phenotypes of the possible offspring.

BBHH BBHh BBHhBbHH BbHH BbHhBbHh BbHh BbHh

Bbhh Bbhh

bbHh bbHh

bbhh

BBhh

bbHH

There is a 9 out of 16 chance that the offspring will have a black wiry coat.

There is a 3 out of 16 chance that the offspring will have a black smooth coat.

There is a 3 out of 16 chance that the offspring will have a wiry tan coat.

There is a 1 out of 16 chance that the offspring will have a smooth tan coat

S-B-8-1_Heredity PowerPoint

Page 39: Heredity

THINK-PAIR-SHARE

In guinea pigs, short hair (H) is dominant to long hair (h) and a short tail (T) is dominant to long tails (t). If a hybrid short hair and tail guinea pig is crossed with a long hair and tail guinea pig, what are the expected genotypic and phenotypic ratios?

S-B-8-1_Heredity PowerPoint

Page 40: Heredity

Step 1:Hybrid parent Purebred Parent

HhTt hhtt

Step 2: Hybrid gametes Purebred gametesHT, Ht, hT, ht ht

Step 3:Step 4: ht

HT Ht hT ht

Step 5:HhTt Hhtt hhTt hhtt

Step 6: HhTt

Hhtt

hhTt

hhtt

There is a 1 in 4 chance that the offspring will have short hair and tailThere is a 1 in 4 chance that the offspring will have short hair and a long tail

There is a 1 in 4 chance that the offspring will have long hair and a short tail

There is a 1 in 4 chance that the offspring will have long hair and tail

S-B-8-1_Heredity PowerPoint