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Genetics. THE STUDY OF HEREDITY LH – Winter 2011. Gregor Mendel. The scientific study of heredity is called GENETICS ! Augustinian Monk Began working on pea plants in his monastery Correctly believed that heritable factors (genes) retain their individuality from generation to - PowerPoint PPT Presentation
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THE STUDY OF HEREDITYLH – Winter 2011
Genetics
Gregor Mendel
• The scientific study of heredity is called GENETICS! • Augustinian Monk• Began working on pea plants in his monastery• Correctly believed that heritable factors (genes) retain their individuality from generation to generation
– i.e. – marbles
Why Pea Plants?
• Mendel chose to study garden peas, because:– They reproduce quickly & have a short life cycle– They have seven distinct & observable traits– They produce many offspring in one cross– Ease in manipulating pollination
• Self-fertilization vs. Cross Pollination
• Mendel produced seeds by cross pollinating• Note: Since pea plants can self-pollinate, they are called true-breeding. This means they will produce offspring
identical to themselves.
Pea Plant Characteristics
Mendel (con’d)
• Mendel worked with his pea plants until he was sure that all were true-breeding varieties
• He could finally begin his studies: What would happen if different characteristics were crossed?– Purple flower x White flower?
• Hybrid the offspring of two different varieties (also called a monohybrid cross)
• P generation (P1) parental generation have offspring called the F1 generation
• If F1 generations cross, their offspring are called the F2 generation
P F1 F2
Mendel’s Experiments
• After studying pea plants, Mendel concluded that:– Traits are passed from one generation to the next through
genes.– Each trait is controlled by a different form of a gene called
an allele– Some alleles are dominant to others called recessive
traits
• New question: Have the recessive alleles disappeared or are they still present in the parents?
Mendel’s Experiments
• Mendel crossed the first generation and saw that the recessive trait showed up in about 1 of 4 plants.
• Conclusion: Law of Segregation! • Segregation of the alleles happens during the formation of gametes. Each gamete will carry one form of the allele.
Using Probability to Predict Offspring
• Punnett square – a diagram that shows the gene combinations that might result from a genetic cross of two parents
• Phenotype – a description of what an individual LOOKS like (tall, red)
• Genotype – a description of the genetic make-up of an individual (TT, Rr)
Important Vocabulary
• Dominant – allele that appears more frequently. It masks the recessive. – Represented by a capitol letter (R=red)
• Recessive – allele that appears less frequently (b/c it is repressed when paired with a dominant allele)– Represented by a lower case letter (r=white)
• AA – HOMOZYGOUS dominant• aa – homozygous recessive• Aa – HETEROZYGOUS one of each allele
• Homozygous – two identical alleles for a trait• Heterozygous – two different alleles for a trait
LH BiologyWinter 2011
Solving Punnett Squares
Punnett squaresStep 1
R = roundr = wrinkled
STEP 1 Define the alleles
If a homozygous round pea plant is crossed with a heterozygous round pea plant, what will their offspring look like?
Step 2
• Define the parents
RR x Rr
If a homozygous round pea plant is crossed with a heterozygous round pea plant, what will their offspring look like?
Step 3Draw the Punnett square
R R
R
r
Step 4
RR RR
Rr Rr
R R
R
r
Cross the parents find the probability of offspring
Step 5
RR RR
Rr Rr
R R
R
r
Genotype: genetic make-up (letters)
Phenotype: physical characteristics
Find the genotype and phenotype of the offspring
Finished Product
RR RR
Rr Rr
R R
R
r
Genotype ratio: 2 RR: 2Rr
Phenotype ratio: 100% Round
R=roundR=wrinkled
RR x Rr
Dihybrid Crosses
• Dihybrid Cross – a cross of parents differing in TWO characteristics – For example: homozygous round & yellow x
homozygous wrinkled & green seeds– RRYY x rryy
• Law of Independent Assortment – each pair of alleles for different traits segregate independently of other pairs of alleles during gamete formation – This explains genetic diversity among organisms
Setting up a dihybrid• #1- list all 4 alleles
– For example: R=round, r=wrinkled, Y=yellow, y=green• #2 – Create the parental genotypes (4 letters each)
– Example: RRYY (Round, yellow) x rryy (wrinkled, green)• #3 – Using the “foil” method, determine the sets of
gametes (up to 4 possibilities)– Example: RRYY RY
RrYy RY, Ry, rY, ry• #4 – Fill in the tops and sides of punnett square with
gamete combinations• #5 - Genotype and Phenotype as usual
Dihybrid Example Problem #1
• Round is dominant over wrinkled• Yellow is dominant over green• Two pea plants produce offspring. One is
round and heterozygous for yellow seed color. The other is wrinkled and heterozygous for yellow seed color.
• Parental genotypes = RRYy x rrYy
Possible gametes
RY, Ry rY, ry
Dihybrid Example Problem #1
• Set up the dihybrid cross
RY Ry rY
ry
RrYY RrYyRrYy Rryy
Dihybrid Example Problem #1
Determine the genotype and phenotype!
RY Ry
Genotype: Phenotype: 1 RrYY: 2 RrYy : 1 Rryy 3 Round, yellow
1 Round, green
rY
ry
RrYY RrYyRrYy Rryy
Dihybrid Example Problem #2
• Key: Black fur is dominant (B) to white fur (b)Long hair is dominant (L) to short hair (l)
• Two guinea pigs mate. The dad is homozygous for black fur and long hair. The mom is also homozygous, but for white fur and short hair. – 1) Determine the possible gametes of each– 2) What is the only gamete possibility for their
offspring?
Dihybrid Example Problem #2Key: Black fur is dominant (B) to white fur (b)Long hair is dominant (L) to short hair (l)
Two guinea pigs mate. The dad is homozygous for black fur and long hair. The mom is also homozygous, but for white fur and short hair. 1) Determine the possible gametes of each
Dad 4 BL Mom 4 bl 2) What is the only gamete possibility for their offspring?
100% BbLl (Black, long-haired)
Incomplete Dominance
• Incomplete Dominance – type of inheritance where the phenotype of a heterozygous (Bb) is intermediate between the phenotypes of two parents (BB & bb)
• Neither allele is dominant • Heterozygous condition shows a blending of genes• Assign capital & lowercase letters for alleles• This is not blending
Incomplete Dominance Problem #1
If a red four o’clock flower is crossed with a pink four o’clock flower what will their offspring look like?
RR = redrr = whiteRr = pink
Parent Genotypes RR x Rr
Perform cross R R
• Genotype ratio: 2 RR : 2 Rr
• Phenotype ratio: 50% Red flowers & 50% Pink!
RR RR
Rr Rr
R
r
KEYRR = redrr =
whiteRr = pink
Codominance
• Both alleles are equally expressed in the organism
• Use capital letters for both alleles
Codominance Example #1• Black feathers and white feathers in chickens are codominant.
In the heterozygous condition the feathers are called “erminette” and appear blue. – BB = black– WW = white– BW = blue
Cross a black chicken with a blue roosterParents = BB x BW
BB BB
BW BW
B B
B
W
Genotype ratio: 2 BB : 2 BWPhenotype ratio:
50% Black feather s50% Blue feathers
Codominance Example #2• Roan is a coat color found in some cows
– RR = red hair– RW = red and white hair (Roan) – WW = white hair
Cross a roan cow with a red cowParents = RW x RR
RR RW
RR RW
R W
R
R
Genotype ratio: 2 RR : 2 RWPhenotype ratio: 50% Roan, 50% Red
Multiple Alleles
• Most genes can be found in more than 2 forms multiple alleles– Example blood types
• There are 3 alleles (A,B,O) • When combined, they create 4 blood
phenotypes: A, B, AB, O• We write the alleles:
– A = IA
– B = IB
– O = i
Multiple Alleles!
Blood Type Key
• AA = homozygous Type A IAIA
• AO= heterozygous Type A IAi• BB= homozygous Type B IBIB
• BO= heterozygous Type B IBi• AB= (codominant AB) IAIB
• OO = Type O ii
• Each parent gives us 1 allele• Because there are 3 alleles, there are SIX total combination
possibilities
Example Problem
• A type AB woman marries a type O man. What are the possible genotypes of their offspring?
AB x OO (phenotype)IAIB x ii
IA IB
B
i
i
IAi IBi
IAi IBi