Transcript Monohybrid and Dihybrid - Ms Kim`s Biology Class

Monohybrid and Dihybrid
Honors Biology-Ms. Kim
What is a genotype?
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A. Brown Hair
B. Freckles
C.Tt
D. All of the above
Homozygous Dominant
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A. aa
B. Aa
C. AA
D. Blue Eyes
If I crossed a 2 True Breeding Plants
with different traits, their offspring
would be…
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A. All Heterozygous
B. Hybrids
C. AA, Aa, aa
D. Both A and B
What is a carrier?
• The heterozygous genotype that does not
express the phenotype when disorders are
caused by recessive alleles
What is the probability that 2 carriers
for Cystic Fibrosis will have a child with
Cystic Fibrosis/
A.
B.
C.
D.
100%
75%
50%
25%
What is Probability?
–The chance that a specific event will
occur
• Probability = number of ways a specific event can occur
number of total possible outcomes
How is Probability used in Genetics?
• Used to explain the chance of an offspring
inheriting a specific trait
• Each box represents ¼ or 25 %
What is a Punnett Square?
• Way to predict ALL possible outcomes of a
cross
How do you read a Punnett square?
• Axes represent possible gametes from each
parent
• Boxes represent possible genotypes for
offspring
Genetics Vocab (pt 2)
• Monohybrid cross  cross where parents
differ in only one trait (Rr x rr)
• Dihybrid cross  cross where parents differ in
two traits (RrHh x rrHH)
• Punnett square – a diagram that shows the
gene combinations that might result from a
genetic cross of two parents
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Monohybrid Cross
• a cross between 2 individuals that looks at
1 trait
– Ex: Just looking at the possibility of getting
freckles
Dihybrid Cross
• a cross between 2 individuals that looks at
the possibilities of inheriting 2
DIFFERENT traits at one time
– Ex: looking at the possibility of getting freckles AND
dimples in the SAME offspring
Monohybrid Punnett Square
Mom’s genotype (Hh) x Dad’s genotype (hh)
Tall
Short
Mom’s allele #1
Dad’s
Allele #1
h
Dad’s
Allele #2
h
Mom’s allele #2
H
h
Hh
hh
Hh
hh
Genotype Outcome (Ratio) vs.
Phenotype Outcome (Ratio)
• Genotype Possibilities = the
GENOTYPE probabilities (expected
results) of offspring
»Ex: 50% Hh and 50% hh (0:2:2)
• Phenotype Possibilities= the
PHENOTYPE probabilities (expected
results) of offspring
»Ex: 50% Tall 50% Short (2:2)
Genetics Vocab (pt 3)
• 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)
• Genotype – a description of the genetic make-up of
an individual (TT, Rr)
• Phenotype – a description of what an individual
LOOKS like (tall, red)
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Genetics Vocab (pt 4)
• Homozygous – two identical alleles for a trait
– AA – HOMOZYGOUS dominant
– aa – homozygous recessive
• Heterozygous – two different alleles for a trait
– Aa – HETEROZYGOUS one of each allele
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Solving Punnett Squares
H Biology
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Punnett squares
Step 1
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?
R = round
r = wrinkled
Why are we using the same letter?
Why not use “R” for round and “W”
for wrinkled?
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Step 2
• Define the parents
If a homozygous round pea plant is crossed with a
heterozygous round pea plant, what will their
offspring look like?
RR x Rr
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Step 3
Draw the Punnett square
R
R
R
r
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Step 4
Cross the parents to find the probability of offspring
1. Bring the top letter down and the side letter over…
R
RR
R
RR
R
r
Rr
Rr
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Step 5
Find the genotype and phenotype of the offspring
R
R
R
RR
RR
r
Rr
Rr
Genotype: genetic make-up
(letters)
Phenotype: physical
characteristics
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Finished Product
Key:
R=round
r=wrinkled
R
R
r
RR
Rr
R
RR
Rr
Genotype: genetic make-up
(letters)
2 RR: 2Rr
50% RR
50% Rr
Phenotype: physical
characteristics
4 (round): 0 (wrinkled)
100% Round
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Practice #2
• In pea plants, round seeds are dominant over
wrinkled. A plant that is homozygous dominant
for round seeds is crossed with a heterozygous
plant.
Key:
R = round
r = wrinkled
Cross:
RR x Rr
Genotype:
Phenotype:
Example:
Heterozygous x Heterozygous
Do the following cross:
Mom’s genotype (Hh) x Dad’s genotype (Hh)
Tall
Tall
Draw a Punnett Square and determine the offspring’s
genotype and phenotype.
Give the probabilities and ratios for each
Example:
Heterozygous x Heterozygous
Mom’s genotype (Hh) x Dad’s genotype (Hh)
Tall
Tall
H
h
H
HH
Hh
h
Hh
hh
Genotype ratio = 25% HH, 50% Hh, 25% hh (1:2:1)
Phenotype ratio = 75% Tall, 25% short (3:1)
Working Backwards…The
Testcross
• Allows us to determine the genotype of
an organism with the dominant
phenotype, but unknown genotype
– Genotype is not obvious…could be HH or Hh
• Cross an individual with the dominant
phenotype with individual that is
recessive for the same trait
• Conduct a test cross, where
the unknown dominant
individual is crossed with the
known recessive .
• H _ ?_ x hh
Test Cross
Mom’s genotype (H?) x Dad’s genotype (hh)
Tall
Short
H
?
h
Hh
?h
h
Hh
?h
If all the offspring are ALWAYS tall…Mom has
to be HH
If some offspring are short…Mom has to be Hh
DIHYBRID
CROSSES:
Assuming genes
follow Mendelian
Genetics
(complete dominance)
Dihybrid Crosses
• crosses involving crossing 2 DIFFERENT traits
at one time
– Example: Mate 2 parents and look at the
probability of seeing 2 traits, such as:
• eye color AND hair color
• freckles AND dimples
How do You Do Dihybrid
Crosses?
• Setting up a complex Punnett Square
OR
• 2 separate monohybrid
• 1 square for EACH trait
• use PROBABILITY RULES and
MULTIPLY
What is a dihybrid cross?
• Cross that shows inheritance of two different traits
– For example: homozygous round & yellow crossed
with a homozygous wrinkled & green seed
– RRYY x rryy
Setting up a Dihybrid
• #1- read the problem & 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:
1. RRYY
2. RrYy
RY
RY, Ry, rY, ry
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Setting up a Dihybrid
• #4 – Fill in the tops and sides of punnett square
with gamete combinations
RY
– Example:
Ry
rY
ry
RY
1. RRYY 2. RrYy
RY
RY, Ry, rY, ry
• #5 - Genotype and Phenotype as usual
RY
RY
RRYY
Ry
RRYy
rY ry
ry
RrYY
RrYy
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Dihybrid Final Product
• R=round, r=wrinkled, Y=yellow, y=green
RY
RY
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RRYY =
RRYy =
RrYY =
RrYy =
RRYY
Ry
RRYy
rY
RrYY
ry
RrYy
Round and Yellow
Round and Yellow
Round and Yellow
Round and Yellow
• So…we can say that all of our offspring (100%) will be round and
yellow!
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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.
STEP 1:
• Parental genotypes = RRYy x rrYy
Possible gametes
RY, Ry
rY, ry
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Dihybrid Example Problem #1
STEP 2:
• Set up the dihybrid cross using the gametes
from before…
RY
rY
ry
RrYY
RrYy
Ry
RrYy
Rryy
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Dihybrid Example Problem #1
STEP 3: Determine the genotype and phenotype!
RY
rY
RY
RrYY
ry
RrYy
Genotype:
Ry
Ry RrYy
Rryy
Phenotype:
1 RrYY: 2 RrYy : 1 Rryy
3 Round, yellow
1 Round, green
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Dihybrid Example Problem #2
• Black fur is dominant to white fur
• Long hair is dominant to short hair
• 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 dominant & recessive traits
– 2) Determine the possible gametes of each parent
– 3) What is the only gamete possibility for their
offspring?
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Dihybrid Example Problem #2
1) 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.
2)
3)
Determine the possible gametes of each
Dad  ALL BL
Mom  ALL bl
What is the only gamete possibility for their offspring?
GENOTYPE: 100% BbLl
PHENOTYPE: Black, long-haired
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Ggbb x ggBb
 FOIL (FIRST, OUTER, INNER, LAST)
Ggbb
X
ggBb
G= Grey hair
g = white hair
B = Black eyes
b = red eyes
Ggbb x ggBb
Gb
gb
gB
GgBb
ggBb
gb
Ggbb
ggbb
Ggbb x ggBb
Gb gb
Gb
gb
gB
1/16
GgBb
1/16
ggBb
1/16
GgBb
1/16
ggBb
gb
1/16
Ggbb
1/16
ggbb
1/16
Ggbb
1/16
ggbb
gB
1/16
GgBb
1/16
ggBb
1/16
GgBb
1/16
ggBb
1/16
Ggbb
1/16
ggbb
1/16
Ggbb
1/16
ggbb
gb
GGbb x ggBb
 GENOTYPE POSSIBILITIES:
GgBb = 4/16 or ¼ = 25%
Ggbb= 4/16 or ¼ = 25%
ggBb= 4/16 or ¼ = 25%
ggbb= 4/16 or ¼ = 25%
 Genotypic ratio:
1 GgBb : 1 Ggbb : ggBb : 1 ggbb
 PHENOTYPE POSSIBILITES:
Grey Hair Black Eyes: 25%
Grey Hair Red Eyes: 25%
White Hair Black Eyes: 25%
White Hair Red Eyes: 25%
 Phenotypic ratio:
1: 1: 1: 1
Practice
• In pea plants, yellow seeds are dominant to
green seeds in peas. Round seeds are
dominant over wrinkled. Cross two plants
that are heterozygous for both traits.
– Write the genotypes for the parents.
– Then use “FOIL” to determine your possible allele
combinations from each parent
– Then set up Punnett Square and fill in the boxes
– Then figure out the genotypic and phenotypic
ratios
YyRr x YyRr
YR Yr
yR
yr
YR
1/16
YYRR
1/16
YYRr
1/16
YyRR
1/16
YyRr
Yr
1/16
YYRr
1/16
YYrr
1/16
YyRr
1/16
Yyrr
yR
1/16
YyRR
1/16
YyRr
1/16
yyRR
1/16
yyRr
yr
1/16
YyRr
1/16
Yyrr
1/16
yyRr
1/16 yyrr
YyRr x YyRr
YR Yr
yR
yr
YR
1/16
YYRR
1/16
YYRr
1/16
YyRR
1/16
YyRr
Yr
1/16
YYRr
1/16
YYrr
1/16
YyRr
1/16
Yyrr
yR
1/16
YyRR
1/16
YyRr
1/16
yyRR
1/16
yyRr
yr
1/16
YyRr
1/16
Yyrr
1/16
yyRr
1/16
yyrr
YyRr x YrRr
GENOTYPE POSSIBILITIES:
YyRr = 4/16 or 1/4 = 25%
YYRr=2/16 or 1/8 =12.5%
YyRR=2/16 or 1/8 =12.5%
Yyrr=2/16 or 1/8 =12.5%
yyRr=2/16 or 1/8 =12.5%
YYrr= 1/16 = 6.25%
YYRR= 1/16 = 6.25%
yyRR=1/16 = 6.25%
yyrr =1/16 = 6.25%
PHENOTYPE POSSIBILITES:
9 yellow, round
3 yellow, wrinkled
3 green, round
1 green, wrinled
Would you like to know
a few SHORT CUTS?
Short Cuts for
MONOHYBRID CROSSES
• Every parent “donates” only 1 allele to each offspring
– Law of Segregation
• When crossing 2 heterozygous individuals in complete
dominance, you will ALWAYS get
– 1:2:1 GENOTYPE ratio
• 1 homozygous dominant: 2 heterozygous: 1 recessive
– 3:1 PHENOTYPE ratio
• 3 dominant phenotype: 1 recessive
Short Cuts for DIHYBRID
CROSSES
• When crossing 2 heterozygous
individuals in complete dominance, you will
ALWAYS get
– 9:3:3:1 PHENOTYPE ratio
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9 dominant, dominant phenotype
3 dominant, recessive phenotype
3 recessive, dominant phenotype
1 recessive, recessive phenotype
NOTE: The genotypes have to
be ALL heterozygous
•Ex: HhFf x HhFf
Now…Let’s do Multicharter
Problems 
• What is the probability of producing an
offspring with the genotype AaBBCcDDeeFf in
a cross between 2 parents with the following
genotypes?
– AABbCcDDeeFf
X
– AaBbCcDdeeFf
• ½ x ¼ x ½ x ½ x 4/4 x ½
• = 4/256 = 1/64 or 1.5% chance
Practice
1) In humans, curly hair is dominant over straight
hair. A woman heterozygous for hair curl marries a
man with straight hair and they have four children.
a) Show the parental cross and the possible
gametes produced.
b) Use a Punnett square to find the
possible genotypes for the F1 generation.
c) What are the phenotypic and genotypic
ratios of the F1?
d) What is the probability that the first child
will have curly hair? What is the probability that
the third child will have curly hair?
Practice
2) The ability to taste the drug phenyl-thio
carbamide (P.T.C.) is due to a dominant gene. A
non-taster man marries a taster woman whose
father was a non-taster.
a) What will be the expected genotypes of their
four children?
b) What would be the expected phenotypes for
ten children?
c) What is the probability that their first child has
the heterozygous genotype?
d) Give the phenotypic and genotypic ratios of
the possible offspring produced
More Practice?
• The dark wizard, Lord Voldemort, is able to speak parseltongue (pp),
which is a recessive trait. Not only can he speak parseltongue, he can also
do dark magic (dd). Him and his followers, the Death Eaters, spend a lot
of time together practicing dark magic. Over the years, Voldemort and
Bellatrix Lestrange fell in love. In order to continue their dream of taking
over the wizarding world, they decided to have children. Bellatrix, who
can do dark magic, cannot speak parseltongue. However, they only want
children if they can speak pareseltongue as well.
– What is Voldemort’s genotype?
– What does Bellatrix’s genotype have to be in order to have a child who can
do black magic and speak parseltongue?
– Create a dihybrid cross using Bellatrix’s genotype of part B, include genotypic
and phenotypic ratios.
– What is the percent chance that they have a child who can speak
parseltongue and do black magic?
Practice
• Determine the results of a cross between two
dolphins. Female is heterozygous dominant
for skin color and has a short tail land the
male is heterozygous for both traits. Grey skin
(G) is dominant to white skin (g) and long tails
(T) are dominant over short tails (t). List all
phenotype ratios and % for the F1 generation.
Practice
In humans, right-handedness (R) is dominant to lefthandedness
and hitchhiker thumb (H) is dominant to straight thumb.
Chuck is homozygous right-handed and has a hitchkiker’s
thumb. His father has a straight
thumb. Emily is left-handed and can’t bend her thumb back.
– What are the phenotype possibilities if Chuck and Emily
have children?
– What is the probability that their baby’s genotype will be
like:
• Chuck?
• Emily?