monohybrid and dihybrid honors biology-ms. kim. what is a genotype? a. brown hair b. freckles c.tt...

Monohybrid and DihybridHonors Biology-Ms. Kim

What is a genotype?

• A. Brown Hair• B. Freckles• C.Tt• D. All of the above

Homozygous Dominant

• A. aa• B. Aa• C. AA• D. Blue Eyes

If I crossed a 2 True Breeding Plants with different traits, their offspring

would be…• 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. 100%B. 75%C. 50%D. 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

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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

Monohybrid Cross

• a cross between 2 individuals that looks at 1 trait – Ex: Just looking at the possibility of getting

freckles

• 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

Dihybrid Cross

Monohybrid Punnett Square Mom’s genotype (Hh) x Dad’s genotype (hh)

Tall Short

Mom’s allele #1 Mom’s allele #2

H hDad’sAllele #1 h

Dad’sAllele #2 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)

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• 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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• 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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H Biology

Solving Punnett Squares

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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?

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

RR x Rr

If a homozygous round pea plant is crossed with a heterozygous round pea plant, what will their offspring look like?

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Step 3Draw the Punnett square

R R

R

r

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Step 4

RR RR

Rr Rr

R R

R

r

Cross the parents to find the probability of offspring1. Bring the top letter down and the side letter over…

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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

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Finished Product

RR RR

Rr Rr

R R

R

r

Genotype: genetic make-up (letters)

Phenotype: physical characteristics

Key: R=roundr=wrinkled

2 RR: 2Rr50% RR50% Rr

4 (round): 0 (wrinkled)100% Round

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:

Cross:

Genotype:

Phenotype:

R = round r = wrinkled

RR x Rr

Example:Heterozygous x Heterozygous

Do the following cross: Mom’s genotype (Hh) x Dad’s genotype (Hh)

Tall TallDraw 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

h

Genotype ratio = 25% HH, 50% Hh, 25% hh (1:2:1)

Phenotype ratio = 75% Tall, 25% short (3:1)

HH Hh

Hh hh

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

h

If all the offspring are ALWAYS tall…Mom has to be HHIf some offspring are short…Mom has to be Hh

Hh ?h

Hh ?h

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

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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– Example:

1. RRYY 2. RrYy

RY RY, Ry, rY, ry

• #5 - Genotype and Phenotype as usual

RY Ry rY ry

RY Ry rY ry ryRRYY RRYy RrYY RrYyRY

RY

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Dihybrid Final Product• R=round, r=wrinkled, Y=yellow, y=green

• RRYY =• RRYy =• RrYY =• RrYy = • So…we can say that all of our offspring (100%) will be round and

yellow!

RRYY RRYy RrYY RrYy RY Ry rY ry

RY

Round and Yellow

Round and Yellow

Round and Yellow

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 =

Possible gametes RY, Ry rY, ry

RRYy x rrYy

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STEP 2: • Set up the dihybrid cross using the gametes

from before…

rY

ry

RrYY RrYyRrYy Rryy

RY Ry

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STEP 3: Determine the genotype and phenotype!

RY Ry

Genotype: Phenotype: 1 RrYY: 2 RrYy : 1 Rryy

rY

ry

RrYY RrYyRrYy Rryy

3 Round, yellow1 Round, green

RY Ry

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• 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 #21) 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) Determine the possible gametes of each

Dad ALL BL Mom ALL bl 3) What is the only gamete possibility for their offspring?

GENOTYPE: 100% BbLl PHENOTYPE: Black, long-haired

Ggbb x ggBb

FOIL (FIRST, OUTER, INNER, LAST)

Ggbb XggBb

G= Grey hairg = white hairB = Black eyesb = red eyes

Ggbb x ggBb

GgBb ggBb

Ggbb ggbb

Gb gb

gB

gb

Ggbb x ggBb

1/16 GgBb

1/16 ggBb

1/16 GgBb

1/16 ggBb

1/16 Ggbb

1/16 ggbb

1/16 Ggbb

1/16 ggbb

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 gb Gb gbgB

gb

gB

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

1/16 YYRR

1/16 YYRr

1/16 YyRR

1/16 YyRr

1/16YYRr

1/16 YYrr

1/16 YyRr

1/16 Yyrr

1/16 YyRR

1/16 YyRr

1/16 yyRR

1/16 yyRr

1/16 YyRr

1/16 Yyrr

1/16 yyRr

1/16 yyrr

YR Yr yR yrYR

Yr

yR

yr

YyRr x YrRrGENOTYPE 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, round3 yellow, wrinkled3 green, round1 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

• 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?

– AABbCcDDeeFfX

– AaBbCcDdeeFf

• ½ x ¼ x ½ x ½ x 4/4 x ½

• = 4/256 = 1/64 or 1.5% chance

Practice1) 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?

Practice2) 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.

In humans, right-handedness (R) is dominant to left-handedness

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?

Practice

monohybrid and dihybrid honors biology-ms. kim. what is a genotype? a. brown hair b. freckles c.tt...

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