Lab 9 Heredity Lab Report PDF

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Lab 9 - Heredity Lab Report 1. Monohybrid Cross Example For this activity we will consider the trait for flower color in Mendel’s pea plants. The allele P is dominant and results in purple flowers. The allele p is recessive and results in white flowers. A Heterozygous Male plant is mated with a Heterozygous Female plant. First Determine the parent Genotypes. Individual

Genotype

Phenotype

Homozygous dominant

PP

Purple

Heterozygous (hybrid)

Pp

Purple

Homozygous recessive

pp

White

Write the parent genotypes ____Pp______ Next, Determine the possible gametes of each parent

__P___ or __p___

X

____Pp______

__P___ or __p____

Put the gametes on the top and left of the punnett square. Combine the gametes in the boxes. P

p

P

PP

Pp

p

Pp

pp

What are the genotype and phenotype ratios of the offsping? 25% PP 3 purple : 1 white 50% Pp 25% pp

2. Monohybrid cross For this Activity consider the gene for plant height. The allele T is dominant and results in tall plants. The allele t is recessive and results in short plants. Fill out the following steps to detemine the result of mating two hybrid parents. First Determine the parent Genotypes.

Edited 8.8.19

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Individual

Genotype

Phenotype

Homozygous dominant

TT

Tall

Heterozygous (hybrid)

Tt

Tall

Homozygous recessive

tt

Short

Write the parent genotypes _____Tt_____ Next, Determine the possible gametes of each parent

___T__ or ___t__

X

____Tt______ __T___ or __t____

Put the gametes on the top and left of the punnett square. Combine the gametes in the boxes. T

t

T

TT

Tt

t

Tt

tt

What are the genotype and phenotype ratios of the offsping? (TT =25%, Tt =50%,tt =25% ) Ratio: 1:2:1 (Tall :Short )

3:1

3. ABO blood groups – Codominance and multiple alleles The following table shows the possible genotypes and phenotypes of human ABO blood groups. Genotype(s)

Phenotype

IAIA or IAiO

A

IBIB or IBiO

B

IAIB

AB

iOiO

O

Complete a cross of a mother with genotype IAiO and a Father with IBIB. 2

Write the parent genotypes

IAi __________

Next, Determine the possible gametes of each parent

IA I _____ or _____

X

I Bi __________ IB I Bi _____ or ______

Put the gametes on the top and left of the punnett square. Combine the gametes in the boxes.

IA I B

IAi

(AB)

(A)

I Bi (B)

ii (O)

What are the genotype and phenotype ratios of the offsping? Genotype: (IA IB),(IAi),(IBi), and (ii) The ratio is 1:1:1:1 Phenotype: (AB),(A), (B), and (O) The ratio is 1:1:1:1

Critical thinking. In the 1940’s Charlie Chaplin was involved in an unknown paternity case. The baby’s blood type was type B, the mother’s type A, and Chaplin’s type O. Legally he was found guilty of being the biological father of the child. Was Charlie Chaplin the father? Use a punnett square to support your conclusion. Note:blood typing can never prove a paternity claim, but that it can disprove certain ones. Genotype(s) A A

A O

B B

B O

I I or I i

Phenotype A

I I or I i

B

iOiO

O

Mother Baby

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Charlie   

The baby’s blood was type B, so its genotype would either be I B I B or I B i O. The mother’s blood type was A so her genotype would be either I A I A or I A I O Charlie Chaplin’s blood type was type O so his genotype would be i O i O.

i

i

IA

IAi (A)

IAi

IB

I Bi (B)

IB i (B)

(A)

4. Andalusiona Chickens – Incomplete dominance In Andalusian chickens feather color is determined genetically. B results in black feathers and b results in white feathers. Heterozygous chickens (Bb) show a blended color phenotype called and are called ‘blue’ chickens. The first cross in the P generation is between a Black chicken and a White chicken. Individual

Genotype

Phenotype

Homozygous dominant

BB

Black

Heterozygous (hybrid)

Bb

Blue

Homozygous recessive

Bb

White

Write the parent genotypes Next, Determine the possible gametes of each parent

BB __________ B B _____ or _____

X

bb __________ b b _____ or ______

4

Put the gametes on the top and left of the punnett square. Combine the gametes in the boxes. B B

b

Bb

Bb

b

Bb

Bb

What are the genotype and phenotype ratios of the offsping?

Genotype: BB=50%, Bb=50% 1:1 Ratio Phenotype: Black = 100% As you can see from the diagram above, all possible outcomes involve ‘B’ so the dominant phenotype will always be expressed in offspring.

Now cross two of the offspring from the previous cross to determine the F2 generation. Write the parent genotypes Bb Bb __________ X __________ Next, Determine the possible gametes of B b B b each parent _____ or _____ _____ or ______ Put the gametes on the top and left of the punnett square. Combine the gametes in the boxes. B b

B

BB

bB

b

Bb

bb

What are the genotype and phenotype ratios of the offsping? 1 trait cross 3 allele combinations

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Genotype

Count

Percent

bB

2

50

BB

1

25

bb

1

25

Phen Co Per otype unt cent [ pb B]

2

50

[ pB ]

1

25

[ pb ]

1

25

5. Sex-linked traits Sex of humans is determined by chromosomes. A female has the genotype XX and a male has the genotype XY. Perform a cross between an male and a female Write the parent genotypes

XX __________

Next, Determine the possible gametes of each parent

XX XX _____ or _____

X

XY __________ XY XY _____ or ______

Put the gametes on the top and left of the punnett square. Combine the gametes in the boxes.

X

X

X

XX

XX

Y

XY

XY

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What are the genotype and phenotype ratios of the offsping? An X chromosome from the mother and an X chromosome from the father - producing a girl (female phenotype from the XX genotype) An X chromosome from the mother and a Y chromosome from the father - producing a boy (male phenotype from the XY genotype) The ratio of female to male offspring is 1:1

Critical thinking. King Henry VIII of England blamed his wives for failing to produce a male heir. Considering the previous punnett square, was he correct to blame his wives? Explain why or why not. According to the above Punnett square, the wife of King Henry VIII can give birth to a male child, and the gender is determined by the male chromosome and not by the female chromosome. So, all in all, King Henry VIII was incorrect.

Hemophilia is a sex linked recessive disorder in humans that results in the inabilty to for blood clots. The allele XH results in a normal phenotype, and the allele Xh results in hemophilia. Use a punnett square to cross a father who is normal (XHY) and a mother who is a carrier for hemophilia (XHXh). Write the parent genotypes Next, Determine the possible gametes of each parent

X HY __________ Y XH _____ or _____

X

X HX h __________ XH

Xh _____ or ______

Put the gametes on the top and left of the punnett square. Combine the gametes in the boxes. Xh

X

XH

XH X

XH Xh

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Y

XY

Xh Y

What are the genotype and phenotype ratios of the offsping?

1. a male child with genotype phenotype hemophilia 2. a male child with genotype phenotype normal 3. a female child with genotype phenotype normal 4. a female child with genotype XX phenotype normal Genotype ratio 1:1:1:1

Critical Thinking. Males and females are affected differently by X-linked disorders. From the previous cross what percentage of sons will have hemophilia? What percentage of daughters will have hemophila?

25% female carrier, 25%female with hemophilia, 25% normal male, and 25% male with hemophilia.

6. Dihybrid Cross Example A dihybrid cross will consider two traits at the same time. This activity will consider the traits for flower color and plant height. The table below will remind you of the alleles and pheontypes of these traits. Trait Allele Phenotype Flower color P Purple flowers Flower color p White flowers Plant height T Tall Plant height t Short A dihybrid cross mates two parents who are heteozygous for both traits. Write the parent genotypes

_____PpTt_____

X

____PpTt______

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Next, Determine the possible gametes of each parent. Use the FOIL method PT or Pt or pT or pt PT or Pt or pT or pt Put the gametes on the top and left of the punnett square. Combine the gametes in the boxes. PT

Pt

pT

pt

PT

PPTT

PpTt

PpTT

PtTt

Pt

PPTt

Pptt

PpTt

Pptt

pT

PpTT

PpTt

ppTT

ppTt

pt

PpTt

Pptt

ppTt

pptt

What are the phenotype ratios of the offsping? Purple and Tall – 9 Purple and short – 3 White and Tall – 3 White and Short -1

7. Labrador Retrievers – Dihybrid Cross The coat color of labs is controlled by two genes. One trait codes for coat color, B is dominant and results in black fur, and b is recessive and results in brown fur. The second trait codes for whether the fur has any pigment at all. E is dominant and codes for pigmented fur, e is recessive and codes for no pigment. Dogs with no pigment (ee) appear as yellow labs no matter what their gene for coat color says. A dihybrid cross mates two parents who are heteozygous for both traits. Write the parent genotypes

_______BbEe____

X

___BbEe__________

Next, Determine the possible gametes of each parent. Use the FOIL method ___BE_ or Be_____ or ___BE__ or ___be___

BE or

Be or

bE

or

be 9

Put the gametes on the top and left of the punnett square. Combine the gametes in the boxes. BE bE Be be

BE

BBEE

BbEE

BBEe

BbEe

bE

BbEE

bbEE

BbEe

bbEe

Be

BBEe

BbEe

BBee

Bbee

be

BbEe

bbEe

Bbee

bbee

What are the phenotype ratios of the offsping?    

Black Labradors can have any genotype with at least one dominant allele at both the B and E loci: BBEE, BBEe, BbEE, or BbEe. Chocolate Labradors will have a genotype with at least one dominant E allele but must have only recessive b alleles: bbEE and bbEe. Yellow Labradors with black skin pigment will have a dominant B allele but must have recessive e alleles: BBee or Bbee. Yellow Labradors with pale or chocolate pigment, or an absence of skin pigment, can have only recessive alleles at both loci: bbee.

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