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Group H: Cystic Fibrosis Genetic problem. 

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Group H: Cystic Fibrosis Genetic problem. 

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Cystic fibrosis is a sex linked disorder. Bob, who has Cystic fibrosis, marries Sue, who is a carrier for Cystic fibrosis.  What are the chances of a child of theirs having the disorder and the chances of any child carrying the disorder?  Using the Punnett square, show the cross, and give the genotypic and phenotypic ratios for the other children.              KEY

H – Normal

h– Cystic fibrosis

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Genotype- 2:2 ratio
Phenotype- 2:2 ratio
There is a 50% chance the children will be Color Blind, and there is a 50% chance that the children will not be Color Blind.

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Mom is Heterozygous for Color Bindness, Gg, and Dad is Homozygous, gg, for not Color Blindness.

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G- Color Blind
g- Not Color Blind

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Color Blind is a sex linked disorder. Bob, who is color blind, marries Sue, who is a carrier for color blindness. What are the chances of a child of theirs having disorder and the chances of any child carrying the disorder?

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G          g                        g   {Gg}  {gg}
g    {Gg}   {gg}

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Interpret Results
Phenotype : 100% No Cystic Fibrosis
Genotype : Every offspring would have Ee

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Punnet Square

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Genotype and Phenotype
Sue is homozygous dominant for EE: No cystic fibrosis
Bob is homozygous recessive for ee: Cystic Fibrosis

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Key
Dominant Allele: E (Sue)
Recessive Allele: e (Bob)
E: No Cystic Fibrosis
e: Cystic Fibrosis

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Cystic fibrosis is a sex linked disorder. Bob, who has Cystic fibrosis, marries Sue, who does not have Cystic fibrosis.  If they were to have a child, what would be the chances of a child having the disorder and the chances of any child carrying the trait?

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Genotype Ee:ee, 2:2
Phenotype: 2/4 Unattached ears and 2/4 Attached ears.
According to the punnet square, there was a 50% chance out of four offspring would come out with unattached or attached ears. The only child, so far out of the three, carries a recessive gene. There is another chance their fourth child will display the recessive gene for attached ears.

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E e
e [ Ee ][ ee ]
e  [ Ee ][ ee ]

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Mom, Sue, is heterozygous for unattached ears, while Dad, Bob, is homozygous for attached ears

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E- Unattached Ears
e- Attached Ears

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In humans, unattached ears are dominant over attached. Bob marries Sue, but Bob has attached ears while Sue has unattached. One of their offspring ended up with attached ears.

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Genotype Bb:bb, 2:2

There was a 50% chance that either of the 3 Childs would come out with or without freckles in this case the both homozygous recessive gene were in two children with no freckles and the only child with freckles has the heterozygous gene for freckles.

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Sue mom is homozygous for no freckles; bb
Bob the dad is heterozygous for freckles; Bb

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B - Freckles
b- no freckles

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In humans, freckles are dominant over no freckles. Bob, who

Has freckles, marries Sue, who has no freckles. One of their  three children has freckles.

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Genotype; Ff 2:2 ratio
Phentotype; ff 2:2 ratio

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Sue heterozygous for not color blind; Ff

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Sue heterozygous for not color blind; Ff

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Bob homozygous for color blind; ff

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F- Not color blind
f- Color blind

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Color Blind is a sex linked disorder. Bob, who is color blind, marries Sue, who is not color blind .What is the chances of a child of theirs having the disorder and the chances of any child carrying the disorder. Using the Punnett square, show the cross, and give the genotypic and phenotypic ratios for the other children.

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Genotype: 100% Heterozygous
Phenotype: 100% Colored Vision

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P

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Darrel homozygous dominant: KK: Not colorblind
Donna homozygous recessive: kk: Colorblind

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K - Not colorblind
k - Colorblind

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Suppose that Donna is colorblind but Darrel is not. If Donna and Darrel have a family, what percentage of their boys will be colorblind?

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Genotype; Bb : B : bb : b
1:1:1:1
Phenotype; no ocular albinism: ocular albinism

There is a 50% chance for the kids to be affected by ocular albinism for both male and female children.

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Bob has ocular albinism which affects his singular X-chromosome; b
Sue is heterozygous for ocular albinism; Bb

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B - no ocular albinism
b - ocular albinism
the mutation for ocular albinism occurs on the X-chromosome making men more susceptible to being affected by the mutation.

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Ocular albinism is a recessive gene which causes a person to lack melanin pigment in the persons eyes only. Bob who is affected by ocular albinism is married to Sue who is a carrier for ocular albinism. What is will be the ratio of ocular albinism to normal children if Sue and Bob were to have kids.

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A- Cleft Chin
a- Smooth Chin

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Genotype 2 : 2 ratio
Phenotype 2 : 2 ratio

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A-Cleft Chin
a-Smooth Chin

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Genotype; Dd:dd
1:1
Phenotype; hair on back of hand: no hair on back of hand

Probability that 50% of children will have hair on the back of the hand while 50% will not have hair on the back of the hand.

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KEY
D hair on back of hand
d no hair on back of hand

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GENOTYPES AND PHENOTYPES genetic problem D

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GENOTYPES AND PHENOTYPES genetic problem D

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Sue heterozygous for hair on the back of the hand; Dd
Bob homozygous for no hair on the back of the hand; dd

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D - hair on back of hand
d - no hair on back of hand

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In humans, hair on the back of the hand is dominant over no hair. Bob, who has no hair, marries Sue, who has hair on the back of the hand. One of their three children, has hair on the back of the hand.

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Bob- Heterozygous dominant for cleft chin; Aa
Sue- Homozygous recessive for smooth chin

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In people, having a cleft chin is a dominant gene and smooth chins are recessive. Bob has a cleft chin, his wife Sue has a smooth chin. Out of their three children one has a smooth chin. Using the Punnet Square it shows the cross between each phenotype and gives the genotype and phenotype ratios.

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Genotype; Ff:ff
Phenotype; not colorblind : colorblind

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Bob homozygous for colorblindness ff
Sue homozygous or heterozygous FF / Ff

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F -- not color blind
f -- color blind

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Bob is color blind and is married to sue who isn't color blind but we do not know if she's a carrier. what is the genetypic and phenotypic of sue (FF/Ff) and bob's (ff) kids who will carry the disorder?

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Genotype- 100%
Phenotype- All would be normal

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Genotype: two heterozygous children, and two homozygous recesive
phenotype; 50 % with hair, and 50% without hair

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Punnet Square

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Genotypic & Phenotypic
Mom is homozygous dominant for cystic fibrosis. "EE"
Dad is homozygous recessive. "ee"

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Genotype: two heterozygous children and two Homozygous recessive

Phenotype: 50% Normal children 50% Children with Cystic Fibrosis

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Bob Is Homozygous recessive Cystic Fibrosis (cc)

Mom is Heterozygous for normal gene (Cc)

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Create Key
Dominant Allele Sue- "E"
Recessive Allele Bob- "e"
E- normal
e- cystic fibrosis

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Cystic fibrosis is a sex linked disorder. Bob, who has Cystic fibrosis, marries Sue, who is a carrier for Cystic fibrosis.

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Cystic fibrosis is a sex linked disorder. Bob, who has Cystic fibrosis, marries Sue, who does not have Cystic fibrosis.  What is the chances of a child of theirs having the disorder and the chances of any child carrying the disorder?

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Cystic fibrosis is a sex linked disorder. Bob, who has Cystic fibrosis, marries Sue, who does not have Cystic fibrosis.  What is the chances of a child of theirs having the disorder and the chances of any child carrying the disorder? 

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Genotype: Bb, bb

Phenotype: Freckles, No Freckles

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Dad:  Homozygous for Cystic Fibrosis: hh

Mom: Heterozygous for Cystic Fibrosis:  Hh

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H: functioning gene
h: mutation

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Sue: Heterozygous for hair
Bob: Homozygous recessive for no hair
Key

D- Hair on back of hand
d- No hair on back of hand

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In humans, freckles are dominant over no freckles. Bob, who Has freckles(B), marries Sue, who has no freckles(b). One of their three children has freckles. 3:1

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Mom: Recessives Homozygous for freckles;bb

Dad: Dominant Heterozygous for freckles;Bb.

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Sue heterozygous for hair; Dd
Bob homozygous no hair; dd

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B: freckles on face            b: no freckles on face

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Sue, the mother has a recessive trait of no freckles, while Bob has a dominant trait for freckles.

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D: Hair on back of hand
d:  No hair on back of hand

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In humans, hair on the back of the hand is dominant over no hair. Bob, who has no hair, marries Sue, who has hair on the back of the hand. One of their three children, has hair on the back of the hand.

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5.   Answer the question

Genotype;     AA : Aa : aa

Phenotype;    Curly :  Straight

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3.     Writing down phenotypes and genotypes                                                         

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3.     Writing down phenotypes and genotypes                                                         

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3. Write out phenotypes and genotypes of all the individuals involved
Ex; Mom Heterozygous for Curly hair; Aa
Ex; Dad Heterozygous for Curly hair; Aa

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2. Dominant Allele and Recessive Allele. Include gender where gender is a factor in inheritance.

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4.  Fill out a Punnett Square to solve your problem

Ex; Mom and dad heterozygous for Curly hair  (If you can upload a pictur of your punnett square, do this, but if you can't describe, the cross. You may describe the punnett by using the terms Quadrant 1 =BB, Quadrant 2= Bb (Reading from left to right, then proceeding to the next row)

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1.  Rewrite the genetic problem