Lab 1 Mendelian Inheritance Practice Problems 4Aug2020-1 PDF

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Lab 1: Mendelian Inheritance Practice Problems-Monohybrid Crosses Biology II Laboratory BSC101

Version: AG BJ NB AN SF JR BJ BE

3Aug2020

Stephanie Ruiz Sydney Payne Group members: ___________________________________ , ____________________________________ , Jada Smith Ana ____________________________________ Rodriguez Manuele _______________________ _____________________ ,______ Work the problems 2-6 below by filling in the boxes. Save the completed document with a new name and submit to Canvas. Simple Monohybrid Cross This type of cross involves traits controlled by two alleles at a single gene. One dominant allele, one recessive.

EXAMPLE PROBLEM WITH SOLUTION. The answers to problem 1 are provided in red. Follow through the solution to problem 1: Simple Monohybrid Cross—this type of cross involves traits controlled by two alleles at a single gene. One allele is dominant, and the other is recessive. 1. A gene for tongue rolling (the ability in humans to roll the tongue at its tip) has two physical expressions: the ability to roll the tongue, and the inability to roll the tongue. The ability to roll the tongue is dominant.

yes

Can you roll your tongue? __________ (yes or no) Using the letters, A and a, identify the following: a. The genotypes (there are two) of a person who can roll their tongue AA Aa b. The genotype of a person who cannot roll their tongue: aa

c. Write out the genotypes and cross for the parents: AA x aa d. What will the genotypes and proportions of the genotypes of their children be? Complete the Punnett square. Most people will draw the following Punnett square:

a a

A A Aa Aa Aa Aa

However, when setting up a Punnett square you need to put all possible gametes of one parent on the top, and all possible gametes of the other parent on the side, then fill in the square with all possible combinations of the gametes. So, the following is better: A a

Aa

e. What will the phenotypes, and proportion (%) of the phenotypes of their children be? All (100%) of the children will be tongue rollers

Lab1: Mendelian Inheritance Practice P roblems

Suppose a person who is true-breeding for tongue rolling (this means the person’s genotype must be AA, or homozygous for the trait) has children with a person that cannot roll their tongue (since the inability to roll the tongue is recessive, the genotype must be aa) .

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2. A gene for white color (t) in tigers is recessive to the normal yellow color (T).

tt a. Write out all possible genotypes for white color tigers: ____________________

TT b. Write out all possible genotype for yellow color tigers: _Tt ___________________ c. If two white tigers are crossed and produce offspring, what are the expected proportions (%) of genotypes and phenotypes of the offspring? (Hint: fill only the part of the square that is necessary, using the possible gamete genotypes). Note: Boxes on the top and left side are for the haploid genotypes of the parent gametes (example: T or t). Larger black boxes are for the diploid offspring genotypes that result from the cross (example: TT, Tt, or tt). (See example on page 1 if needed.)

t t ____ ____ t _ ____ tt ____ tt ___ t tt ____ tt ____ ____ 100% are tt F1 genotype(s) and their proportions (%) ________________________________________

Lab1: Mendelian Inheritance Practice P roblems

100% are white tigers F1 phenotype(s) and their proportions (%) ________________________________________

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3. At the Tampa Zoo, a yellow tiger named Martin, comes from a family of tigers that has always shown the yellow color (yellow is true breeding in his family). Martin is crossed with a white tiger named Lizzy and they have two offspring, Lady and Gent. Lady is crossed with another white tiger named Sam.

homozygote _ a) What kind of genotype (homozygote, heterozygote) must Martin have if he is true breeding? ___________________ TT b) Write out the genotype for Martin__________ tt c) Genotype for Lizzy__________

Tt d) Genotype for Lady__________ Tt e) Genotype for Gent__________ tt f) Genotype for Sam__________ g) What are the expected proportions of genotypes and phenotypes of the offspring from Lady and Sam? Fill in the Punnett square below. (Hint: fill only the part of the square that is necessary, using the possible gamete genotypes).

T ____ t ____ t _ ____ Tt ____ tt ___ t Tt ____ tt ____ ____ 50% Tt 50% tt F1 genotype(s) and their proportions (%) ________________________________________

Lab1: Mendelian Inheritance Practice P roblems

50% yellow 50% white F1 phenotype(s) and their proportions (%) __________________________________________________

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4. The most common form of the phenotype for albinism is due to a single gene and the alleles for albinism are recessive. To show the trait, individuals must have two recessive alleles (genotype aa). If two heterozygous parents (P generation, genotype Aa) choose to have children (F1). Fill in the Punnett square to answer the questions below.

A ____ a ____ AA ____ Aa A _ ____ ___

a ____ aa Aa ____ ____ What is the probability of having:

25 a) An albino child? ________ % 75 b) A child without albinism? ________ % 5. Short hair in chihuahuas (allele S) is dominant to long hair (allele s). A true-breeding long-haired animal (genotype SS) is crossed with a purebred short-haired animal (genotype ss). This is the parental generation. a) Given the genotypes of the P generation above, what is (are) the genotype(s) and phenotype(s) of the F1 generation? Fill in the Punnett square then answer the questions below: (Hint: fill only the part of the square that is necessary, using the possible gamete genotypes).

S ____ S ____ Ss ____ Ss s _ ____ ___ Ss ____ Ss s ____ ____

Ss 100%

F1 genotype(s) and their proportions (%) ________________________________________

b) Given the genotypes of the F1 generation above, what is (are) the expected genotype(s) and phenotype(s) of the F2 generation? Fill in the Punnett square and answer the questions below:

S ____ s ____ SS ____ Ss S _ ____ ___

Ss ____ s ss ____ ____

SS 25% Ss 50% ss 25%

F2 genotype (s) and their proportions (%) ____________________________________________________________

short hair 75% long hair 25% F2 phenotype (s) and their proportions (%) ___________________________________________________________

Lab1: Mendelian Inheritance Practice P roblems

hair 100% F1 phenotype(s) and their proportions (%) _Short _______________________________________

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6. Cystic fibrosis is a genetic disorder caused by a single gene. An individual that expresses cystic fibrosis is homozygous for recessive alleles (genotype ff). The trait is recessive, so heterozygous individuals do not express the disease and are not affected. Heterozygous individuals are carriers, because they do not express the phenotype, but they can carry it from one generation to the next. A heterozygous individual fathers a child with a woman who does not have cystic fibrosis nor is she a carrier.

Ff

carrier, but no disease

a) What is the genotype of the man? __________ What is the phenotype of the man? _________________________

No disease

FF b) What is the genotype of the woman? __________ What is the phenotype of the woman? _______________________ c) Fill in the Punnett square, based on a cross between the man and the woman, to answer the following questions. (Hint: fill in only the part of the Punnett square that is necessary, using the possible gamete genotypes.)

F ____ f ____ F _ ____ FF ____ Ff ___ F FF ____ Ff ____ ____ 0% d) What is the probability that they will produce a child with the disease? (%) __________ 50 % e) What is the probability that they will produce a child who is a carrier? (%) __________ (A carrier has one allele for the disease but does not express the disease phenotype. Note that the father is a carrier.) f) What is the probability that they will produce a child that does not have the disease AND is not a carrier?

50% (%) __________

Lab1: Mendelian Inheritance Practice P roblems

Remember that the Punnett square predicts probabilities—not certainties. Just because statistics predict that the probability of a specific event occurring is 50%, that does not mean that it will always happen 50% of the time. For example, just because statistics predict that ½ of the children a couple has will be boys and ½ will be girls, this does not always happen in real life. Genetics might predict that a couple had 2 out of 4 chances of producing a child with a disease that is inherited with a single dominant allele, such as Huntington’s disease. Suppose this couple then had four children—all of whom were normal. We would say they were lucky—they beat the odds.

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