Solve by filling out a Punnett square. What will be the probable genotypic and phenotypic ratios of their offspring?

Monohybrid Crosses

1. Free ear lobes in humans is dominant (F) to attached earlobes (f). Show the cross between two individuals that are heterozygous for free earlobes. Solve by filling out a Punnett square. What will be the probable genotypic and phenotypic ratios of their offspring?

Insert Punnett Square Photo below:

 

Genotypic Ratio: Click or tap here to enter text.

Phenotypic Ratio: Click or tap here to enter text.

2. Tongue rolling in humans is a dominant trait (R) to non-rolling (r). If two parents are unable to roll their tongue, what will be the genotypic and phenotypic ratios of their offspring? Solve by filling out a punnett square

Insert Punnett Square Photo below:

Genotypic Ratio: Click or tap here to enter text.

Phenotypic Ratio:  Click or tap here to enter text.

Test Cross

3. An animal breeder wants to know if a black guinea pig is homozygous or heterozygous for coat color. He test crosses the black guinea pig with a homozygous recessive white guinea pig. If one of the offspring is white, what is the genotype of the black parent guinea pig? Click or tap here to enter text.

Pedigree Study

4. The ability to taste a chemical called PTC (phenolthiocarbimide) is controlled by a dominant gene (T). Use (t) for non-tasters. Shaded individuals cannot taste PTC (non-tasters). Record the genotype for all designated individuals using the table below the pedigree. For unknown dominant phenotype individuals use (T?).

 

Generation	Individual	Genotype (1/2 point each)
I.	1.	Click or tap here to enter text.
II.	4.	Click or tap here to enter text.
II.	5.	Click or tap here to enter text.
III.	2.	Click or tap here to enter text.

 

Dihybrid Crosses

5. If a purebred thorny plant (T) with red flowers (R) were crossed with a plant that was purebred for yellow flowers (r) and no thorns (t), what would be the phenotypic ratio for the F₂ Generation (like the Mendel dihybrid cross for pea shape and color in notes and textbook)?

 

Insert F₂ Punnett Square Photo below:

F₂ Phenotypic Ratio: Click or tap here to enter text.

6. In fruit flies long wing is dominant to vestigial (underdeveloped) wing and brown body is dominant to black body. Cross a fruit fly that is homozygous dominant for long wing and heterozygous for brown body with a fruit fly that is heterozygous for brown body and homozygous for vestigial wing.

Insert Punnett Square Photo below:

Phenotypic Ratio: Click or tap here to enter text.

 

Incomplete Dominance

7. Snapdragon flowers can be red, white, or pink. Pink flowers result when red and white flowers cross-pollinate. Red snapdragons are homozygous red (RR) and white snapdragons are homozygous white (R’R’). Show the cross between a red snapdragon and a pink snapdragon.

Insert Punnett Square Photo below:

Genotypic Ratio: Click or tap here to enter text.

Phenotypic Ratio: Click or tap here to enter text.

 

Codominance

8. In Erminette chickens, one allele produces black feathers (B) and the other produces white feathers (W). Heterozygous individuals have some black feathers and some white feathers, giving them a speckled appearance. If a black rooster and a white hen mate, what would be the ratio of genotypes and phenotypes in the F2 generation?

Insert Punnett Square Photo below:

 

Genotypic Ratio: Click or tap here to enter text.

Phenotypic Ratio: Click or tap here to enter text.

 

Blood Typing

9. A man with type A+ blood and a woman with type B+ blood have a baby with type O- blood. Be sure to include Rh factor (+/-). Determine the genotype for each:

Man = A+: Click or tap here to enter text.

Woman = B+: Click or tap here to enter text.

Baby =O-: Click or tap here to enter text.

 

10. There is a mix up at the hospital between parents and babies. As the administrator in charge, you MUST send the correct baby home with it proper parents. Match proper parents with their baby.

 

Baby R is A+

Baby S is B+

Baby T is B-

Baby U is O-

 

Proper Match:

Parents #1 are blood type AB- and O-	Baby: Click or tap here to enter text.
Parents #2 are blood type A- and B-	Baby: Click or tap here to enter text.

 

Sex-linked

11. Hemophilia in humans is caused by a recessive mutation on the X-chromosome. What will be the outcome (genotypic and phenotypic ratio) of mating between a homozygous normal female (X^HX^H) and a hemophilic male (X^hY)? See lecture notes for proper recording of ratios for each sex.

Insert Punnett Square Photo below:

Genotypic Ratio: Click or tap here to enter text.

Phenotypic Ratio: Click or tap here to enter text.