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1. suppose that a dominant allele (p) codes for a polka-dot tail and a recessive allele (p) codes for a solid colored tail. in addition, suppose that a dominant allele (l) codes for long eyelashes and a recessive allele (l) codes for short eyelashes. if two individuals heterozygous for both traits (tail color and eyelash length) mate, what's the probability of the phenotypic combinations of the offspring?

a. 9: 3: 3: 1 ratio (9 polka-dot tails and long eyelashes, 3 polka-dot tails and short eyelashes, 3 solid tails and long eyelashes, 1 solid tail and short eyelash)
b. 1: 1 ratio (half are polka-dot tails and long eyelashes; half are solid tails and short eyelashes)
c. 1: 1 ratio (half are polka-dot tails and short eyelashes; half are solid tails and long eyelashes)
d. 9: 3: 3: 1 ratio (9 polka-dot tails and short eyelashes, 3 polka-dot tails and long eyelashes, 3 solid tails and short eyelashes, 1 solid tail and long eyelash)

2. which of the following is correct about the hershey-chase experiment?

a. they discovered that the protein from the virus entered the cell, while the dna stayed behind.
b. they infected viruses with bacterial cells.
c. they determined that infected bacteria contained tags from the virus’s dna.
d. they found no way for hereditary material to be transferred between viruses and bacteria.

3. suppose that a dominant allele (p) codes for a polka-dot tail and a recessive allele (p) codes for a solid colored tail. if two heterozygous individuals with polka-dot tails mate, what's the probability of the phenotype and genotype combinations of the offspring?

a. 75 percent chance of a polka-dot tail and 25 percent chance of a solid colored tail (25 percent pp; 50 percent pp; 25 percent pp)
b. 75 percent chance of a polka-dot tail and 25 percent chance of a solid colored tail (75 percent pp; 25 percent pp)
c. 50 percent chance of a polka-dot tail and 50 percent chance of a solid colored tail (50 percent pp; 50 percent pp)
d. 25 percent chance of a polka-dot tail and 75 percent chance of a solid colored tail (25 percent pp; 50 percent pp; 25 percent pp)

1. A.
2. C.
3. A.

1. For the first question, you have two heterozygous individuals for both traits so the genotype of the parents will PpLl and PpLl. So to get the alleles that we're going to use for the Punnett square we use the FOIL method on the genotype of each parent. 
 PpLl
F = PL      ;O=Pl     ; I=pL      ;and L=pl

We use the same pairs for both because the problem says that they are BOTH heterozygous for both traits. Your Punnett square will then look like this:
          PL          Pl          pL          pl
PL    PPLL    PPLl      PpLL      PpLl
Pl     PPLl     PPll       PpLl       Ppll
pL    PpLL    PpLl      ppLL      ppLl
pl     PpLl     Ppll       ppLl       ppll

The next step will be to list down all phenotypic combinations and then count how many genotypes in the Punnett Square will fall under that category. Remember that dominant traits (Capital letters) will always be expressed. 
    Phenotypes                           Genotypes                  Frequency
Polka-dot;Long lashes;           PPLL, PpLl, PPLl                  9
Polka-dot; Short lashes;         PPll , Ppll                              3
Sold; Long lashes;                  ppLl, ppLL                            3
Solid; Short lashes;                 ppll                                       1

So the ratio is 9:3:3:1

2. As for number 2, Hershey and Chase concluded in their experiment that it was DNA that entered the bacteria cells and not the protein. So it was the DNA that was the transformation principle in the cell that is used to produce more of the virus. 

3. Again, we use a Punnett square and this one is much simpler because we are just considering the tail color and not the eyelashes, as we did in the first. 

So again, both individuals are heterozygous polka-dots. So we know that the genotype of each is Pp. So we put that into our Punnet square:
      P     p
P  PP    Pp
p  Pp    pp

The question is the probability of the genotype and phenotype combinations. In this Punnett we only have 4 squares. So unlike the first, we will do genotype first, then phenotype, just so you can get the probability of all possible genotypes. There are 3 possible genotypic combinations:

Genotype        Phenotype                                     Frequency          Probability
     PP                Polka-dot (Homozygous)                    1                        1/4 = 25%
     Pp                Polka-dot (Heterozygous)                  2                       2/4= 50%
     pp                Solid color (Homozygous)                  1                         1/4= 25%

So we know that the probability for polka-dot tails is 25%+50% = 75%. Broken down it is 25%PP, 50%Pp, and 25%pp.