ONE TRAIT —  DOMINANT/ RECESSIVE

1.  Mendel crossed pure tall-stemmed pea plants with short-stemmed plants and all of the offspring were tall-stemmed.
    a.  Using T for tall and t for short, write the genotyypes for all of the plants mentioned in this problem.
                            TT (pure tall)    tt(pure short)   Tt(offspring)
    b.  How many different types of genes would the tall parent have to offer its offspring?
                          one:  T
    c.  How many different types of genes would the short parent have to offer its offspring?
                          one:  t

2.  If two F1 plants from problem #1 were crossed, out of 100 F2 offspring, how many would you expect to be tall?
                          75 plants (this is the expected phenotypic percentage)

3.  Suppose you crossed a tall-stemmed plant with a short one and the offspring were 51 tall and 49 short.  How would you explain these results?
51 and 49 are statistically close enough to 50:50 ratio.  This is expected when you cross a heterozygous tall with a homozygous short.
 

X	T	t
t	Tt	tt

4.  Pluto and Esmaralda both can roll their tongues.  They marry and have a child who cannot roll his tongue.
    a.  Is Pluto homozygous or heterozygous?
                 heterozygous  (For both parents to have the trait but not the child, the parents must be heterozygous and the child homozygous)
    b.  Is Esmaralda homozygous or heterozygous? 

5.  Sword swallowing is recessive to non-sword swallowing.  a male non-sword swallower, whose father was a sword swallower marries a sword-swallower, both of whose parents were not sword swallowers.  They have three children, two of whom were not able to swallow swords and one who runs away to join the circus just like his grandfather—because of his great sword-swallowing ability.  Make a pedigree showing all phenotypes and genotypes.
           A=dominant non sword swallowing gene      a=recessive sword swallowing gene

                                 aa (grandfather)  sword swallower
                                  |
                       aa  (father) sword swallower —  Aa/A  (mother)  non-sword swallower              Aa  (mother) non sword swallower  —      Aa    (father) same
                                                                     |                                                                                                                                       |
                                                                   Aa  (male)  non-sword swallower  —————————————aa  (female)  sword swallower
                                                                                       |                                                |                                                        |
                                                                     aa sword swallower                         Aa  non sword swallower                        Aa non sword swallower

6.  In dogs, wire hair is due to a dominant gene, smooth hair to its recessive allele.  Two wire-haired dogs produce a pup which is wire-haired.  To determine whether he carries the gene for smooth hair he should be mated to what kind of female?
              homozygous recessive (smooth haired):  if he is heterozygous he will have some (50% probability) smooth haired pups, if he is homozygous dominant he will produce no smooth haired pups.
 

TWO TRAITS, DOMINANT/ RECESSIVE

7.  In horses, black is dependent on a dominant gene, B, chestnut on its recessive allele b.  The trotting gait is due to a dominant gene T, its recessive allele, t, produces the pacing gait.  If a homozygous black pacer is mated to a homozygous chestnut trotter, what will be the phenotypes of the F1?  what are the F1 genotypes?
            All offspring will be BbTt and will be black trotters.

8.  If two F1 individuals from problem #7 were mated, what kinds of offspring would you get and in what proportions?
              1/16:  BBTT  (black trotters)                2/16:  BbTT  (black trotters)                1/16:  bbTT  (chestnut trotters)
                2/16:  BBTt  (black trotters)                 4/16:  BbTt    (black trotters)                2/16:  bbTt   (chestnut trotters)
                1/16:  BBtt   (black pacers)                   2/16:  Bbtt     (black pacers)                  1/16:  bbtt     (chestnut pacers)

                9/16:  phenotypic black trotters
                3/16:  phenotypic black pacers
                3/16:  phenotypic chestnut trotters
                1/16:  phenotypic chestnut pacers

9.  In rabbits, black is due to a dominant gene B, and brown to its recessive allele, b.  Short hair is due to a dominant gene S, long hair to its recessive allele, s.  A cross between a homozygous brown, long-haired female and a homozygous black, short hiared male will produce what kinds of offspring in the F1 and F2?  Give all phenotypes and genotypes.
              All F1  offspring will be BbSs  and will be black short haired.

   F2         1/16:  BBSS  (black short)                2/16:  BbSS  (black short)                   1/16:  bbSS  (brown short)
                2/16:  BBSs  (black short)                 4/16:  BbSs    (black short)                  2/16:  bbSs   (brown short)
                1/16:  BBss   (black long)                   2/16:  Bbss     (black long)                  1/16:  bbss     (brown long)

                9/16:  phenotypic black short
                3/16:  phenotypic black long
                3/16:  phenotypic brown short
                1/16:  phenotypic brown long
 
 

ONE TRAIT, CO-DOMINANCE

10.  Yellow guinea pigs crossed with thite ones always produce cream-colored offspring.  Two creams, when crossed, produce yellow/cream/white in the ratio of 1:2:1.  Give all genotypes and phenotypes.
           YY=yellow    Yy=cream      yy=white
 

11.  A cream guinea pig is mated to a yellow.  What phenotypes would you expect in the offspring?  Give the expected percents.
        50% Yy  cream       50%  yy  white
 

MULTIPLE ALLELES WITH CO-DOMINANCE

12.  Blood types are sometimes used in paternity suits to establish the possible identity of the father.  What blood type would the father have to have if the child has type AB and the mother B?
           A blood type

13.  What if the child had blood type B and the mother AB?  Is blood typing a reliable way to establish paternity?  Why or why not?
        The father could have blood type, A, B, O or AB.  Blood typing is only reliable to discount paternity or if neither parent has AB blood type.

SEX-LINKED

14.  Color blindness is sex-linked, recessive trait.  Two people with normal color vision have a color-blind son.  What are the genotypes of the parents?  What are the chances of their next child being a color blind daughter?
        The father must be X^NY to have normal color vision.  The mother must be X^NXⁿ if they are to pass the trait on to the son.  The son must be XⁿY and there is no chance of having a color blind daughter.
 

15.  A man with normal color vision marries a woman with normal color vision.  Her father is color blind.  they have two daughters who grow up and marry.  The first daughter has five sons all with normal color vision.  The second daughter has two daughters with normal color vision and a son who is color blind.  Diagram the family history and give all genotypes.
        The man must be X^NY to have normal color vision.  The woman must be X^NXⁿ if her father was color blind (he could only give an Xⁿ to his daughter and the other X^N  had to come from the mother).  The first daughter must be X^NX^N(normal female) if she is to have five sons with normal color vision (statistically speaking).  The second daughter must be X^NXⁿ if she is to pass the trait on to the son who is XⁿY colorblind.  Her daughters could be either X^NX^N (normal female)  or  X^NXⁿ  (female carrier) depending on which X allele they received from their mother.  Their father must be X^NY  if they are not color blind.