Download Review Genetics Problems with Answers

Review Genetics Problems with Answers
1. If a tall plea plant is crossed with a short pea plant, what would be the expected genotypic and phenotypic ratios
if:
a. the tall plant was homozygous
b. the tall plant was heterozygous
2. The “blue” Andalusian variety of chicken (really a gray color) is produced by a cross between the black and white
varieties of this chicken. Only a single pair of alleles is involved in this character.
a.
b.
c.
What color chickens, and in what proportions, would you expect to get if you crossed two blue?
What is you crossed a blue and a black?
What type of inheritance is this?
3. In Jimson weed, the allele that produces violet petals is dominant over that for white petals, and the allele that
produced prickly capsules is dominant of that for smooth capsules. A plant with white petals and prickly capsules
was crossed with one that had violet petals and smooth capsules. The F1 generation was composed of 47 plants with
white petals and prickly capsules, 45 white and smooth plants, 50 violet and prickly plants, and 46 violet smooth
plants.
What were the genotypes of the parents?
4. In cocker spaniels, the genotype A B (means AA or Aa with BB or Bb) is black, aaB is liver colored (yech!), A
bb is red and aabb is lemon. A black cocker spaniel is mated to a lemon cocker and produces a lemon pup. If this
original black dog is again mated to another of its own genotype, what proportions of the different colors would be
expected in the offspring?
5. In humans hemophilia A or B is caused by an X-linked recessive gene. A woman who is a non-bleeder had a
father who was a hemophiliac (“bleeder”). She marries a non-bleeder, and they plan to have children. Determine
the probability of hemophilia in their female and male offspring. What proportion of the children (in general) will
have hemophilia?
6. An Rh- (dd) woman had three children by a heterozygous Rh+ husband. What is the total probability that all
three children are Rh positive? Remember, the total probability of independent events happening in sequence is
equal to the product of all probabilities, while the total probability of one OR another event happening is the sum of
probabilities.
7. A child with type A blood is born to a woman with type B blood. What are the possible genotypes of the father?
8. In mugwumps, wings are either green (wild type) or white (recessive). Antennae are either long (wild type) or
short (recessive). A green winged (heterozygous), long antenna (heterozygous) mugwump is mated with a
white short mugwump.
a.
b.
What are the predicted genotypic and phenotypic ratios?
The offspring from this cross were actually in the following proportions: 191 green long, 199 white short, 6
green short, and 4 white long. Is this what you predicted? If not, what is the most likely explanation for
these results?
Answers
1.
Will define T= tall and t= short
a.
b.
2.
TT x tt yields 4TT:0tt, or all tall offspring (100% tall)
Tt x tt yields 2Tt:2tt:0TT, or 1 tall:1 short
Will define black = CB and white as Cw. Alternately, you could use B= black or b= white.
a. Two blues crossed is CB Cw x CB Cw, which yields 1 CB CB:2 CB Cw:1 Cw Cw, or
1 black: 2blue: 1white. This could also be written as 25% black, 50% blue,
25% white or 1/4black, 1/2 blue, 1/4 white.
b. A blue and a black crossed is CB Cw x CB CB, which yields 2 CB CB:2 CB Cw:0 Cw Cw, or 2 black: 2blue:
0 white. This could also be written as 50% black, 50% blue, 0% white or 1/2black and 1/2 blue
c. This type of inheritance is incomplete dominance, where the heterozygote has a phenotype that is
intermediate between the phenotypes of the other two homozygotes.
3.
Will define V=violet petals , v=white petals and P = prickly capsules while p= smooth capsules. At the
start of the problem we know that the parental cross is : VvP_ x V_pp, where the blanks (_) mean that the
allele could be either dominant or recessive for that trait (we don't know yet). The genotypes of the
offspring can be determined as follows:
a. 47 white prickly vvP_
b. 45 white smooth vvpp
c. 50 violet prickly V_P_
d. 46 violet smooth V_pp
The offspring are present in a 1:1:1:1 ratio, and you could try to get these results by plugging in alleles into
the parental genotypes and working out the Punnett Squares. An easier way is to realize that every
offspring plant inherits one gene for every characteristic from each of its two parents. The White Smooth
offspring (vvpp) tells us that both parents must have a little v in their genotype as well as a little p. This
means that the parental genotypes must be VvPp xVvpp.
4.
The genotypes of the parents are laid out in the question: A_B_ = black, aaB_ = liver, a_bb = red, and
aabb= lemon. When a black pup is mated with a lemon cocker (A_B_ x aabb) and the results is a lemon
pup (aabb), this means that the black parent must be AaBb. This deduction arises from knowing that the
lemon pup inherited one little "a" from one parent and one little "a" from the other; the same is true for the
b genes. Thus the black parent must carry an "a" and "b".
a. When two black dogs of this type are mated (AaBb x AaBb), the gametes for each parent are: AB,
Ab, aB, and ab. If these gametes are set up in a Punnett Square and tabulated, you should get:
9/16 black
3/16 liver
3/16 red
1/16 lemon
or 9 black: 3 liver: 3 red: 1 lemon
5.
Will define H= normal blood clotting and h= hemophilia (bleeder). Hemophilia is an X-linked trait, so sex
linkage symbology must be used to solve this problem. A woman who is a non bleeder (XHX-) must be
XHXh if her father (XhY)was a hemophiliac. This is because fathers pass their one and only X chromosome
to each of their daughters. If this women then has children by a non-bleeding male, the cross is: XHXb x
XHY. The resultant offspring proportions are 1X HXH: 1XHXh:1XHY: 1XhY. Now, to answer the question.
Of all the children, only 25% or 1/4 will be hemophiliacs. If the question ask what proportion of the boys
would have hemophilia, the answer is 50%.
6.
If D= Rh positive and d= Rh negative, then the cross is dd x Dd. The offspring ratio is: 2DD:2dd. The
probability of a single child being DD (Rh positive) is 50%. The probability of three children together
being Rh positive is 50% x 50% x 50% (or 1/2 x1/2 x1/2) = 0.124 or 1/8.
7.
There are three alleles for the ABO blood types, I A= Type A blood, IB= Type B blood and i= Type O
Blood. IA and IB are codominant to each other and dominant over the recessive gene "i". To begin with, the
child must be IA_, where the blank must be IA or i. The woman with Type B blood must be IB_, where the
blank is IB or i. In looking at the child then, one can rule out IAIA because the mother cannot possibly carry
a IA to give to the child (since she has type B blood). If the child is I Ai, then the father must be the
contributor of the IA gene. The father's possible genotypes are then I AIA, IAi, or IAIB. His possible blood
types then are Type A or Type AB.
8.
If we define G= green wings (wild type) and g= white wings, and L= long wings (wild type) and l= short
wings, then the cross is GgLl x ggll. The phenotypic ratios expected are 1 green long: 1 green short: 1
white long: 1 white short. But the offspring are mostly all green long and white short. These two
phenotypes are identical to the phenotypes of the parents. Thus, the combinations of genes seen in the
parents are not sorting independently. They are sticking together, and so are linked (carried on the same
chromosome).