9 BSC 2011 MENDELIAN GENETICS PROBLEMS Due October 10

9 bsc 2011 mendelian genetics problems due october 10 the following problems are provided to develop your skill and test your un

9
BSC 2011
MENDELIAN GENETICS PROBLEMS Due October 10
The following problems are provided to develop your skill and test
your understanding of solving problems in the patterns of inheritance.
They will be most helpful if you solve them on your own. However, you
should seek help if you find you cannot answer a problem. Most of
these problems are fairly simple, yet mastering their solutions will
provide the background to solve many genetic puzzles and will
strengthen your understanding fundamental principles of genetics.
The completion of this set of problems constitutes an assignment worth
10 points. To receive credit, you must work each of the problems to
its correct solution. I have provided the answers, I want you to show
how to arrive at them. You must show how you solved each problem, even
if you can do it in your head. Number the problems and make your work
neat enough for me to read. You must turn in your worked problems by
October 10 (the last class before the exam) to receive any credit.
NOTE: The points for this assignment will be awarded on an ALL-OR-NONE
basis. That means you must work ALL problems to their correct solution
in order to receive ANY credit. You may ask me or your TAs for help on
any of these problems. I recommend that you make a copy of your work
before you hand it in so that you can use it while studying for the
exam.
A. PROBABILITY
1. You and your spouse have no children. You stand to inherit a
sizeable fortune from your crazy Uncle Irving if you can produce three
daughters in your family of three children. What is the probability of
doing just that?
2. If you could convince Uncle Irving that simply having three
children all of the same sex would do, then what would be the
probability of your receiving the inheritance?
3. In quest of the family stipulated in #2 above, you produce a boy as
your first child. Now what is the probability of inheriting the
fortune?
4. Why are the answers to #2 and #3 the same?
5. Finally, you have convinced Uncle Irving that you will agree to try
for at least three girls out of four children. How likely are you to
become wealthy given those conditions?
B. MONO-, DI-, AND POLYHYBRID CROSSES; DOMINANCE AND RECESSIVENESS
In all of the following problems, capital letters will be used to
denote a dominant trait, and lower-case letters will be used for the
recessive trait.
6. In peas, seeds may be round (R) or wrinkled (r). What proportion of
the offspring in the following crosses would be expected to be
wrinkled?
a. RR x rr b. Rr x Rr c. Rr x rr
7. In peas, seeds may be yellow (Y) or green (y). What proportion of
the offspring in the following crosses would be expected to be yellow?
a. YY x Yy b. Yy x Yy c. yy x yy
8. In peas (again), the stem length may result in a tall (T) or dwarf
(t) plant. What proportion of the offspring in the following crosses
would be expected to be tall, and what proportion dwarf?
a. TT x tt b. TT x Tt c. Tt x Tt d. tt x Tt
9. What proportion of the plants from the following crosses would be
tall with yellow, wrinkled seeds?
a. TtYYRr x ttYYrr b. TTYyRr x TtYyRr
c ttYyrr x ttyyRr d. TtYyRr x TtYyRr
10. From the crosses TTYyRr x TtYyrr, what proportion of the offspring
would be expected to be
a. tall plants with round, yellow seeds
b. tall plants with round, green seeds
c. dwarf plants with round, green seeds
d. tall plants with yellow, wrinkled seeds
e. tall plants with green, wrinkled seeds
11. For the purpose of this problem assume that in humans the gene for
brown eyes is dominant to that for blue eyes.
a. A brown-eyed man marries a blue-eyed woman, and they have eight
brown-eyed children.
What are the genotypes of all the individuals in the family?
b. What is the probability that the first child produced in parents
who are both heterozygous for brown eyes will be blue-eyed?
c. If the first child is a brown-eyed girl (same parents as in b),
what is the probability that the second child will be a blue-eyed boy?
d. Again referring to the marriage in b, what is the probability that
the first three children will be blue-eyed girls and the fourth a
brown-eyed boy?
12. Eye color in certain species of flies is controlled by a single
pair of genes. A white-eyed fly, both of whose parents had white eyes,
was crossed with a red-eyed fly, and all of their offspring (both male
and female) were red-eyed.
a. Is the gene for red eyes or that for white eyes dominant? Proof?
b. What was the genotype of the white-eyed parents?
c. What was (were) the genotype(s) of the red-eyed offspring?
d. If one of the red-eyed offspring was mated with the white-eyed
parent, what would be the expected ratio of offspring, with respect to
eye color?
e. If two of the red-eyed offspring are mated, how many genetically
different kinds of zygotes, with respect to eye color, will be formed,
and what will the proportions be?
13. In cattle, the gene for hornless (H) is dominant to the gene for
horned (h), the gene for black (B) is dominant to that of red (b), and
the gene for white face (or Hereford spotting) (S) is dominant to that
for solid color (s). A cow with the genotype BbHhSs is inseminated by
a bull of the genotype bbhhSs. What is the probability of the calf's
being:
a. a black, hornless cow with Hereford spotting
b. a red, horned bull with solid color
c. a red, hornless bull with Hereford spotting
14. Assume that D, E, F, G, H, and I are autosomal genes on different
chromosomes. From the mating
DdeeFfGGHhIi x DdEEFFGgHhii:
a. What is the probability that one of the offspring will have the
genotype DdEeFFGghhIi?
b. What is the probability that one of the offspring will be
heterozygous for each allele?
c. What is the probability that one of the offspring will have the
genotype DDEEFfGGhhii?
C. INCOMPLETE DOMINANCE
15. In cattle, RR = red, Rr = roan, and rr = white. What are the
predicted color phenotypes and their frequencies for the offspring
from crosses between:
a. a red bull and a white cow
b. a red bull and a roan cow
c. a roan bull and a roan cow
16. Given the following information about the inheritance of
characteristics in pea plants, answer the questions below:
Y (yellow) is dominant to y (green)
R (round) is dominant to r (wrinkled)
B (bitter) is dominant to b (sweet)
S (smooth) is dominant to s (hairy)
L (long pod) shows incomplete dominance to 1 (short pod) (Ll is medium
in length)
Given this cross: (P1) Yy Rr Bb SS Ll (male) x yy RR Bb Ss Ll (female)
a. How many different gametes can be formed by the female plant?
b. How many different genotypes are possible in the F1 offspring?
c. How many different phenotypes are possible in the F1 offspring?
d. What percent of the F1 individuals will be
green, bitter, and smooth _______________
hairy, medium, and sweet ______________
round, bitter, and long _________________
D. MULTIPLE ALLELES
17. In humans, the ABO blood groups are controlled by three alleles
(only two of which occur in any one individual): the alleles for A and
B type blood are co-dominant toward each other, and both are dominant
to the allele for O type blood.
a. If a person with type AB blood marries someone with type O blood,
what are the possible phenotypes of their offspring?
In the following, determine the genotypes of the parents:
b. One parent has type A and the other has type B, but all four blood
groups are represented in the children.
c. Both parents have type A, but 3/4 of the children are A and 1/4 are
O.
d. One parent has type AB and the other has type B, but of the
children 1/4 have type A, 1/4 have type AB, and 1/2 have type B.
18. In the following cases of disputed paternity, determine the
probable parent.
a. Mother is type B, child is type O. Father #1 is A; father #2 is AB.
b. Mother is type B, child is type AB. Father #1 is A; father #2 is B.
c. Mother is type O and bears non-identical twins, one type A and one
type B. Father #1 is type A; father #2 is type B.
19. Two babies in a maternity ward have lost their identity bands, and
there is some confusion about their footprint records. Baby #1 is type
A; baby #2 is type B. If you are one of the mothers and your blood
type is O, which one of the following statements applies.
a. Neither baby could be yours.
b. The type A baby is yours.
c. The type B baby is yours.
d. Either baby could be yours.
20. A woman with type A blood has parents who are both type AB and a
husband who is a type B. What is the probability that their first
child will be a son with type O blood?
21. In a local court, a woman is suing a male acquaintance for
financial support of her recently born child. If the woman is blood
type B, Rh+, and the baby is type O, Rh-, and the man is blood type
AB, Rh-, what are her chances of success in the lawsuit?
22. In the organism under consideration, r* acts like r allele, except
when homozygous (r*r*). From the information given below, work out the
phenotypic and genotypic ratios for each of the crosses.
Given: RR (red) x rr (white)  Rr (pink); and
rr* x rr*  3/4 white, 1/4 dead zygotes
a. Rr x Rr b. Rr x rr c. Rr* x Rr* d. Rr* x rr*
23. In rabbits, fur color is determined by a set of multiple alleles
at one locus (gene) that have the following relationship:
C+ (agouti) is dominant to all other alleles
ch (himalayan) is dominant to ca (albino)
cu (chinchilla) shows incomplete dominance with regard to ch and ca
The genotypes cuch and cuca are light-grey phenotypes
a. What breeding stock (parents) would you select if you wished all of
the offspring to be chinchilla?
b. In one of the matings of rabbits, the litter contained 4 grey
bunnies, 2 albino bunnies, and 2 himalayan bunnies. What were the
genotypes of the parents?
c. In another mating, the litter contained 3 agouti bunnies and 3
light-grey bunnies. What were the genotypes of the parents of this
litter?
E. MULTIPLE GENES
24. In cocker spaniels, the following genotypes and phenotypes are
found:
AABB = white A-bb = red aabb = lemon AaB- = black aaB- = liver AABb =
grey
a. A red female is mated with a liver-colored male, and one of the
pups produced is lemon-colored. What are the genotypes of the parents?
b. What proportion of these offspring would be expected to be black?
c. A black male is mated with a liver-colored female, and they produce
the following pups:
3/8 black 1/8 red
3/8 liver-colored 1/8 lemon-colored
What are the genotypes of the two parents?
25. If two cocker spaniels of the genotypes below are mated, and eight
pups are born, what is the most likely distribution of coat colors in
that litter?
P1 AaBb x AABb
______white ______red ______lemon ______black ______liver ______ grey
26. A dominant gene, A, causes yellow color in rats. The dominant
allele of another independent gene, R, produces black coat color. When
the two dominant genes occur together (A-R-), they interact to produce
grey coat color. Rats of the double recessive genotype are
cream-colored. If a grey male and a yellow female are mated and
produce approximately 3/8 yellow, 3/8 grey, 1/8 cream, and 1/8 black,
what were the genotypes of the parents?
F. SEX LINKAGE
27. In humans, dark hair (B) is dominant over blondness (b), and color
blindness (c) is a sex-linked recessive trait. A women has a blond
brother, a blond mother, and a dark-haired father. Her brother and her
parents have normal vision. She bears the following three children by
her blond, normal-visioned husband:
a dark-haired son with normal vision
a dark-haired daughter with normal vision, and
a dark-haired color-blind son
a. Make a pedigree of the entire family showing the probable genotypes
of all individuals.
b. What is the probability that her next (fourth) child will be a
color-blind boy?
c. If her fourth child is a boy, what is the probability that he will
have dark hair?
d. What is the chance that her next four children will all be girls?
28. Your research in hematology has led you to the discovery of a new
type of inherited anemia, which exists in two forms--mild and severe.
The information you have gathered on the familial patterns for the
disease is summarized below. From these data determine a mechanism for
the inheritance of these traits that is consistent with all of the
information.
a. normal female x normal male = all normal children
b. normal female x anemic male = 1/2 mild females
= 1/2 normal males
c. mild female x normal male = 1/4 normal males
= 1/4 anemic males
= 1/4 normal females
= 1/4 mild females
d. mild female x anemic male = 1/4 normal males
= 1/4 anemic males
= 1/4 mild females
= 1/4 anemic females
e. anemic female x normal male = 1/2 mild females
= 1/2 anemic males
f. anemic female x anemic male = all anemic children
29. In chickens and other birds, the chromosomal basis of inheritance
is the opposites of that in man; i.e., in birds, XX individuals are
males, and XY individuals are females.
In chickens, barred plumage is dominant to nonbarred plumage; the gene
is sex-linked.
Suppose that you were a poultry breeder and that you needed large
numbers of barred males and nonbarred females. Describe a breeding
stock that you could assemble for this purpose, which would produce
only barred males and nonbarred females.
Be certain that you show the genotypes of both the roosters and the
hens in your breeding stock and also the genotypes of all the
offspring that this stock will yield.
30. The inheritance of color blindness in humans is due to a recessive
gene located on the X chromosome (X linked).
X+ (normal) > Xc (color blind)
a. If a color-blind boy is born to parents both of whom have normal
vision, what are the genotypes of the three individuals?
b. What is the probability that the second child born to that couple
will be a color-blind daughter?
c. If this couple has four children, what is the probability that the
first two children will be color-blind boys and the last two children
will be girls with normal vision?
31. A particular inherited abnormality in humans has been shown to be
an X-linked recessive trait. In one family seeking professional help
from a genetic counselor, the following information is known:
1. The mother shows the abnormality.
2. Their only child is a daughter who is normal.
a. If their next child is a son, what is the probability that he will
show the abnormality?
b. What is the probability that the next child born will be a daughter
showing the abnormality?
G. HUMAN GENETICS
32. Research has shown that a particular eye defect is represented in
a family pedigree as follows:













A B




C

a. On the basis of this data, which of the following mechanisms of
inheritance are POSSIBLE? autosomal dominant, autosomal recessive,
sex-linked dominant, sex-linked recessive, Y-linked.
b. What is the most PROBABLE mechanism of inheritance?
c. What is the genotype of female A?
d. What is the genotype of male B?
e. What is the probability that a child from marriage C will show this
eye defect?
33. The following pedigree shows the presence of individuals bearing
extra fingers and toes--exhibiting "polydactyly." Explain the genetic
mechanism for the inheritance of this trait in the family.














34. Given the family history below for the inheritance of a skin
abnormality in humans, what is the correct mechanism of inheritance
for this abnormality?



a. X-linked dominant c. autosomal dominant e. y-linked
b. x-linked recessive d. autosomal recessive
35. The inheritance of a particular facial hair pattern in a given
family is indicated below. Answer the following questions about this
family tree.




B


C



a. Which is the correct mechanism of inheritance?
autosomal dominant autosomal recessive y-linked
x-linked dominant x-linked recessive
b. Using the symbols "F" and "f" to indicate dominance and
recessiveness, show the genotypes of the parents at B.
c. In the marriage at C, what is the probability that the next child
born will show the trait?
36. Because you have some basic knowledge of genetics, a young woman
(A, below) has come to you for help. She has a brother who is mentally
retarded, but her parents and all of her husband's family are normal.
She has a two-year-old daughter who shows the retardation of her
brother. She is now pregnant and is considering a therapeutic abortion
because of her fear that the next child will also be retarded. Can you
help her by answering the questions below?



A


B ?
a. What is the probability that her second child will be mentally
retarded?
b. If the second child is a boy, what is the probability that he will
be mentally retarded?
H. GENERAL REVIEW QUESTIONS
37. In domesticated cats, the following genetic patterns have been
described as independently assorting. Normal ears (T) is dominant to
tufted ears (t); curved whiskers (C) is dominant to straight whiskers
(c); the presence of six toes (S) is dominant to five toes (s); and
the gene for hair length is an X-linked codominant. The three
phenotypes for hair length are long (XHXH), medium (XHXh), and short
(XhXh); medium is the heterozygous condition. Sex determination in
cats is the same as for humans (XX = female).
Given two parent cats:
Tt cc SS XHXh x Tt Cc Ss XHY
a. How many different gametes could be formed in the female cat with
respect to these four traits?
b. How many phenotypes are possible in the offspring from this mating?
c. What fraction of the offspring will have tufted ears, curved
whiskers, six toes, and long hair?
d. In a litter of 10 kittens, six are male. How many of the males
would be expected to have tufted ears, six toes, straight whiskers,
and medium-length hair?
38. Assume that the cell below is from an insect testis and is about
to undergo spermatogenesis (gamete production by meiosis). The letters
represent dominant or recessive alleles at particular gene positions
on each chromosome.
a. What is the haploid chromosome number of the insect?
b. How many genetically different kinds of gametes could be formed
from this cell (assume no crossing over)?
c. How many pairs of homologous chromosomes would you observe in a
muscle cell from this insect?
d. What is the probability that a gamete produced by this cell will be
completely recessive for the alleles shown?
e. Which of Mendel's laws tells us that a gamete containing a
chromosome with an "A" allele will not necessarily contain a
chromosome with a "B" allele?
f. Which of Mendel's laws tells us that a gamete cannot contain both
an "A" allele and an "a" allele?
I. CROSSOVER
39. In the common bluebell, two linked, autosomal genes control flower
color and plant height, as follows:
B (blue flowers) > b (white flowers)
T (tall plants) > t (dwarf plants)
If the crossover frequency between these two loci (genes) in 26%, what
will be the expected F1 phenotypes and their frequencies from the
following cross:
P1 BT//bt x bt//bt
40. Genes A and B are located 20 map units apart on the X chromosome
in humans. Thus they are X-linked. The two genes show simple dominance
to their recessive alleles, a and b, respectively. Given the following
genotypes for the parents:
XAB//Xab x XAb//Y
a. What will be the frequency of offspring that are male and show both
dominant traits (A and B)?
b. What will be the frequency of offspring that are female showing
both dominant traits (A and B)?
41. In the fruit fly, Drosophila melanogaster, crossing over is absent
in the males. Suppose you are interested in the relationship between
two linked genes on chromosome 2. The genes are for black body (b) and
for curved wings (c), each of which is recessive to the normal (B)
body color and normal (C) wing shape. In your laboratory, females that
are heterozygous at the body color and wing shape loci are mated with
a black-bodied male with curved wings. The offspring of these matings
were counted, with the following results:
367 normal body, curved wing
131 normal body, normal wing
139 black body, curved wing
363 black body, normal wing
a. What is the cross-over frequency between these two loci?
b. How many map units separate these two loci?
c. Show the pattern of linkage (coupling or repulsion) in the female
flies of this mating?
ANSWERS



1. 1/8
2. 1/4
3. 1/4
4. Because the sex of the first child is irrelevant.
5. 5/16
6. a. none
b. 1/4
c. 1/2
7. a. all
b. 3/4
c. none
8. a. all tall
b. all tall
c. 3/4 tall; 1/4 dwarf
d. 1/2 tall; 1/2 dwarf
9. a. 1/4
b. 3/16
c. 0
d. 9/64
10. a. 3/8
b. 1/8
c. 0
d. 3/8
e. 1/8
11. a. female bb; male BB; children all Bb
b. 1/4
c. 1/8
d. 3/4096
12. a. red
b. homozygous recessive
c. heterozygous
d. 50% red-eyed; 50% white-eyed
e. 3; 1/4 homozygous, 1/2 heterozygous,
1/4 homozygous recessive
13. a. 3/32
b. 1/32
c. 3/32
14. a. 1/64
b. 1/32
c. 0
15. a. all roan
b. 1/2 red, 1/2 roan
c. 1/4 red, 1/2 roan, 1/4 white
16. a. 8
b. 72
c. 12
d. 37.5%
0
18.75%
17. a. 1/2 A; 1/2 B
b. AO and BO
c. both AO
d. AB and BO
18. a. type A father (1#)
b. type A father (#1)
c. either the real father must be AB or both men are the fathers, one
of each twin. (There is documented evidence for such an occurrence.)
19. d; either baby could be yours
20. zero
21. zero
22. a. 1/4 red (RR), 1/2 pink (Rr), 1/4 white (rr)
b. 1/2 pink (Rr), 1/2 white (rr)
c. 1/3 red (RR), 2/3 pink (Rr*); (r*r* dead)
d. 2/3 pink (Rr and Rr*), 1/3 white (rr*); (r*r* dead)
23. a. cucu x cucu
b. cuca x chca
c. various correct answers
e.g. c+ca x cucu
c+cu x caca
c+ch x cucu
24. a. female = Aabb
male = aaBb
b. 1/4
c. female = aaBb
male = AaBb
25. 1 white, 3 black, 2 red, 0 liver, 0 lemon, 2 grey
26. male = AaRr; female = Aarr
27. a. BbXCY bbXCXc
bbXCY BbXCXc bbXCY
BbXCy BbXCX? BbXcY
b. 1/4
c. 1/2
d. 1/16
28. Normal (A) is incompletely dominant to anemic (a);
allele is sex-linked
AA = normal
Aa = mild anemia
aa = severe anemia
Genotype of crosses are:
a. AA x Ay d. Aa x aY
b. AA x aY e. aa x AY
c. Aa x AY f. aa x aY
29. P1 (male) XbXb x XBY (female)
F1 XBXb -- all barred
Xby female -- all nonbarred
30. a. father = X+Y b. zero
mother = X+Xc c. 1/64
son = XCY
31. a. 1
b. zero
32. a. could be sex-linked recessive or autosomal recessive
b. sex-linked recessive (why?)
c. XEXe
d. XEY
e. 1/4; probability of mother’s being XEXe is 1/2; father is XeY, so
half of their children will be affected; therefore, overall
probability is 1/2 x 1/2 = 1/4
33. The trait is dominant and probably carried on the sex (X)
chromosome because
x  but not
x  both and
34. c; autosomal dominant
35. a. autosomal dominant
b. Ff x Ff
c. 1/2
36. a. 1/4
b. 1/4
37. a. 4 c. 1/16
b. 16 d. zero
38. a. 3 d. 1/8
b. 8 e. independent assortment
c. 3 f. segregation
39. Blue, tall = . 37 White, tall = .13
Blue, dwarf = .13 White, dwarf = .37
40. a. 0.20 b. 0.25
41. a. 27%
b. 27
c. Cb//cB