MCB142/IB163, Thomson
Mendelian and Population Genetics
Sep 9, 2003
Practice questions for lectures 1-5
1.(a). If an albino individual has one albino and one normally pigmented parent, write out the
genotypes of the parents (albinism is a completely penetrant autosomal recessive trait). What
ratio of albino to normal would be expected among subsequent offspring?
(b). If two normally pigmented parents produce an albino offspring, what ratio of albino to
normal would be expected among subsequent offspring?
2. Achondroplasia is a form of dwarfism inherited as a simple monogenic trait with complete
penetrance. Two achondroplastic dwarfs have a dwarf child and later have a second child who is
normal.
(a). Is achondroplasia produced by a recessive or dominant allele?
(b). What are the genotypes of the two parents in this mating?
(c). Write down all possible genotypes that can be produced by such a mating.
3. Mom and Dad both find the taste of a chemical called, phenylthiocarbamide (PTC), very
bitter. Of their four children, two find the chemical tasteless. Assuming the inability to taste to be
a monogenic and fully penetrant trait, is it dominant or recessive?
4. In Drosophila, yellow body is an X-linked recessive trait (females are XX and males are XY).
If a female Drosophila with a yellow body is crossed to a wild-type male, denote the genotypes
and phenotypes and their proportions in the F1 males and females, and in the F2 generation
formed when the F1 are crossed among themselves.
5. A true-breeding strain of blue-eyed bats is developed. In bats, as in humans, the male is the
heterogametic (XY) sex. Males of this true breeding blue eyed strain are mated to females of a
true-breeding brown-eyed strain of bats. The F1 are all brown-eyed. What would be the
phenotypes of the F1 bats in the reciprocal cross (true breeding blue eyed females to true
breeding brown eyed males) if:
(a). The blue-eyed character is an X-linked recessive? (Show your work)
(b). The blue-eyed character is an autosomal recessive? (Show your work)
Pedigree problems
For all the following pedigree questions, unless otherwise stated, assume:
(a) The trait is due to the action of a single gene, which has, for the purposes of our study, only
two phenotypically distinguishable alleles (one for the trait, one for wild-type if the trait is
rare).
Page
1
(b) The trait shows complete penetrance and there are no age of onset effects, i.e., the trait is
expressed at birth.
(c) The trait shows uniform expressivity.
(d) There are no new mutations in any of the individuals in the pedigrees.
(e) Mendelian segregation has occurred.
(f) If the trait is rare, assume individuals marrying into the pedigree are not carriers, unless there
is evidence from the pedigree to the contrary.
(g) For maternally (mitochondria) inherited traits, assume there is no heteroplasmy, i.e.,
individuals are homogeneous for the mitochondrial variation.
(h) Roman numerals (I, II, III etc.) are used to distinguish each generation (starting from the top)
and numbers (1, 2, 3, etc.) to depict each individual in a generation (starting from the left).
(i) For rare recessive traits assign as heterozygous carriers the minimum number compatible
with the pedigree, e.g., (1) assume individuals marrying into the pedigree are homozygous
for the wild type allele unless they have an affected child in which case they are an obligate
heterozygous carrier, and (2) if a first cousin marriage produces an affected child, then one of
the grandparents they share must have been a heterozygous carrier, however we don’t know
(unless there is additional information to indicate) whether it was the grandmother or
grandfather, so we note that one of them is homozygous wild type and the other a carrier.
6. Consider the following pedigrees of a rare autosomal recessive disease.
(a). Using the allele notation of T for unaffected and t for the disease, for the following
individuals in generations I and II, indicate the genotype if there is only one possibility, and the
genotypes and their respective probabilities if there is more than one possibility:
individual I-1:
individual I-2:
individual II-2 (A):
individual II-3:
individual I-3:
individual II-4 (B):
individual II-5:
(b). If individuals A and B have a child, what is the probability that the child will have the
disease?
(c). If individuals C and D have a child, what is the probability that the child will have the
disease?
Page
2
7. A couple without a family history of Tay-Sachs disease (a rare autosomal disease) have two
normal children and an infant affected with Tay-Sachs. The sister of the male parent wants to
mate with the brother of the wife. In such a mating, what would be the probability of their first
child having Tay-Sachs disease?
8. The following pedigree shows the inheritance of attached earlobes (black symbols) and
unattached earlobes (white symbol). Both alternative phenotypes are quite common in human
populations.
For the following possible modes of inheritance for attached earlobes, indicate which is (are)
compatible with the pedigree, and which are incompatible with the pedigree, and in this case
indicate one individual who is incompatible with that mode of inheritance and why they are
incompatible.
(a)
(b)
(c)
(d)
autosomal dominant
autosomal recessive
X-linked dominant
X-linked recessive
9. Huntington's disease is an autosomal dominant with a variable age of onset, but most cases
occur after the age of 35. A young woman whose great-grandfather had Huntington's disease
marries and is considering having a child. Both her father and grandfather were killed in wars
while in their twenties. If she consulted you, how would you advise her about the risks to her
child? (Make sure you give her probability values for your predictions.)
Page
3
10. The following pedigree depicts the inheritance of a rare hereditary disease affecting muscles:
For the following possible modes of inheritance, indicate which is (are) compatible with the
pedigree, and which are incompatible with the pedigree, and in these cases indicate one
individual who is incompatible with that mode of inheritance and why they are incompatible.
(a) autosomal dominant
(b) autosomal recessive
(c) X-linked dominant
(d) X-linked recessive
(e) Y-linked inheritance
(f) Maternal (mitochondrial inheritance)
11. A young woman is worried about having a child because her mother's only sister had a son
with Duchenne muscular dystrophy (DMD). The young woman has no brothers or sisters. (DMD
is a rare X-linked recessive disorder.)
(a) Draw the relevant parts of the pedigree of the family described above. (Be sure to include the
grandmother, the three women mentioned, and all their mates.)
(b) State the most likely genotype of everyone in the pedigree.
(c) Calculate the probability that the young woman's first child will be a boy with DMD.
12. Drosophila eyes are normally red. Several purple-eyed strains have been isolated as
spontaneous variants (mutants) and the purple phenotype has been shown to be inherited as a
Mendelian autosomal recessive in each case. Two purple-eyed pure strains were crossed.
(a). If the purple mutations in this case are in the same gene (that is they are allelic), the F1 is
expected to have what genotype and phenotype.
(b). If the purple mutations are in different genes (that is they complement each other), the F1 is
expected to have what genotype and phenotype (assign appropriate symbols for the alleles and
show all genotypes and phenotypes).
Page
4
13. In a hypothetical plant species, flower color is determined by a gene with 2 incompletely
dominant alleles, such that AA is red, AB is pink, and BB is white, while plant height is
determined by a gene on a separate chromosome which assorts independently from the flower
color gene, again with incomplete dominance, such that CC is short, CD is medium, and DD is
tall.
For a cross between pure breeding parents AA DD x BB CC, determine
(a). the phenotypes of the parents, and the genotype and phenotype of the F1 generation.
(b). The F1 generation is selfed, what are the genotypes and phenotypes, and their respective
frequencies, of the F2’s?
14. In chickens the dominant allele Cr produces the creeper phenotype (having extremely short
legs). However the creeper allele is lethal in the homozygous condition. The genotype rr pp
gives chickens a single comb, R- P- gives a walnut comb, rr P- gives a pea comb, and R- pp
gives a rose comb. If a creeper with a pea comb (genotype rr Pp), is mated with a creeper with a
rose comb (genotype Rr pp), what are the genotypes, phenotypes and their frequencies of the
living progeny? Assume the 3 genes (Cr, R, and P) segregate independently; they are all
autosomal genes.
15. In the purple penguin, a series of alleles occurs at the p locus on an autosome. All alleles
affect the color of feathers: pd = dark-purple, pm = medium-purple, pl = light-purple, and pvl=
very pale purple. The order of dominance is pd > pm > pl > pvl. This means that pd is dominant to
pm and pl and pvl, and pm is dominant to pl and pvl, and pl is dominant to pvl. If a light-purple
female, heterozygous for very pale purple, is crossed to a dark-purple male, heterozygous for
medium-purple, what is the ratio of genotypes and phenotypes expected among the baby
penguins?
16. Four babies were born in a hospital on the same night. ABO and MN blood typing have been
done on all the babies and parents. Assign the babies to the correct families. (Each family has
only one child.) (In the ABO blood group system, alleles A and B are codominant, and allele i is
recessive for the O blood group, while the MN blood group system is codominant.)
Mother
Father
Child no.
Child
Family no.
1
O, M
O, N
1
O, N
2
AB, MN
O, N
2
AB, M
3
A, M
B, MN
3
A, MN
4
O, N
O, N
4
O, MN
Page
5
17. Three genes, each represented by a normal (dominant) and a mutant (recessive) allele, are on
separate chromosomes and assort independently of one another. The F2 of a cross of the
homozygous (AA ; BB ; CC) normal strain with the triply mutant strain (aa ; bb ; cc) will
produce 3/4 normal to 1/4 mutant for each character. Using these figures, and without a
checkerboard (Punnett square), determine what fractions of the F2 will be of the phenotypes
indicated below using the branched line approach. Show the component probabilities.
(a) A- ; B- ; cc
(b) aa ; bb ; cc
(c) A- ; B- ; C(d) aa ; B- ; cc
18. In mice the allele for color expression is C (c = albino). Another gene determines color (B =
black and b = brown). Yet another gene modifies the amount of color so that D = normal amount
of color and d = dilute (milky) color. Two mice that are Cc ; Bb ; Dd are mated. What proportion
of progeny will be dilute brown (assume complete dominance at each locus and independent
assortment of the genes)?
19. (a). In the cross Mm ; Nn ; Oo x mm ; nn ; oo, what proportion of the offspring will have
the genotype Mm ; Nn ; Oo? (Capital letters are dominant to lower case.) The three traits assort
independently.
(b) In the cross Mm ; Nn ; Oo x Mm ; Nn ; Oo, what proportion the offspring will have the
genotype MM ; n ; Oo?
20. In Drosophila, vermillion eye color is due to a recessive allele (v) located on the X
chromosome (females are XX, and males are XY). Curved wings is due to a recessive allele (cu)
located on an autosome, and ebony body is due to a recessive allele (e) located on another
autosome. A vermillion male (wildtype for the other two traits) is mated to a curved, ebony
female (wildtype for eye color), and the F1 males are phenotypically wildtype. If these males
were backcrossed to curved, ebony females, what proportion of the offspring will be wildtype
males?
21. In pigs, when a pure-breeding red is crossed to a pure-breeding white the F1 are all red.
However, the F2 shows 9/16 red, 1/16 white, and 6/16 are a new color, sandy.
(a) Provide a genetic hypothesis for this observation
(b) How would you test this hypothesis?
Page
6
22. Color in onions is conditioned by two genes as follows:
R- = red, rr = yellow; C- = color, cc = no color can be expressed (clear)
(a) For the following cross write out all genotypes and phenotypes that will be produced and
their frequencies (the two genes are on separate chromosomes and assort independently):
Rr ; Cc (red) x rr ; cc (clear).
(b) What are the phenotypic ratios in the F2 generation starting with the parental cross
RR ; CC (red) x rr ; cc (clear).
23. In some people, the carbohydrate A and B antigens of the ABO blood groups not only are
found on the surface of the red blood cells, but also are secreted into the saliva and other body
fluids. Secretion of the antigens is due to a dominant allele of another genetic locus called
secretor. People of genotype Se Se or Se se secrete the A and B antigens, but those of genotype
se se do not. There is epistasis between the genetic systems, because people of blood group O are
nonsecretors regardless of their secretor genotype. The ABO and secretor genes are on different
chromosomes and therefore show independent assortment. In crosses of the following genotype,
what is the expected ratio of secretors to nonsecretors?
AO ; Se se x BO ; Se se
24. In a plant species, stalk height is determined by four independently assorting genes, A, B, C,
and D, each segregating two alleles; with each gene one allele, denoted by a 1, e.g., A1, adds one
centimeter to the basic stalk height of 10 centimeters (assume no environmental influences on
height). If all the alleles of these genes act additively to determine stalk height:
(a). What is the phenotype of a plant with the genotype A0A1 B0B1 C0C1 D0D1, and
(b). If this plant is selfed, what fraction of its offspring will be 10 centimeters tall?
Page
7
Answers to Practice questions for lectures 1-5
1.(a). Albino parent is aa, normally pigmented parent must be Aa, since an albino child was
produced. Therefore, expected ratio of albino to normal offspring is 1:1 (1/2 aa : ½ Aa).
(b). Parents must be both Aa. Expected offspring ¼ AA (normal), ½ Aa (normal carriers), ¼ aa
(albino), therefore, ratio of albino to normal expected is 1:3
2.(a). Dominant, since if the disease were recessive, two affected individuals could not have an
unaffected child.
(b). Dd and Dd.
(c). ¼ dd (normal), ½ Dd (dwarf), ¼ DD (dwarf, possibly lethal (although there is no way to tell
this from the information given)).
3. If tasting were recessive, Mom and Dad would have to be homozygous and then all their
children would taste too. Some children do not taste, tasting must be dominant, Mom and Dad
must be heterozygous, and nontasting is recessive.
y
y
+
4. Parental (P) generation: female X X (yellow) x male X Y (wildtype)
y +
y
F1 generation: females are all X X (wildtype), males are all X Y (yellow)
y
+
F2 generation: female gametes from F1 are ½ X and ½ X ,
y
male gametes from F1are ½ X and 1/2Y
y +
y y
F2 offspring, females are ½ X X (wildtype) and ½ X X (yellow)
+
y
and males are ½ X Y (wildtype) and ½ X Y (yellow)
5.(a). X-linked recessive trait: We are given that pure breeding blue eyed males mated to pure
breeding brown eyed females results in F1 individuals all with brown eyes, males and females. If
the trait is X-linked recessive, then blue is recessive and denoting the allele by b, this cross was
female (X+X+) (brown eyes) x male XbY (blue eyes) with F1 females (X+Xb) (brown eyes) and
F1 males (X+Y) (brown eyes). In the reverse cross the female is pure breeding blue eyed
(XbXb), and the male is pure breeding brown eyed (X+Y), so that the F1 females will all be
(XbX+) (brown eyes) and the males will all be (XbY) (blue eyes).
5.(b). Autosomal recessive trait: The original cross would be female (++) (brown eyes) x male
(bb) (blue eyes) with all F1 individuals +b (brown eyes). In the reverse cross the female is bb
(blue eyes) and the male is ++ (brown eyes) and all the F1 individuals will be as in the first
cross, i.e., all +b (brown eyes).
Page
8
6.(a).We know that individual II-3 is tt, thus individuals I-1 and I-2 must both be Tt and
individual II-2 (A) is TT with probability 1/3 and Tt with probability 2/3, while individuals III-14 (C) are all TT if A is TT, and TT or Tt each with probability ½ if A is Tt. (since the disease is
very rare, and there is no evidence to the contrary, I-4, II-1, and II-6 are all TT.) Individual I-3 is
tt, thus individuals II-4 (B) and II-5 must both be Tt, individuals III-5(D) and III-6 and -7 are
either TT or Tt each with probability ½. Answer is:
individual I-1: Tt
individual I-2: Tt
individual II-2 (A): TT (probability 1/3), Tt (probability 2/3)
individual II-3: tt
individual I-3: tt
individual II-4 (B): Tt
individual II-5: Tt
(b). Male B must be Tt, and female A has a 2/3 chance of being Tt (both must be heterozygous to
have an affected child, in which case ¼ on average of their children will be affected).
Answer is:
The overall chance of an affected child is 2/3 × 1/4 = 2/12 = 1/6.
(c). Again both parents, male C and female D, must be heterozygotes (Tt)
to have an affected child,
in which case on average ¼ of their children will be affected
If A is a heterozygote (Tt) (probability 2/3),
then C will be a heterozygote (Tt) with probability ½;
individual II-5 (the mother of D) must be Tt,
hence D has a ½ probability of being Tt.
Answer is:
The overall chance of an affected child is 2/3 x ½ x 1 x ½ x ¼ = 1/24.
7. The couple who had the affected child are each heterozygous so in both sets of their
phenotypically normal parents (the grandpararents of the affected child) there must have been
one heterozygote, i.e., they were Tt x TT (we assume since the disease is rare that only 1 was a
heterozygote carrier) and the brother and sister each have a 1/2 chance of being Tt (and ½
chance of being TT). So, in the proposed mating, there is a ½ x ½ chance of them both being
heterozygous carriers, in which case ¼ on average of their children would be affected, so overall
probability = (1/2)(1/2)(1/4) = 1/16.
8.(a). autosomal dominant, compatible.
(b). autosomal recessive, incompatible, II-2 would also have to have attached earlobes if this trait
is recessive, since both parents have the trait.
(c). X-linked dominant, incompatible, II-2 would have to have the trait since her father has the
trait.
(d). X-linked recessive, incompatible, II-2 would have to have the trait since both parent have the
trait.
Page
9
9. Apparently the father and grandfather did not show symptoms of Huntington's disease, but
since it is a late-onset disease, they probably would not have expressed it in their early twenties.
There is a 1/2 chance that the grandfather inherited the allele, a 1/2 chance that the father
received it from him, a further 1/2 chance that the woman received it, and then a further 1/2
chance that her future child would receive it. The overall chance that the woman's child will
develop the disease is 1/2 × 1/2 × 1/2 × 1/2 = 1/16
10.
(a) autosomal dominant, incompatible, either I-1 or I-2 would have to be affected since they
have two affected children, and II-2 would have to be affected since she has two affected
children.
(b) autosomal recessive, compatible, although II-1 marrying in would have to be a heterozygous
carrier. The fact that all affected individuals are male is slightly more in favor of an X-linked
recessive, but autosomal recessive cannot be excluded.
(c) X-linked dominant, incompatible, I-1 would have to be affected since she has two affected
sons, and all the daughters of II-6, i.e., III-4-7 would have to be affected.
(d) X-linked recessive, compatible.
(e) Y-linked inheritance, incompatible, I-2 would have to be affected, also II-3 and III-8-9, and
III-2-3 would not be affected.
(f) Maternal (mitochondrial inheritance), incompatible, individual I-1 would have to be affected
to have two affected sons, and also all of her children (II-2-6) would then have to be affected,
etc.
11.
(c) The grandmother must have been Dd. There is a 1/2 chance that the mother of the young
woman is Dd and, if so, a further 1/2 chance that the woman herself is Dd. If she is, 1/2 of her
sons will have DMD. Since the probability of a son is also 1/2, the overall probability is
1/2 × 1/2 × 1/2 × 1/2 = 1/16.
Page
10
12. (a). Denoting the two alleles as p1 and p2 in the same gene,
the parental cross is p1p1 x p2p2 (purple x purple)
so the F1 are all p1p2 and will show the recessive phenotype of all purple.
(b). Denoting the recessive allele for purple at the first gene by a1, and at the second gene by b2,
the parental cross is a1/a1 ; +/+ x +/+ ; b2/b2,
so the F1 are all a1/+ ; +/b2
and since they are now heterozygote at both the genes, they will show the dominant
phenotype of red.
13. (a). The parents are red and tall (AA DD) and white and short (BB CC)
(b). The F1 generation is all pink and medium (AB CD),
The F2 generation has 9 genotypes and phenotypes, with the frequencies listed in
parentheses:
AA CC (red, short, 1/16), AA CD (red, medium 2/16), AA DD (red, tall, 1/16)
AB CC (pink, short, 2/16), AB CD (pink, medium 4/16), AB DD (pink, tall, 2/16)
BB CC (white, short, 1/16), BB CD (white, medium 2/16), BB DD (white, tall, 1/16)
Check that total frequency adds to 1.
14. Consider the creeper and comb genotypes, phenotypes and frequencies separately, and then
do the branched line approach
Cr+ x Cr+ (creeper x creeper) yields ¼ CrCr (lethal), ½ Cr+, ¼ ++,
so of the living progeny 2/3 will be creeper Cr+ (= (1/2)/(3/4)) and 1/3 non-creeper (++).
rr Pp x Rr pp yields ¼ Rr Pp (walnut), ¼ rr Pp (pea), ¼ Rr pp (rose), ¼ rr pp (single)
Combining these two sets of outcomes gives the following expected, genotypes, phenotypes,
and frequencies:
Cr+ Rr Pp (creeper, walnut, 2/12), Cr+ rr Pp (creeper, pea, 2/12),
Cr+ Rr pp (creeper, rose, 2/12), Cr+ rr pp (creeper, single, 2/12)
++ Rr Pp (non-creeper, walnut, 1/12), ++ rr Pp (non-creeper, pea, 1/12),
++ Rr pp (non-creeper, rose, 1/12), ++ rr pp (non-creeper, single, 1/12)
Check sum of frequencies = 1
15. The cross is pl/pvl x pd/pm, so that the 4 equally likely genotypes and their phenotypes in
parentheses are pl/pd (dark purple), pl/pm (medium purple), pvl/pd (dark purple), and pvl/pm
(medium purple), so that the phenotypic ratio is 1: 1 dark to medium purple.
16.
Family no.
Child no.
1
4
2
3
3
2
4
1
This is the only way all 4 families and children can be correctly assigned.
Page
11
17.
(a) ¾ x ¾ x ¼ = 9/64
(b) ¼ x ¼ x ¼ = 1/64
(c) ¾ x ¾ x ¾ = 27/64
(d) ¼ x ¾ x ¼ = 3/64
18. We need to determine the probability of a C- ; bb ; dd offspring from this cross = ¾ x ¼ x ¼
= 3/64.
19. (a). 1/8. MmNnOo individual produces gametes MNO with probability ½ x ½ x ½.
(b). 1/32 (¼ x ¼ x ½).
20. P: ++ ; ++ ; XvY x cu cu ; ee ; X+X+
F1: males are cu + ; e + ; X+Y (wildtype)
These males are backcrossed to the females in the P generation, the proportion of the offspring
that will be +- ; +- ; X+Y is ½ x ½ x ½ = 1/8
21.
(a) The presence of a dominant allele at either of two loci (A- ; bb or aa ; B-) gives sandy, where
red is determined by the dominant alleles of both loci (A- ; B-), and white by the recessive
genotype at both loci (aa ; bb).
(b) Do testcrosses of a large number of the sandy offspring with the white offspring. Under this
hypothesis, the sandy offspring are AA ; bb (1/16) (gametes Ab) or Aa ; bb (2/16) (gametes
½ Ab and ½ ab) or aa ; BB (1/16) (gametes aB) or aa ; Bb (2/16) (gametes ½ aB and ½ ab),
i.e., three types of gametes with equal frequencies: Ab, aB, and ab, so the offspring from the
testcrosses with (aa ; bb) should be equal proportions of
Aa ; bb (sandy)
aa ; Bb (sandy)
aa ; bb (white),
i.e., 2/3 sandy and 1/3 white.
(One should also perform other crosses to confirm the model, including testcrosses and
crosses of the putative genotypes of specific sandy offspring.)
22. (a). There are 4 equally frequent gametic types from parent 1 (RC, Rc, rC, and rc, and only 1
gametic type (rc) from parent 2, giving the following genotypes and their frequencies
Gametes parent 1
RC
Rc
rC
rc
Parent 2
rc
RrCc
Rrcc
rrCc
rrcc
Frequencies:
1
:
1
:
1
:
1
phenotypes:
red
clear
yellow
clear,
thus the phenotypic ratios in the offspring are 1:2:1 red: clear: yellow.
(b). The phenotypic ratios in the F2 from this cross are 9:3:4 red: yellow: clear.
Page
12
23. The ABO segregation produces the following four genotypes in equal frequencies: AB, AO,
BO, and OO. The first three are potential secretors, depending on the secretor genotype, whereas
the genotype OO is a nonsecretor regardless of the genotype at the secretor locus. Segregation of
the secretor gene produces ¾ Se-: ¼ se se. The former are potential secretors depending on the
ABO genotype, whereas the se se is a nonsecretor regardless of the ABO genotype. Because
these segregations are independent, ¾ x ¾ = 9/16 are secretors, while the remainder are
nonsecretors (7/16).
24. (a). The phenotype of the quadruple heterozygote should be the basic height of (10cm) plus
the contribution of each of the alleles with a 1 instead of a zero (4 x 1cm), so total height is
14cm.
(b). Among the progeny of the selfed plant, only those that are homozygous for all the alleles
with a zero will manifest the basic phenotype of 10 cm. These quadruple zero-homozygotes will
have an expected frequency of 1/4x1/4x1/4x1/4 = 1/256.
Page
13