This document contains genetics concept questions and their answers related to topics like Mendelian inheritance, sex determination, extensions of Mendelian inheritance, and blood types. Some example questions include the probability of offspring genotypes from given parental crosses, determining heterozygotes vs homozygotes, inheritance patterns like sex-influenced traits, and solving genetics problems using rules of probability. The responses provide detailed explanations and use techniques like Punnett squares, product rule, and binomial/multinomial expansions to calculate the requested probabilities.

1. Genetics
Concept question
วราลี สินธุวา
ศูนย์ สอวน. ชีววิทยา โรงเรียนเตรียมอุดมศึกษาพัฒนาการ

2. MENDELIAN INHERITANCE
trickquestion

3. Pascal's triangle

4. 1. A heterozygous pea plant that is tall with
yellow seeds, TtYy, is allowed to self-fertilize.
What is the probability that an offspring
will be either tall with yellow seeds, tall with
green seeds, or dwarf with yellow seeds?

5. 1 Answer: This problem involves three mutually
exclusive events, and so we use the sum rule to
solve it.
tall with
yellow
seeds, tall
with green
seeds, or
dwarf with
yellow
seeds?
tall with
yellow seeds
:3x3 = 9
, tall with
green seeds:
3x1=3, or
dwarf with
yellow seeds
1x3=3
(9+3+3)/16

6. 2. For an individual expressing a dominant trait, how can
you tell if it is a heterozygote or a homozygote?

7. 2Answer: One way is to conduct a testcross with an
individual that expresses the recessive version of the same
character. If the individual is heterozygous, half of the
offspring will show the recessive trait, but
if the individual is homozygous, none of the offspring will
express the recessive trait.
Dd × dd or DD × dd
1Dd : 1 dd All Dd
(dominant trait) (recessive trait) (dominant trait)
Another way to determine heterozygosity involves a more
careful examination of the individual at the cellular or
molecular level.

8. 3. As described in this chapter, a human disease known as
cystic fibrosis is inherited as a recessive trait. Two unaffected
individuals have a first child with the disease. What is the
probability that their next two children will not have the
disease?

9. 3 Ans. Because the parents are unaffected with the
disease, we know that both of them must be
heterozygous carriers for the recessive disease-
causing allele.
The probability of a single unaffected offspring is
Punaffected = 3/(3 + 1) = 3/4
To obtain the probability of getting two unaffected
offspring in a row
(i.e., in a specified order), we must apply the product
rule.
3/4 × 3/4 = 9/16 = 0.56 = 56%
The chance that their next two children will be
unaffected is 56%.

10. 4. A pea plant is heterozygous for three genes (Tt Rr
Yy), where T = tall, t = dwarf, R = round seeds, r =
wrinkled seeds, Y = yellow seeds, and y = green
seeds. If this plant is self-fertilized, what are the
predicted phenotypes of the offspring, and what
fraction of the offspring will occur in each category?

11. 4 Ans.

12. 5. In dogs, black fur color is dominant to white. Two
heterozygous black dogs are mated. What would be the
probability of the following combinations of offspring?
A. A litter of six pups, four with black fur and two with
white fur.
B. A first litter of five pups, four with black fur and one with
white fur, and then a second litter of seven pups in which the
firstborn is homozygous, the second born is black, and the
remaining five pups are three black and two white

13. 5 Ans.
A. Because this is an unordered combination of events,
we use the
binomial expansion equation, where n = 6, x = 4, p = 0.75
(probability of black), and q = 0.25 (probability of white).
The answer is 0.297, or 29.7%, of the time.
B. The order of the litters is specified, so we need to use
the product rule to multiply the probability of the first litter
times the probability of the second litter. We use the
binomial

14. expansion equation to determine the probability of the first
litter. The probability of the second litter is a little more
complicated. The firstborn is homozygous. There are two
mutually exclusive ways to be homozygous, BB and bb. We
use the sum rule to determine the probability of the first pup,
which equals 0.25 + 0.25 = 0.5.
The probability of the second pup is 0.75, and we use the
binomial expansion equation to determine the probability of
the remaining pups.
(binomial expansion of first litter)([0.5][0.75][binomial
expansion of second litter])
For the first litter, n = 5, x = 4, p = 0.75, q = 0.25. For the last
five pups in the second litter, n = 5, x = 3, p = 0.75, q = 0.25.
The answer is 0.039, or 3.9%, of the time.

15. 6. A cross is made between two heterozygous tall
plants with axial flowers (TtAa), where tall is
dominant to dwarf and axial is dominant to terminal
flowers. What is the probability that a group
of five offspring will be composed of two tall plants
with axial flowers, one tall plant with terminal
flowers, one dwarf plant with axial flowers, and one
dwarf plant with terminal flowers?

16. the binomial expansion equation is used when only two
phenotypic outcomes are possible. When more than two
outcomes are possible, we use a multinomial expansion
equation to solve a problem involving an unordered number of
events. A general expression for this equation is
where P = the probability that the unordered number of events
will occur.
n = total number of events
( p is the likelihood of a, q is the likelihood of b, r is the
likelihood of c, and so on)
For example, this formula can be used to solve
problems involving an unordered sequence of events in a
dihybrid experiment.

17. 6 Ans.
Step 1. Calculate the individual probabilities of each
phenotype. The phenotypic ratios are 9 tall with axial flowers,
3 tall with terminal flowers, 3 dwarf with axial flowers, and 1
dwarf with terminal flowers.
The probability of a tall plant with axial flowers is
9/(9 + 3 + 3 + 1) = 9/16. p
The probability of a tall plant with terminal flowers is
3/(9 + 3 + 3 + 1) = 3/16. q
The probability of a dwarf plant with axial flowers is
3/(9 + 3 + 3 + 1) = 3/16. r
The probability of a dwarf plant with terminal flowers is
1/(9 + 3 + 3 + 1) = 1/16. s

18. Step 2. Determine the number of each type of event
versus the total number of events.
n = 5 (ต้น)
a = 2
b = 1
c = 1
d = 1

19. Step 3. Substitute the values in the multinomial
expansion equation.
P = 0.04 = 4%
This means that 4% of the time we would expect to
obtain five
offspring with the phenotypes described in the
question.

20. 7. A recessive allele in mice results in an abnormally long
neck. Sometimes, during early embryonic development, the
abnormal neck causes the embryo to die. An experimenter
began with a population of true-breeding normal mice and
true-breeding mice with long necks. Crosses were made
between these two populations to produce an F1 generation
of mice with normal necks. The F1 mice were then mated
to each other to obtain an F2 generation. For the mice that
were born alive, the following data were obtained:
522 mice with normal necks
62 mice with long necks
What percentage of homozygous mice (that would have
had long necks if they had survived) died during embryonic
development?

21. 7 Ans. We would expect a ratio of 3 normal : 1 long
neck. In other words, there should be 1/3 as many
long-necked mice as normal mice. If we multiply
522 times 1/3, the expected value is 174. However,
we observed only 62. Therefore, it appears that 174
– 62, or 112, mice died during early embryonic
development; 112 divided by 174 gives us the
percentage that died, which equals 0.644, or 64.4%.

22. SEX DETERMINATION AND SEX
CHROMOSOMES

23. 8. An unaffected woman (i.e., without disease
symptoms) who is
heterozygous for the X-linked allele causing
Duchenne muscular
dystrophy has children with a man with a normal
allele. What are
the probabilities of the following combinations of
offspring?
A. An unaffected son
B. An unaffected son or daughter
C. A family of three children, all of whom are affected

24. 8 Answer: The first thing we must do is construct a
Punnett square to
determine the outcome of the cross . D represents the
normal allele,
and d is the recessive allele causing Duchenne muscular
dystrophy. The
mother is heterozygous, and the father has the normal
allele

25. A. There are four possible children, one of whom is an
unaffected
son. Therefore, the probability of an unaffected son is
1/4.
B. Use the sum rule: 1/4 + 1/2 = 3/4.
C. You could use the product rule because there would
be three
offspring in a row with the disorder: (1/4)(1/4)(1/4) =
1/64 =
0.016 = 1.6%.

26. 9. Hemophilia is a X-linked recessive trait in
humans. If a heterozygous
woman has children with an unaffected man,
what are the
odds of the following combinations of children?
A. An affected son
B. Four unaffected offspring in a row
C. An unaffected daughter or son
D. Two out of five offspring that are affected

27. A.1/4
B.(3/4)(3/4)(3/4)(3/4) = 81/256
C.3/4
D.The probability of an affected offspring is 1/4
and the probability of an unaffected offspring is 3/4.
For this problem, you use the binomial expansion
equation where x = 2, n = 5, p = 1/4, and q = 3/4.
The answer is 0.26, or 26%, of the time.
9 Ans.

28. 10. Incontinentia pigmenti is a rare, X-linked dominant
disorder in humans characterized by swirls of pigment in
the skin. If an affected female, who had an unaffected
father, has children with an unaffected male, what would
be the predicted ratios of affected and unaffected sons
and daughters?

29. 10 Answer: 1 affected daughter : 1 unaffected
daughter : 1 affected son : 1 unaffected son

30. 11. A cross was made between female flies with
white eyes and miniature
wings (both X-linked recessive traits) to male flies
with red eyes and normal wings. On rare ccasions,
female offspring were produced with white eyes.
If we assume these females are due to errors in
meiosis, what would be the most likely
chromosomal composition of such flies? What
would be their wing shape?

31. 11 Answer: These rare female flies would be
XXY. Both X chromosomes would carry the
white allele and the miniature allele. These
female flies would have miniature wings
because they would have inherited both X
chromosomes from their mother.

32. EXTENSIONS OF MENDELIAN
INHERITANCE

33. 12. In Ayrshire cattle, the spotting pattern of the
animals can be either red and white or mahogany
and white. The mahogany and white pattern is
caused by the allele M. The red and white phenotype
is controlled by the allele m. When mahogany and
white animals are mated to red and white animals,
the following results are obtained:
Genotype Phenotype
Females Males
MM Mahogany and white Mahogany and white
Mm Red and white Mahogany and white
mm Red and white Red and white
Explain the pattern of inheritance.

34. 12 Answer: The inheritance pattern for this trait
is sex-influenced inheritance.
The M allele is dominant in males but recessive
in females,
whereas the m allele is dominant in females but
recessive in males.

35. 13. The following pedigree involves a single
gene causing an inherited disease. If you
assume that incomplete penetrance is
not occurring, indicate which modes of
inheritance are possible.
(Affected individuals are shown as filled
symbols.)

36. A. Recessive
B. Dominant
C. X-linked, recessive
D. Sex-influenced, dominant
in females
E. Sex-limited, recessive in
females

37. 13 Answer:
A. It could be recessive.
B. It could not be dominant (unless it was
incompletely penetrant),because affected offspring
have two unaffected parents.
C. It could not be X-linked recessive because
individual IV-2 does not have an affected father.
D. It could not be sex-influenced dominant in
females because individual II-3 (who would have to
be homozygous) has an unaffected mother (who
would have to be heterozygous and affected).
E. It could not be sex-limited because individual II-3
is an affected male and IV-2 is an affected female.

38. 14. Pattern baldness is an example of a sex-
influenced trait that is dominant in males and
recessive in females. A couple, neither of
whom is bald, produced a bald son. What are the
genotypes of the
parents?

39. 14 Answer: Because the father is not bald, we
know he must be homozygous,
bb. Otherwise, he would be bald.
A female who is not bald can be either Bb or
bb. Because she has produced a bald son, we
know that she must be Bb in order to pass the
B allele to her son.

40. 15. Two pink-flowered four-o’clocks were
crossed to each other. What
are the following probabilities for the offspring?
A. A plant will be red-flowered.
B. The first three plants examined will be white.
C. A plant will be either white or pink.
D. A group of six plants contain one pink, two
whites, and three reds.

41. 15 Answer: The first thing we need to do is construct a
Punnett square to determine the individual probabilities for
each type of offspring. Because flower color is
incompletely dominant, the cross is Rr and Rr.

42. The phenotypic ratio is 1 red to 2 pink to 1 white.
In other words, 1/4 are expected to be red, 1/2 pink,
and 1/4 white.
A. The probability of a red-flowered plant is 1/4,
which equals 25%.
B. Use the product rule.
1/4 x 1/4 x1/4 = 1/64 = 1.6%
C. Use the sum rule because these are mutually
exclusive events. A given plant cannot be both white
and pink.
1/4 + 1/2 = 3/4 = 75%

43. D. Use the multinomial expansion equation. In this case,
three phenotypes are possible.
where
n = total number of offspring = 6
a = number of reds = 3
p = probability of reds = (1/4)
b = number of pinks = 1
q = probability of pink = (1/2)
c = number of whites = 2
r = probability of whites = (1/4)

44. If we substitute these values into the equation,
This means that 2.9% of the time we would
expect to obtain six plants,
three with red flowers, one with pink flowers,
and two with white flowers.

45. 16. A type A woman is the daughter of a type
O father and type A
mother. If she has children with a type AB
man, what are the
following probabilities?
A. A type AB child
B. A type O child
C. The first three children with type AB
blood
D. A family containing two children with
type B blood and one
child with type AB

46. 16. Answer:
A. 1/4
B. 0
C. (1/4)(1/4)(1/4) = 1/64
D. Use the binomial expansion equation:
( )!
!( )!
3, 1/ 4, 1/ 4, 2
3/64 0.047, or 4.7%



   
 
x n xn
P p q
x n x
n p q x
P

47. 17. Propose the most likely mode of inheritance
(autosomal dominant,
autosomal recessive, or X-linked recessive) for the
following pedigree. Affected individuals are shown
with filled (black) symbols.

48. 17 Answer:
X-linked recessive
(unaffected mothers transmit the trait to
sons)

49. 18. In a species of plant, two genes control flower color.
The red allele
(R) is dominant to the white allele (r); the color-producing
allele
(C) is dominant to the non-color-producing allele (c). You
suspect that either an rr homozygote or a cc homozygote
will produce white flowers. In other words, rr is epistatic
to C, and cc is epistatic to R. To test your hypothesis, you
allowed heterozygous plants (RrCc) to self-fertilize and
counted the offspring. You obtained the following data: 201
plants with red flowers and 144 with white flowers. Conduct
a chi-square analysis to see if your observed data are
consistent with your hypothesis.

50. 18 Answer: In this cross, we expect a 9:7 ratio between
red and white flowers. In other words, 9/16 will be
red and 7/16 will be white. Because
there are a total of 345 plants, the expected
values are
9/16  345 = 194 red
7/16  345 = 151 white
2
2
2 2
2
2
( )
(201 194) (144 151)
194 151
0.58
O E
E





 
 


With 1 degree of freedom, our chi square value is
too small to reject our hypothesis.
Therefore, we accept that it may be correct.

51. GENETIC LINKAGE AND
MAPPING IN EUKARYOTES

52. 19. In the garden pea, orange pods (orp) are recessive
to green pods (Orp), and sensitivity to pea mosaic
virus (mo) is recessive to resistance to the virus (Mo).
A plant with orange pods and sensitivity to the virus
was crossed to a true-breeding plant with green pods
and resistance to the virus. The F1 plants were then
testcrossed to plants with orange pods and sensitivity
to the virus. The following results were obtained:
160 orange pods, virus-sensitive
165 green pods, virus-resistant
36 orange pods, virus-resistant
39 green pods, virus-sensitive
400 total

53. A. Conduct a chi square analysis to see if these
genes are linked.
B. If they are linked, calculate the map distance
between the two
genes.

54. 19 Answer:
A. Chi square analysis.
1. Our hypothesis is that the genes are not linked.
2. Calculate the predicted number of offspring based
on the hypothesis. The testcross is:

55. The predicted outcome of this cross under our
hypothesis is
a 1:1:1:1 ratio of plants with the four possible
phenotypes. In other
words, 1/4 should have the phenotype orange pods,
virus-sensitive; 1/4
should have green pods, virus-resistant; 1/4 should
have orange pods,
virus-resistant; and 1/4 should have green pods, virus-
sensitive. Because
a total of 400 offspring were produced, our hypothesis
predicts 100
offspring in each category.

56. 3. Calculate the chi square.

57. 4. Interpret the chi square value. The calculated chi
square value is quite large. This indicates that the
deviation between observed and expected values is very
high. For 1 degree of freedom in Table 3.2 , such a large
deviation is expected to occur by chance alone less than
1% of the time. Therefore, we reject the hypothesis that
the genes assort independently. As an alternative, we
may infer that the two genes are linked.

58. B. Calculate the map distance.
Map distance = (Number of recombinant offspring)
= 18.8 mu
The genes are approximately 18.8 mu apart.

59. 20. Two recessive disorders in mice—droopy ears
and flaky tail—are caused by genes that are located
6 mu apart on chromosome 3. A true-breeding
mouse with normal ears (De) and a flaky tail (ft)
was crossed to a true-breeding mouse with droopy
ears (de) and a normal tail (Ft). The F1 offspring
were then crossed to mice with droopy ears and
flaky tails. If this testcross produced 100 offspring,
what is the expected outcome?

60. 20 Answer: The testcross is:
The nonrecombinant offspring are
Dede ftft Normal ears, flaky tail
dede Ftft Droopy ears, normal tail
The recombinant offspring are
dede ftft Droopy ears, flaky tail
Dede Ftft Normal ears, normal tail

61. Because the two genes are located 6 mu apart on
the same
chromosome, 6% of the offspring will be
recombinants. Therefore, the
expected outcome for 100 offspring is
3 droopy ears, flaky tail
3 normal ears, normal tail
47 normal ears, flaky tail
47 droopy ears, normal tail

62. 21. The following X-linked recessive traits are
found in fruit flies: vermilion eyes are recessive
to red eyes, miniature wings are recessive to long
wings, and sable body is recessive to gray body.
A cross was made between wild-type males with
red eyes, long wings, and gray bodies to females
with vermilion eyes, miniature wings, and sable
bodies. The heterozygous females from this cross,
which had red eyes, long wings, and gray bodies,
were then crossed to males with vermilion eyes,
miniature wings, and sable
bodies.

63. The following outcome was obtained:
Males and Females
1320 vermilion eyes, miniature wings, sable body
1346 red eyes, long wings, gray body
102 vermilion eyes, miniature wings, gray body
90 red eyes, long wings, sable body
42 vermilion eyes, long wings, gray body
48 red eyes, miniature wings, sable body
2 vermilion eyes, long wings, sable body
1 red eyes, miniature wings, gray body
Calculate the map distance between the three
genes.

64. 21 Answer:
The first step is to determine the order of the three genes.
We can do this by evaluating the pattern of inheritance in
the double crossovers. The double-crossovers occur with
the lowest frequency. Thus, the double crossovers are
vermilion eyes, long wings, and sable body,and red eyes,
miniature wings, and gray body. Compared with the
nonrecombinant pattern of alleles (vermilion eyes,
miniature wings, sable body and red eyes, long wings, gray
body), the gene for wing length has been reassorted. Two
flies have long wings associated with vermilion eyes and
sable body, and one fly has miniature wings
associated with red eyes and gray body. Taken together,
these results indicate that the wing length gene is found in
between the eye color and body color genes

65. Eye color—wing length—body color
We now calculate the distance between eye color and
wing length, and between wing length and body color.
To do this, we consider the data according to gene pairs:
vermilion eyes, miniature wings = 1320 + 102 = 1422
red eyes, long wings = 1346 + 90 = 1436
vermilion eyes, long wings = 42 + 2 = 44
red eyes, miniature wings = 48 + 1 = 49

66. The recombinants are vermilion eyes, long wings and red
eyes, miniature wings. The map distance between these two
genes is
(44 + 49)/(1422 + 1,436 + 44 + 49) × 100 = 3.2 mu
Likewise, the other gene pair is wing length and body color.
miniature wings, sable body = 1320 + 48 = 1368
long wings, gray body = 1346 + 42 = 1388
miniature wings, gray body = 102 + 1 = 103
long wings, sable body = 90 + 2 = 92
The recombinants are miniature wings, gray body and long
wings, sable
body.

67. The map distance between these two genes is
(103 + 92)/(1368 + 1388 + 103 + 92) × 100 = 6.6
mu
With these data, we can produce the following
genetic map:

69. Pascal's triangle

70. 22. Which of the following can children only
inherit from their mother?
A mutation:
A. on the X chromosome.
B. on the Y chromosome.
C. in the mitochondrial genome.
D. in a maternally imprinted gene.
E. in the hypervariable region of an antibody gene.
http://www.ibo2013.org/ibo2013/exams/

71. 22. Which of the following can children only
inherit from their mother?
A mutation:
A. on the X chromosome.
B. on the Y chromosome.
C. in the mitochondrial genome.
D. in a maternally imprinted gene.
E. in the hypervariable region of an antibody gene.
http://www.ibo2013.org/ibo2013/exams/

72. 23. The direction of shell coiling in the snail
Limnaea peregra is either dextral or sinistral.
Coiling direction is determined by a pair of
autosomal alleles. The allele for dextral (S+ )
is dominant over the allele for sinistral (s).
Experimental results of two reciprocal
monohybrid crosses are shown below.

74. What is the genetic phenomenon that explains
the inheritance pattern for coiling direction?
A. Cytoplasmic inheritance.
B. Epistasis.
C. Genetic imprinting.
D. Maternal effect.
E. Sex-limited inheritance.

75. What is the genetic phenomenon that explains
the inheritance pattern for coiling direction?
A. Cytoplasmic inheritance.
B. Epistasis.
C. Genetic imprinting.
D. Maternal effect.
E. Sex-limited inheritance.

76. 24. Some fruit flies (Drosophila
melanogaster) have a mutation that makes
them shake. These fruit
flies are called ―shakers‖.
An experimental cross is shown below:

78. What kind of inheritance best explains the
inheritance pattern for the shaker gene?
A. Somatic dominant.
B. Somatic recessive.
C. X-linked dominant.
D. X-linked recessive.
E. Y-linked dominant.

79. What kind of inheritance best explains the
inheritance pattern for the shaker gene?
A. Somatic dominant.
B. Somatic recessive.
C. X-linked dominant.
D. X-linked recessive.
E. Y-linked dominant.

80. http://highered.mheducation.com/sites/0073525332/st
udent_view0/index.html

81. 25. (2 points) The fruit fly Drosophila melanogaster
has a XX(female)-XY(male) system of sex
determination. The Y chromosome determines
maleness in humans, but not in Drosophila. Instead,
sex determination in Drosophila depends on the ratio
of the number of X chromosomes to the number
of autosomal haploid sets in an individual fly.
The table below describes five mutants whose
sex-chromosome complements and haploid sets of
autosomes differ from the normal condition.
Indicate with a checkmark (√) the sex phenotype of
all the mutant flies.

83. 25 Ans