Chapter 14 Study Guide Key
I.
Interactive Questions:
14.1
a.
b.
c.
d.
e.
f.
g.
h.
i.
j.
k.
l.
14.2
a. all tall (Tt) plants
b. 1:1 tall (Tt) to dwarf (tt)
14.3
a. tall purple plants
b. TtPp
c. TP, Tp, tP, tp
d.
sperm
14.4
14.5
R
r
F1 Generation
Rr
F2 Generation
R
r
Rr
Rr
rr
3 round: 1 wrinkled
1 RR: 2 Rr: 1 rr
TTPP
TTPp
TtPp
TtPp
TTPp
TTpp
TtPp
Ttpp
TtPP
TtPp
ttPP
ttPp
TtPp
Ttpp
ttPp
ttpp
e. 9 tall purple: 3 tall white: 3 dwarf purple: 1 dwarf white
f. 12:4 or 3:1 tall to dwarf; 12:4 or 3:1 purple to white
You could determine this in two ways. The probability of getting both recessive
Alleles in a gamete is ¼, and for two such gametes to join is ¼ x ¼, or 1/16. An
easier way to treat each gene involved as a monohybrid cross. The probability of
it obtaining a homozygous cross is ¼. The probability of the offspring being
homozygous recessive for both genes is ¼ x ¼, or 1/16.
a. Consider the outcome for each gene as a monohybrid cross. The probability v
that a cross of Aa x Aa will produce A_ offspring is ¾. The probability that a
cross of Bb x bb will produce a B_ offspring is ½. The probability that a cross of
a cc x CC will produce a C_ offspring is 1. To have all these events occur
simultaneously, multiply their probabilities: ¾ x ½ x 1= 3/8
b. Offspring could be A_bbC_, aaB_C_, or A_B_C_. The genotype A_B_cc is
not possible.
Probability of A_bbC_ = ¾ x ½ x 1 = 3/8
Probability of aaB_C_ = ¼ x ½ x 1 = 1/8
Probability of A_B_C = ¾ x ½ x 1 = 3/8
Probability of off spring showing at least two dominant traits is the total of these
Independent probabilities, or 7/8. The probability of one dominant trait is
(aabbCc) is ¼ x ½ x 1 = 1/8
14.6 a. A: IAIA and IAi
b. B: IBIB and IBi
c. AB: IAIB
d. O: ii
14.7 The ratio of offspring from this Mm Bb x Mm Bb cross would 9:3:4, a common
ratio when one gene is epistatic to the other.
Phenotype
Genotype
Ratio
Black
M_B_
¾ x ¾ = 9/16
Brown
M_bb
¾ x1/4 = 3/16
White
Mm__
¼ x 1 =¼ or 4/16
14.8
a. The parental cross is produced 25-cam tall F1 plants, al AaBbCc plants with 3
units of 5 cm added to the base height of 10 cm.
b. As general rule in polygenic inheritance of quantitative character, the numbe
of phenotypic classes resulting from a cross of heterozygous equals the number of
alleles involved plus one. In this case, 6 alleles (AaBbCc) + 1 = 7. So, there will
be 7 different phenotypic classes in the F2 among the 64 possible combinations of
the 8 types of F1 gametes. Thse 7 classses will go from 6 dominant alleles (49
cm), 5 dominant (35 cm), 40 dominant (30 cm), and so on, to all 6 recessive
alleles(10 cm).
14.9 a. This trait is recessive. If it where dominant, the albinism would be present in
every generation, and it would be impossible to have albino children with
nonalbino (homozygous recessive) parents.
b. father = Aa; mother = Aa, because neither parent is albino and they have
albino offspring (aa)
c. mate 1= AA (probably); mate 2 = Aa; grandson 4 = Aa
d. The genotype of son 2 could be AA or Aa. If his wife is AA, then he could be
Aa (both his parents are carriers) and the recessive allele never would be
expressed in his offspring. Even if he and his wife where both carriers
(heterozygotes), there would be a 243/1024(3/4 x ¾ x ¾ x ¾ x ¾) or 24% chance
that all five children would be normally pigmented.
14.10 a. ¼.
b. 2/3. There is probability that ¾ of the offspring will have normal phenotype. Of
these 2/3 would be predicted heterozygotes, and, thus, carriers of the recessive
allele.
14.11 Both sets of prospective grandparents must have been carriers. The parents do not
Have the disorder so they are not homozygous recessive. Thus each has a 2/3
Chance of being a heterozygote carrier. The probability that both parents are
Carriers is 2/3 x 2/3 = 4/9; the chance that two heterozygotes will have a recessive
homozygous child is ¼. The overall chance that a child will inherit the disease is
4/9 x 4/9 = 1/9. Should this couple have a baby that has the disease, this would
establish that they are both carriers, and the chance that a subsequent child would
have the disease is ¼.
SUGGESTED ANSWERS TO STRUCTURE YOUR KNOWLEDGE
1.
Mendel’s law of segregation occurs in anaphase I, when alleles segregate as
Homologous chromosomes move to opposite poles of the cell. The two cells
formed this division have one-half the number of chromosomes and one copy
of each gene. Mendel’s law of independent assortment relates to the lining up
of synapsed chromosomes at the equatorial plate in a random fashion during
metaphase I. Genes on different chromosomes will assort independently into
gametes.
2.
All alleles operate independently within a cell, coding for their gene products,
Usually enzymes, as specified by their particular sequence of DNA
nucleotides. If both alleles code for functional enzyme s or products, then the
alleles may become dominant with both traits expressed in the heterozygote,
as illustrated by MN or AB blood types. If the recessive allele codes for a
nonfunctional enzyme, then the resulting recessive trait may be obvious only
when homozygous (such as 0 blood types in which no carbohydrate molecules
are attached to the cell membrane). If one copy of an allele produces sufficient
product that the dominant allele is expressed, then we have complete
dominance where the heterozygote phenotype is indistinguishable from the
dominant homozygote (for example, smooth peas in which the enzyme
converts sugar to starch). With incomplete dominance the heterozygote is
distinguishable from both parental types. Genotypes translate into phenotypes
by way of biochemical pathways, which are controlled by enzymes coded for
by genes. Thus molecular processes produce physiological, and behavioral
results.
ANSWERS TO TEST YOUR KNOWLEDGE
Matching:
1. 1
3. F
5. L
7. H 9. C
2. 2
4. E
6. J
8. D 10. K
Multiple Choice:
1. c
5. b
9. b
13. d
2. a
6. d
10. c
14. d
3. e*
7. c
11. d
15. c
4. c
8. d
12. a
16. b
*There are three different ways to get this outcome: HHT, HTH, THH. Each outcome has
a probability of 1/8.
Genetics Problems
1. Summer squash are either white or yellow. To get white squash, at east one of the
parental plants must be white. The allele for which color is dominant?
White alleles are dominant to yellow alleles
2. For the following crosses, determine the probability of obtaining the indicated
genotype in an offspring
a. ¼
b. 1/8
c. ½
d. 1/32
3. True-breeding tall red-flowered plants are crossed with dwarf white-flowered plants.
The resulting F1 generation consists of all tall pink-flowered plants. Assuming that height
and flower color are each determined by a single gene locus on different chromosomes,
predict the results of an F1 cross of dihybrid plants. Choose appropriate symbols for the
alleles of the height and flower color genes. List the phenotypes and predicted ratios for
the F2 generation.
Flower colors of red & white – shows incomplete dominance (pink – F1 generation).
Symbols CR and CW will be used to represent the alleles
-F1 generations: all tall and pink
-Traits are on separate chromosomes (not linked) – assume mendelian genetic cross
-expect to see a 9:3:3:1 result from a dihybrid cross
Possible phenotypes:
Tall red -> T_CRCR = 3/16
Tall pink-> T_CRCW = 3/8
Tall white->T_CWCW = 3/16
Dwarf red->tt_CRCR =1/16
Dwarf pink->tt_CRCW = 1/8
Dwarf White-> tt_CWCW = 1/16
4. Blood typing has been used as evidence in paternity cases, when the blood type of the
mother an child may indicate that a man alleged to be the father could not possibly have
father the child. Or the following mother and child combinations, indicate which blood
groups of potential fathers would be exonerated.
a. no groups exonerated
b. A or O
c. A or O
d. AB only
E. B or O
5. In rabbits, the homozygous CC is normal, Cc results in rabbits with deformed legs, and
cc is lethal. For a gene for coat color, the genotype BB produces black, Bb brown, and bb
a white coat. Give the phenotypic ratio of offspring from a cross of a deformed leg,
brown rabbit with a deformed leg, white rabbit.
CcBb (Deformed, brown) x Ccbb (deformed, white) -> phenotypic ratio of offspring
-all cc offspring will die (lethal gene)
- no BB are possible
-only four phenotypes are possible
normal brown (CCBb) ¼ x ½ = 1/8
normal white (CCbb) ¼ x ½ = 1/8
deformed brown (CcBb) ½ x ½ = ¼ (or 2/8)
deformed white (CcBbb) ½ x ½ = ¼ (or 2/8)
Ratio is 1:1:2:2
6. Polydactyly (extra fingers and toes) is due to a dominant gene. A father is a polydactyl,
the mother has the normal phenotype, and they have had one normal child. What is the
genotype of the father? Of the mother? What is the probability that a second child will
have a normal number of digits?
ANSWER: Father’s genotype must be Pp since polydactyly is dominant and he has had
one normal child. Mother’s genotype is pp. The chance of the next child having normal
digits is ½ or 50% because the mother can only donate a p allele and there is a 50%
chance that the father will donate a p allele.
7. In dogs, black (B) is dominant to chestnut (b), and solid color (S) is dominant to
spotted (s). What are the genotypies of the parents in a mating that produced 3/8 black
solid, 38 black spotted, 1/8 chestnut solid, and 8 chestnut spotted puppies? (Hint: First
determine what genotypes the offspring must have before you deal with the fractions).
ANSWER: The genotypes of the puppies were 3/8 B_S_, 3/8 B_ss, 1/8 bbS_, and 1/8
bbss. Because recessive traits show up in offsring, both parents had to have had at least
one recessive allele for both genes. Black: chestnut occurs in a 6:2 or 3:1 ratio, indicating
a heterozygous cross. Solid:spotted occurs in a 4:4 or 1:1 ratio, indicating a cross
between a heterozygote and a homozygous recessive. Parental genotypes were
BbSs x Bbss
8. When hairless hamsters are mated normal-haired hamsters. About one-half are normal.
When hairless hamsters are crossed with each other, the ratio of normal-haired to hairless
is 1:2. How do you account for the results of the first cross? How would you explain the
unusual ratio obtained in the second cross?
a. hairless hamsters x normal haired - > 50% hairless 50% normal.
1:1 indicates a cross between a homozygous with heterozygous
b. hairless x hairless -> normal 1 hairless 2
*hairless hamsters cannot have a homozygous genotype because then only
hairless hamsters will be produced
if hairless are Hh or HH the second cross will produce a heterozygote Hh x Hh ->
1:2:1
But with a 2:1 ratio suggests that the homozygous recessive may be lethal with
embryos that are hh never develop.
9. Two pigs whose tails are exactly 25 cm in length are bred over 10 years and they
produce 96 piglets with the following tail lengths: 6 piglets at 5 cm, 25 at 20 cm, 37 at 25
cm, 23 at 30 cm, and 5 at 35 cm.
a. How any pairs of genes are regulating the tail length character? Hint: Count the numbr
of phenotypic classes, or determine he sum of the ratios of the classes. In a monohybrid
cross, the F2 ratios add up to 4 (3:1 or 1:2:1). In a dihybrid cross, the F2 ratios ad up to
16 (9:3:3:1 or some variation if the genes re epistatic or quantitative).
b. What offspring phenotypes would you expect from mating between a 15-cm and a 30cm pig?
Two pig tails are 25 cm bred – 10 years
96 piglets with following tail lengths:
6 @ 15 cm
25 @ 20 cm
37 @ 25 cm
23 @ 30 cm
5 @ 35 cm
-appears to be a quantitative character
-always take one mius the number of classes
-5 classes so 4 alleles are assumed (2 gene pairs)
-phenotypic ratio: 1:4:6:4:1
ratio base is 16 indicates a dihybrid cross/each dominant gene adds cm in tail
10. Fur color in rabbits is determined by a single gene locus for which there are four
alleles. Four phenotypes are possible: black, Chinchilla (gray color caused by hite hairs
with black tips), Himalayan (white with black patches on extremities), and white. The
black allele (C ) is dominant over other alleles, the Chinchilla allele (C^ch) is dominant
over Himalayan (C^h) and the white allele (c ) is recessive to all others.
a) A black rabbit is crossed with a Himalayan, and the F1 consists of a ratio of 2 black to
2 chinchilla. Can you determine the genotypes of the parents?
b) A second cross was done between a black rabbit and a Chinchilla. The F1 contained a
ratio of 2 black to 1 chinchilla to 1 himalayan. Can you determine the genotypes of the
parents of this cross?
Rabbit fur color – one gene, 4 alleles -> 4 phenotypes
Phenotypes:
Black = C
Chinchilla = C^ch
Himalayan = C^h
White = c
A. black x Himalayan
F1 2:2
***parents: black CC^h Himalayan ChChor Chc
b. black x chinchilla
F1 = 2:1
***Parents: can definitely determine but you can eliminate:
-black parents cannot be CC or CCch because both the C and Cch alleles are
dominant to Ch and Himalayan offspring were reduced
-the chinchilla cannot be CchCch also. One or both of the parents have to have Ch
as a 2nd allele.
11. In Labrador retriever dogs, the dominant gene B determines black coat color and bb
prouces brown. A separate gene E, however, shows dominant epistasis over the B and b
alleles, resulting in a golden coat color. The recessive e allows expression of B and b. A
breeder wants to know the genotypes of her three dogs, so she breeds them and makes
note of the offspring of several litters. Determine the genotypes of the three dogs.
a. golden female (dog 1) x golden male (dog 2) offspring: 7 golden, 1 black, 1 brown
b. black female (dog 3) x golden male (dog 2) offspring: 8 golden, 5 black, 2 brown
black = B
brown = bb
Owner wants to know the genotypes of her three dogs
Dog 1: bbEe
Dog 2: BbEe
Dog 3: Bbee
12. The ability to taste phenylthiocarbamide (PTC) is controlled in humans by a single
dominant allele (T). A woman nontaster married a man taster, and they had three
children, two boy tasters and a girl nontaster. All the grandparents were tasters. Create a
pedigree for this family for this trait. (Solid symbols should signify nontasters (tt)).
Where possible, indicate whether tasters are TT r Tt.
13. Two true-breeding varieties of garden peas are crossed. One parent had red, axial
flowers and the other had white, terminal flowers. All F1 individuals had red, terminal
flowers. If 100 F2 offspring were counted, how many of them would you expect to have
red, axial flowers?
All F1 = red, terinal flowers (dihybrids) RrTt x RrTt
F2 = 100 offspring; how many would you expect to have red axial flowers?
Expect to see a 9:3:3:1 ratio
3/16 of 100 = 19
19 are expected to be red and axial
14. You cross true-breeding red-flowered plants with true-breeding white-flowered
plants, and the F1 are all red-flowered plans. The F2, however, occur in ratio of 9 red: 6
pale purple : 1 white. How any genes are involved in the inheritance of this color
character? Explain Explain why the F1 are all red and how the 9:6:1 ratio of phenotypes
in the F occurred.
(AABB) True red x (aabb) true white
F1: all red flowered (AaBb) red color requires one dominant allele of both genes
F2: ratio 9 red, 6 pale purple, 1 white
2 genes are involved
A_bb or aaB_
Test Your Knowledge
1.
2.
3.
4.
5.
6.
7.
8.
Codominance (I) both alleles are fully expressed in heterzygote
Homozygote (A) true-breeding variety
Heterozygote (F) genotype with two different alleles
Phenotype (E) the physical characterstics of an individual
Polygenic inheritance (L) two or more genes with additive effect on phenotype
Pleiotropy (J) single gene with multiple phenotypic effects
Epistasis (H) one gene influences the expression of another gene
Testcross (D) cross with recessive homozygote to determine genotype of
unknown
9. Dihybrid cross (C ) cross between hybrids that are heterozygous for two genes
10. incomplete dominance (K) heterozygote intermediate between phenotypes of
homozygotes
Multiple Choice
1.According to Mendel’s Law of segregation,
C – allele pairs separate in gamete formation (p. 254)
2. After obtaining two heads from two tosses of a coin, the probability of tossing the coin
and obtaining a head is
a – ½ (p. 259)
3. The probability of tossing three coins simultaneously and obtaining two heads and one
tail is
e – 3/8 (p. 259)
4. A multifactorial disorder
c – has both genetic and environmental causes (p. 268)
5. The F2 generation
B – is the result of the self fertilization or crossing of F1 individuals (p. 257)
6. The bse height of the dingdong plant is 10 cm. Four genes contribute to the height of
the plant, and each dominant allele contributes 3 cm to height. If you cross a 10 cm plant
(quadruply homozygous recessive) with a 34-cm plant, how many phenotypic classes will
there be in F2?
D – 9 (p.257)
7. A 1:1 phenotypic ratio in a testcross indicates that
C – the dominant phenotype parent was a heterozygote (p.257)
8. Carriers of a genetic disorder
D – are heterozygotes for the gene that can cause the disorder (p.265)
9. If both parents are carriers of a lethal recessive gene, he probability that their child will
inherit and express the disorder is
B – ¼ (p.265)
10. Which phase of meiosis is most directly related to the law of independent assortment?
C – metaphase I (p. 258)
11. You think that two alleles for coat color in mice show incomplete dominance. What is
the best and simplest cross to perform in order to support your hypothesis?
D – a cross of two true-breeding mice of different olors to look for an
intermediate phenotype in the F1 (p. 263)
12.Which of the following human diseases is inherited as a simple recessive trait?
A – Tay Sachs Disease (p. 265)
13. In a dihybrid cross of heterozygotes, what proportion of the offspring will be
phenotypically dominant for both traits?
D – 9/16 (p. 265)
14. A other with type B blood has two children, one with type A blood and one with type
O blood. Her husband has type O blood. Which of the following could you conclude
from this information?
D – the husband could be he father of the child with ype O blood, but not the type
A blood (p. 262)
15.In guinea pigs, the brown coat color allele (B) is dominant over red (b), and the solid
color allele (S) is dominant over spotted (s). The F1 offspring of a cross between true
breeding brown, solid colored guinea pigs and red, spotted pigs are crossed. What
proportion of their offspring (F2) would be expected to be red and solid colored?
C – 3/16 (p. 259)
16. A dominant allele P causes the production of purple pigment; pp individuals are
white. A dominant allele C is also required for color production; cc individuals are white.
What proportion of offspring will be purple from a ppCc x PpCc cross?
B – 3/8 (p.259)