Download CHAPTER 14:MENDEL AND THE GENE IDEA

Answer Section
c. Homologous pairs line up independently
at
metaphase plate
d. Homologous pairs of chromosomes separate
and homologs move toward opposite poles, sister chromatids remain attached at centromere
e. Haploid set of chromosomes, each consisting
of two sister chromatids, aligns at metaphase
467
plate; sister chromatids
not identical due to
crossing over
f. Sister chromatids separate and move to opposite poles as chromosomes
3.
results in
both involve
cells with
haploid set of
chromosomes
results in
in anaphase
of mitosis
produc~g
then in
anaphase II
of
_....:.....
__
L..
-.:=::....----------
ProdUCing-----------.
••IL..._fo_u_r_
cells
ANSWERS TO TEST YOUR KNOWLEDGE
Multiple Choice:
1. b
2 d
3. e
4. c
5. e
6. b
7. a
10. b
11. a
12. b
8. c
9. b
13. c
14. d
15. e
16. b
17. d
18. b
19. e
20. b
CHAPTER 14: MENDEL AND THE GENE IDEA
INTERACTIVE
QUESTIONS
14.1 a. R
b. r
c. Fl Generation
d. Rr
e. F2 Generation
f. R
14.2 a. all tall (Tt) plants
g. r
h. Rr
i. Rr
j, rr
G
G
e
k. 3 round:l wrinkled
1. 1RR:2Rr:lrr
b. 1:1 tall (Tt) to dwarf (tt)
You can use a Punnett square to determine the
expected outcome of a test cross, but a shortcut is to use only 1 column for the recessive
individual's
gametes, since they produce only
one type of gamete.
14.3 a. all tall purple plants
b. TtPp
c. TP, Tp, tP, tp
468
Answer Section
sperm
d.
Phenotype
1/4@ l/ll'» ll/@ 1/l!iJ
1/4@
ITPP
ITPp
TtPP
TtPp
lIlt»
ITPp
ITpp
TtPp
Ttpp
l/lii!J
TtPP
TtPp
l/l!iJ
TtPp
Ttpp
Genotype
Ratio
Black
M_B_
%X%=%6
Brown
M_bb
% X 1/4 = 3/16
White
mm -
% X 1 = 14 Od'I6
<JJ
00
00
Q)
e. 9 tall purple:3 tall white:3 dwarf purple:1
dwarf white
£. 12:4 or 3:1 tall to dwarf;
12:4 or 3:1 purple to white
14.4 Dihybrids produce four types of gametes. The
probability of getting both recessive alleles in a
~amete is 7'4, and for two such gametes to join is
Y4 X 7'4, or %6'
14.5 a. Consider the outcome for each gene as a
monohybrid cross. The probability that a cross of
Aa X Aa will produce an A_ offspring is %. The
probability that a cross of Bb X bb will produce a
B_ offspring is 7'2' The probability that a cross of cc
X CC will produce a C_ offspring is 1. To have all
of these events occur simultaneously, multiply
their probabilities: % X 7'2 X 1 = %.
b. Offspring could be A_bbC--, aaB_C--, or
A_B_C. The genotype A_B_cc is not possible. Can
y.ou see why?
Probability
of A_bbC_
= % X 7'2
Probability
of aaB_C_
=
Probability
of A_B_C_
X
1
=%
% X % X 1 = 1/8
= % X 7'2 X 1 = %
Probability of offspring showing at least two
dominant traits is the sum of these independent
probabilities, or 7/8,
c. There is only one type of offspring that can
show only one dominant trait (aabbCc). Can you
see why? Its probability is 7'4 X 7'2 X 1 = lis.
14.6 a. A: [A[A and [Ai
b. B: [B[B and [Bi
c. AB: [A[B
d. 0: ii
14.7 The ratio of offspring from this MmBb X MmBb
cross would be 9:3:4, a common ratio when one
gene is epistatic to another. All epistatic ratios
are modified versions of 9:3:3:1.
-~
-
--~
-
--
14.8 a. The parental cross produced 25-cm tall -=-.
plants, all AaBbCc plants with 3 units of 5 c::::
added to the base height of 10 cm.
b. As a general rule in the polygenic inheri
of a quantitative
character, the number
phenotypic classes resulting from a cro
heterozygotes
equals the number of all
involved plus one. In this case, 6 all
(AaBbCc) + 1 = 7. So, there will be 7 dill
phenotypic
classes in the F2 among .
64 possible combinations of the 8 types of -=gametes. (See why you wouldn't want to "'through a Punnett square to figure that
These 7 classes will go from 6 dominant
(40 cm), 5 dominant (35 cm), 4 dominant (30
and so on, to all 6 recessive alleles (10 crn).
14.9 a. This trait is recessive. If it were domina
then albinism would be present in every 0
0en:~
tion, and it would be impossible to have
children with two nonalbino
(homozvgees
recessive) parents.
b. father Aa; mother Aa, because neither
is albino and they have albino offspring (aa
c. mate 1 AA (probably); mate 2 Aa; gran
4Aa
d. The genotype of son 3 could be AA or
his wife is AA, then he could be Aa (both his
ents are carriers) and the recessive allele
would be expressed in his offspring. Even ::
and his wife were both carriers (heterozyg .
there would be a 243/1024 (% X % X % X % X 1"
24% chance that all five children would
normally pigmented.
14.10 a.
b.
7'4
%.
Of offspring with a normal pheno . i:
be predicted to be heterozygotes
thus, carriers of the recessive allele.
14.11 Both sets of prospective
grandparents
ill
have been carriers. The prospective paren
not have the disorder so they are not horn
gous recessive. Thus, each has a % chance of
ing a heterozygote carrier. The probability
both parents are carriers is % X % = %;
chance that two heterozygotes
will have
recessive homozygous child is 7'4' The 0,
chance that a child will inherit the disease ..:
% would
Answer Section
% X 14= %. 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 14,
VGGESTED ANSWERS
OUR KNOWLEDGE
law of segregation occurs in
anaphase I, when alleles segregate as homologs move to opposite poles of the cell. The
two cells formed from this division have onehalf the number of chromosomes and one copy
of each gene (although sister chromatids still
have to separate in anaphase II). Mendel's law
of independent assortment relates to the lining
up of attached homologous chromosomes at
the equatorial plate in a random fashion during metaphase 1. Genes on different chromosomes will assort independently into gametes.
2. Aa = 2 different gametes
AaBb = 4 gametes
AaBbCc = 8 gametes
AABbCc = 4 gametes
A general formula is Z", where n is the number
of gene loci that are heterozygous.
3. Always look to the Fl heterozygote. If alleles
show complete dominance/recessiveness, then
the heterozygote will have a phenotype identical to one parent. In incomplete dominance, the
Fl phenotype will be intermediate between that
of the parents. If codominant, both alleles will
be expressed in the heterozygote.
ANSWERS TO GENETICS PROBLEMS
1. White alleles are dominant to yellow alleles. If
yellow were dominant, then you should be able
to get white squash from a cross of two yellow
heterozygotes.
b.
c.
d.
% [12(to get AA) X 12(bb)]
% [% (aa) X % (BB)]
% [1 (Aa)
X
12(Bb)
X
1 (Cc)]
%2 [% (aa) X % (bb) X % (cc)]
3. Since flower color shows incomplete dominance, use symbols such as CR and CW for those
alleles. There will be six phenotypic classes in
the F2instead of the normal four classes found
in a 9:3:3:1 ratio. You could find the answer
with a Punnett square, but multiplying the
probabilities of the monohybrid crosses is more
efficient.
Tall red
Tall pink
%X
%X
14 = 3/16
12 = % or
LCwCw
tt CRCR
tt CRCW
tt CwCw
14 = %6
% X % = %6
%X
14 X 12 = 18or 116
14 X 14 = 116
4. Determine
TO STRUCTURE
1. Mendel's
2. a.
Tall white
Dwarf red
Dwarf pink
Dwarf white
469
%6
the possible genotypes of the
mother and child. Then find the blood groups
for the father that could not have resulted in a
child with the indicated blood group.
a. no groups exonerated
d. AB only
b. A or 0
e. B or 0
c. AorO
5. The parents are CcBb and Ccbb. Right away you
know that all cc offspring will die and no BB
black offspring are possible because one parent
is bb. Only four phenotypic classes are possible.
Determine the proportion of each type by applying the law of multiplication.
Lethal (cc _)
Normal brown (CCBb)
Normal white (CCbb)
Deformed brown (CcBb)
Deformed white (Ccbb)
Ratio of viable offspring:
1/4 of embryos do
not develop
14 X 12 = %
14 X 12 = Ys
12 X 12 = 14
12 X 12 = 14
1:1:2:2
6. 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 % 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.
% B_S-,
B_ss, % bbS-, and % bbss. Because recessive
traits show up in the offspring, 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.
7. The genotypes of the puppies were
%
8. The 1:1 ratio of the first cross indicates a cross
between a heterozygote and a homozygote.
The second cross indicates that the hairless
hamsters cannot have a homozygous genotype,
because then only hairless hamsters would be
produced. Let us say that hairless hamsters are
Hh and normal-haired are HH. The second
cross of heterozygotes should yield a 1:2:1
genotype ratio. The 1:2 ratio indicates that the
homozygous recessive genotype may be lethal,
with embryos that are hh never developing.
470
Answer Section
9. a. Taillength appears to be a quantitative character. A general rule for the number of alleles in
a cross involving a quantitative character is one
less than the number of phenotypic classes.
Here there are five phenotypic classes, thus
four alleles or two gene pairs. In this cross, the
phenotypic ratio is approximately 1:4:6:4:1.The
ratio base of 16 also indicates a dihybrid cross.
Each dominant gene appears to add 5 em in tail
length. The lS-cm pigs are double recessives;
the 3S-cm pigs are double dominants.
b. A cross of a lS-cm and a 30-cm pig would be
equivalent to aabb X AABb. Offspring would
have either one or two dominant alleles and
would have tail lengths of 20 or 25 cm.
10. a. The black parent would be CCh. The
Himalayan parent could be either ChCh or Chc.
First list the alleles in order of dominance: C,
c. To approach this problem you should
write down as much of the genotype as you can
be sure about for each phenotype involved. The
parents were black (C_) and Himalayan (Ch J.
The offspring genotypes would be black (C_)
and Chinchilla (cch_). Right away you can tell
that the black parent must have been cc= because half of the offspring are Chinchilla; cch is
dominant over Ch and c and could not have been
present in the Himalayan parent. The ~enotype
of this parent could be either
or C c.A testcross with a white rabbit would be necessary to
determine this genotype.
b. You cannot definitely determine the genotype of the parents in the second cross. You can,
however, eliminate some genotypes. The black
parent could not be CCor cc=. because both the
C and Ch alleles are dominant to
and
Himalayan offspring were produced. The Chinchilla parent could not be cchCch for the same
reason. One or both parents had to have Ch as
their second allele;one, but not both, might have
c as their second allele.
crc:
cc:
c,
11. a. First figure out possible genotypes: __ E_ =
golden (any B combination with at least 1 E),
B .ee = black, bbee = brown. All you know at the
start is that both parents are __ E_. Since you
see black and brown offspring, you know that
both parents had to be heterozygous for Ee, and
at least one was heterozygous for Bb (to get
black and brown offspring). Since black
and brown are in a 1:1 ratio (like a testcross
ratio), then one dog was Bb and the other was
bb. You need to consider the results from the
second cross to know which dog was Bb.
b. For the second cross, you now know that
Dog 2 is ?bEe and Dog 3 is B?ee. A ratio of approximately 3:1 black to brown looks like a
cross of heterozygotes, so both parents must be
Bb. So Dog 3 is Bbee, and Dog 2 must be BbEe.
That means that Dog 1 must be bbEe.
12.
(At least one of these
grandparents must
have been Tt)
13. Since the parents were true-breeding for two
characters, the Fl would be dihybrids. Since all
F, had red, terminal flowers, those two traits
must be dominant and their genotypes could
be represented as RrTt. One would predict an
F2 phenotypic ratio of 9 red, terminal:3 red,
axial:3 white, terminal:1 white, axial. If 100 offspring were counted, one would expect
approximately 19(%6 X 100) plants with red,
axial flowers.
14. Because the F2generation is a ratio based on 16,
one can conclude that two genes are involved
in the inheritance of this flower color character.
The parent plants were probably double homozygotes (AABB for red and aabb for white).
The all-red Fl must be AaBb, and the red color
must require at least one dominant allele of
both genes. The 9:6:1 ratio of the F2 indicates
that 9 plants had at least one dominant allele of
both genes to produce the red color; 1 plant was
doubly homozygous recessive and thus white;
and the 6 pale purple were caused by having at
least one dominant allele of only one of the two
genes involved in flower color (A_bb or aaBj
ANSWERS TO TEST YOUR KNOWLEDGE
Matching:
1. I
2. A
3. F
4. E
5. L
6. J
7. H
8. D
9. C
10. K
Multiple Choice:
1.
2.
3.
4.
c
a
c
d*
5.
6.
7.
8.
b
d
c
d
9.
10.
11.
12.
b
c
d
a
13.
14.
15.
16.
d
d
c
b
* There are three different ways to get this outcome:
HHT, HTH, THH. Each outcome has a probability of ~
Answer Section
CHAPTER 15: THE CHROMOSOMAL
INTERACTIVE
possibly caused by a total nondisjunction
gamete formation.
b. The genetic balance caused by the trison
would be more disrupted
than that of <
organism with a complete extra set of chrom
somes.
QUESTIONS
15.1
I F2 Generation I
Xw+
ova
xw+ !Xw+xw
BASIS OF INHERITANCE
y
sperm
Xw+y
15.7 translocation,
Xw
xr x:
XWy
Phenotype
red-eyed female red-eyed female red-eyed male white-eyed male
Genotype
XW+Xw+
15.2 The gene is sex-linked, so a good notation is XN,
Xn, and Y so that you will remember that the
Y does not carry the gene. Capital N indicates
normal sight. Genotypes are:
1. XNy
2. XNXn
,3. XNXN or xNxn (probably
are X'Y)
4. XNXn
5. XNy
6. Xny
7. XNXn
15.3 a. tall, purple-flowered
flowered
b. tall, white-flowered
flowered
XNXN since four sons
and
and
dwarf,
dwarf,
whitepurple-
15.4 If linked genes have their loci close together on
the same chromosome,
they travel together
during meiosis and more parental offspring are
produced. Recombinants are the result of crossing over between nonsister chromatids
of homologous chromosomes.
15.5 Solving a linkage problem is often a matter of
trial and error. Sometimes it helps to layout the
loci with the greatest distance between them and
fit the other genes between or on either side by
adding or subtracting map unit distances.
--------15-------m---9-j---6-1-6---k
---------12--------15.6 a. An organism with a trisomy (2n + 1) has an
extra copy of one chromosome, usually caused
by a nondisjunction
during meiosis. A triploid
~rganism (3n) has an extra set of chromosomes,
deletion,
and inversion
15.8 Aneuploidies
of sex chromosomes
appear
upset genetic balance less, perhaps because
atively few genes are located on the Y chro
some and extra X chromosomes are inactiva
as Barr bodies.
1
SUGGESTED ANSWERS TO STRUCTURE
YOUR KNOWLEDGE
1. Genes that are not linked assort independend
and the ratio of offspring from a testcross,
.
dihybrid heterozygote should be 1:1:1:1 (Aa aabb gives AaBb, Aabb, aaBb, aabb offspring
recombination
frequency of 50%-50%
of
spring are recombinants). Genes that are . ,
and do not cross over should produce a 1:1
in this testcross (AaBb and aabb because a
erozygote derived from a parental cross of _'"1..X aabb produces
only AB and ab gametes
crossovers between distant genes almost ah
occur, the heterozygote
will produce
quantities of AB, Ab, aB, and ab gametes, and
genotype ratio of offspring will be 1:1:1:1~
same as it is for unlinked genes. Because ea
recombination frequency is equal to 1map
this measurement
ceases to be meaningful <I
relative distances of 50 or more map
..
However, crosses with intermediate
genes
the chromosome
could establish both tha
genes A and B are on the same chromosome Ch
that they are a certain map unit distance a~
2. A cross between a mutant female fly and a
mal male should produce all normal female
(who get a wild-type allele from their fa
and all mutant male flies (who get the mu
allele on the X chromosome from their mo
X Xm+y produces
Xm+Xmand X
spring (normal females and mutant males).
xrx:
3. The serious phenotypic
effects that are associ
ated with these chromosomal
alterations in .
cate that normal development
and functio ..
are dependent
on genetic balance. Most gene
472
Answer Section
appear to be vital to an organism's existence,
and extra copies of genes upset genetic balance.
Inversions and translocations, which do not
disrupt the balance of genes, can alter phenotype because of the effect of neighboring genes
on gene functioning.
2. c, a, b, d
3. The genes appear to be linked because the
parental types appear most frequently in the
offspring. Recombinant offspring represent 10
out of 40 total offspring for a recombination frequency of 25%, indicating that the genes are map units apart.
4. One of the mother's X chromosomes carries the
recessive lethal allele. One-half of male fetuses
would be expected to inherit that chromosome
and spontaneously abort. Assuming an equal
sex ratio at conception, the ratio of girl to boy
children would be 2:1,or six girls and three boy
ANSWERS TO GENETICS PROBLEMS
1. a. The trait is recessive and probably sex-
linked. Two sets of unaffected parents in the
second generation have offspring with the trait,
indicating that it must be recessive. More males
than females display the trait. Females #3 and
#5 must be carriers since their father has the
trait, and they each pass it on to a son.
b. Using the symbols XT for the dominant
allele and x' for the recessive:
5. a. For female chicks to be black, they mus
have received a recessive allele from the male
parent. If all female chicks are black, the male
parent must have been ZbZb. If the male parent
was homozygous recessive, then all male offspring will receive a recessive allele, and the female parent would have to be ZBW to produce
all barred males.
1. xty
2. XTXt
3. XTXt
4. XTy
b. For female chicks to be both black and
barred, the male parent must have been ZBZb. If
the female parent were ZBW, only barred male
chicks would be produced. To get an equal
number of black and barred male chicks, the female parent must have been Z~.
5. XTXt
6. XTXt or XTXT
7. XTXt
c. Individual #6 has a brother who has the trait
and her mother's father had the trait, so her
mother must be a carrier of the trait, meaning
there is a 12 probability that #6 is a carrier. Since
she is mated to a phenotypically normal male,
none of her daughters will show the trait (0probability). Her sons have a 12 chance of having the
trait if she is a carrier. There is a 12 X % = 14
probability that her sons will have the trait. There
is a % chance that a child would be male, so the
probability of an affected child is 12 X %, or 18.
ANSWERS TO TEST YOUR KNOWLEDGE
Multiple Choice:
1.
2.
3.
4.
e
a
b
d
5.· a
6. e
7. b
8. c
9.
10.
11.
12.
a·
e
d
b
13.
14.
15.
16.
c
a
e
d
17.
18.
19.
20.
e
d
c
d
CHAPTER 16: THE MOLECULAR BASIS OF INHERITANCE
INTERACTIVE
QUESTIONS
16.1 a. They grew T2 with E. coli with radioactive
sulfur to tag phage proteins.
b. T2 was grown with E. coli in the presence of
radioactive phosphorus to tag phage DNA.
c. Radioactivity was found in the supernatant,
indicating that the phage protein did' not enter
the bacterial cells.
d. In the samples with the labeled DNA, most
of the radioactivity was found in the bacterial
cell pellet.
e. They concluded that viral DNA is injected
into the bacterial cells and serves as the hereditary material for viruses.
16.2 a.
b.
c.
d.
e.
sugar-phosphate backbone
3' end of chain
hydrogen bonds
cytosine
guanine
f. adenine (purine)
g. thymine (pyrimidine)
h. 5' end of chain