Download Chapter 5 Answer Keys

Hyde Chapter 5 – Solutions
23
5
MODIFICATIONS TO
MENDELIAN PATTERNS
OF INHERITANCE
CHAPTER SUMMARY QUESTIONS
2.
In codominance, the heterozygote expresses the phenotypes of both homozygotes. In
incomplete dominance, the heterozygote has a phenotype that is intermediate
between those of the two homozygotes.
4.
Penetrance is the proportion of individuals of a particular genotype that shows the
appropriate phenotype; expressivity is the degree to which a trait is expressed.
6.
The disease is recessive at the level of the individual but incompletely dominant at
the enzymatic level.
8.
The IA and IB alleles produce glycosyl transferase enzymes that each add a different
substance to the H structure, whereas the i allele produces a functionless protein that
adds nothing. Thus, for example, the IAi heterozygote converts all the H structures to
the A antigen making IA dominant to i. IAIB heterozygotes convert the H structures to
approximately equal numbers of A and B antigens.
10. The complementation test, or cis–trans test, is used to determine if different
recessive mutations are allelic. Mutations that are allelic (that is, in the same gene)
will produce a mutant phenotype when they are in the trans configuration. Mutations
that are non-allelic (that is, in different genes) will produce a wild-type phenotype
when they are in the trans configuration.
EXERCISES AND PROBLEMS
12. We see three phenotypes in an approximate 1:2:1 ratio (3:7:4). One of the
phenotypes (short) is intermediate between long ears and no ears. Therefore, we have
incomplete dominance. The cross is between two heterozygotes: L1L2 L1L2, where
L1 = long ears and L2 = no ears.
24
Hyde Chapter 5—Solutions
14. Crosses can be IAIA IBIB, IAIA IBi, IAi IBIB, or IAi IBi.
IAIA IBIB
IAIA IBi
IAIB
All type AB
1/2 IAIB:1/2 IAi
1/2 AB:1/2 A
IAi IBIB
IAi IBi
1/2 IAIB:1/2 IBi
1/2 AB:1/2 B
1/4 IAIB:1/4 IAi:1/4 IBi :1/4 ii
1/4 AB:1/4 A:1/4 B:1/4 O
16. Constraints are set by the length of the gene (number of mutable sites) and the
different possible phenotypic effects that can result from alteration of the amino acid
sequence of the protein product of the gene.
18. The cross is
Cc T1T2
(purple, medium)
Cc T1T2
(purple, medium)
This problem can be worked out using the branched-line approach. To calculate the
probability of each class produced, we can independently calculate the probability of
each trait and apply the product rule.
1/4 CC
2/4 Cc
1/4 T1T1
1/16 CC T1T1
purple, tall
2/4 T1T2
2/16 CC T1T2
purple, medium
1/4 T2T2
1/16 CC T2T2
purple, dwarf
1/4 T1T1
2/16 Cc T1T1
purple, tall
2/4 T1T2
4/16 Cc T1T2
purple, medium
1/4 T2T2
2/16 Cc T2T2
purple, dwarf
Hyde Chapter 5 – Solutions
1/4 cc
25
1/4 T1T1
1/16 cc T1T1
white, tall
2/4 T1T2
2/16 cc T1T2
white, medium
1/4 T2T2
1/16 cc T2T2
white, dwarf
Therefore, the phenotypic ratio is
3/16 C– T1T1
purple, tall
6/16 C– T1T2
purple, medium
3/16 C– T2T2
purple, dwarf
white, tall
1/16 cc T1T1
2/16 cc T1T2
white, medium
1/16 cc T2T2
white, dwarf
20. a.
If gene A is on an autosome, there will be 10 different genotypes and five
different phenotypes.
Genotypes
A1A1; A1A2; A1A3; A1A4
A2A2; A2A3; A2A4
A3A3
A3A4
A4A4
Phenotypes
A1
A2
A3
A3A4
A4
b. If gene A is on the X chromosome, the alleles can now be represented as XA1, XA2,
XA3, and XA4. There will be more genotypes and phenotypes, because there is a
different sex chromosome composition between males and females.
Genotypes
XA1XA1; XA1XA2; XA1XA3; XA1XA4
XA2XA2; XA2XA3; XA2XA4
XA3XA3
XA3XA4
XA4XA4
XA1Y
XA2Y
XA3Y
XA4Y
Phenotypes
A1 females
A2 females
A3 females
A3A4 females
A4 females
A1 males
A2 males
A3 males
A4 males
26
Hyde Chapter 5—Solutions
22. The 119:32:9 ratio is very close to a 12:3:1 ratio indicating two loci with dominant
epistasis. For example:
P1
red (AABB) white (aabb)
F1
red (AaBb) self
F2
119 (A–B– + A–bb)
32 (aaB–)
9 (aabb)
CRCW SLSS CRCW SLSS
(purple, ovoid)
(purple, ovoid)
This problem can be worked out using the branched-line approach. To calculate the
probability of each class produced, we can independently calculate the probability of
each trait and apply the product rule.
24. The cross is
1/4 CRCR
2/4 CRCW
1/4 CWCW
1/4 SLSL
1/16 CRCR SLSL
red, long
2/4 SLSS
2/16 CRCR SLSS
red, ovoid
1/4 SSSS
1/16 CRCR SSSS
red, spherical
1/4 SLSL
2/16 CRCW SLSL
purple, long
2/4 SLSS
4/16 CRCW SLSS
purple, ovoid
1/4 SSSS
2/16 CRCW SSSS
purple, spherical
1/4 SLSL
1/16 CWCW SLSL
white, long
2/4 SLSS
2/16 CWCW SLSS
white, ovoid
1/4 SSSS
1/16 CWCW SSSS
white, spherical
26. The F1 indicates lazy is dominant; therefore, a worker must be a recessive
homozygote. If one gene is involved, the cross of the F1 female worker male
(Ww ww) should produce 1 worker:1 lazy, for this is a testcross. This result is not
seen, so we must have more than one gene involved. Perhaps a worker can result
Hyde Chapter 5 – Solutions
27
from more than one gene. Let A–B– = lazy, and A–bb, aaB–, or aabb =
workers. The original worker is aabb, and the cross is
aabb
AABB
AaBb
(lazy)
aabb
(worker)
1/4 A–B–:1/4 A–bb:1/4 aaB–:1/4 aabb
lazy
worker worker worker
The observed results (3 workers:1 lazy) are consistent with this explanation.
If AaBb
AaBb
9/16 A–B–
lazy
3/16 A–bb
3/16 aaB–
1/16 aabb
workers
Thus, the cross of two F1 lazy individuals will produce 9 lazy:7 hard workers.
28. In each cross, we see that all ratios could result from a single gene (3:1, 1:2:1, and
1:1). We can easily explain the red and silver if we assume that red is dominant over
silver. Thus, the first cross could be RR rr or Rr rr. Similarly, the second cross
could be RR R– or Rr Rr. But we cannot explain any of the white progeny with
only two alleles. Since all ratios are single gene ratios, we must propose a third allele
that, when homozygous, produces white. Red must be dominant to white. Note that a
given diploid individual can have only two alleles. To list all the possible genotypes
and their phenotypes, let R = red and r = nonred. RR, RrS, Rrw = red (R is dominant
to all other alleles), rSrS = silver, and rwrw = white.
What is the phenotype of an rSrw individual? Assume silver is dominant; that is,
assume rS is dominant to rw. Thus, rSrw = silver.
Now go back to the results. By our model, rwrw is white. Therefore, in order to
get white progeny, each parent must have at least one rw allele to contribute. The
cross of red by silver to produce white offspring must be
Rrw
rSrw
1/4 RrS:1/4 Rrw :1/4 rSrw :1/4 rwrw
(red) (red) (silver) (white)
30. The cross is
CBCY Ii
(green)
CBCY Ii
(green)
28
Hyde Chapter 5—Solutions
3/4 I–
3/16 CBCB I–
blue
1/4 ii
1/16 CBCB ii
colorless
3/4 I–
6/16 CBCY I–
green
1/4 ii
2/16 CBCY ii
colorless
3/4 I–
3/16 CYCY I–
yellow
1/4 ii
1/16 CYCY ii
colorless
1/4 CBCB
2/4 CBCY
1/4 CYCY
Thus, the phenotypic ratio is 6 green:3 blue:3 yellow:4 colorless (white).
32. The first two crosses indicate that wild type is dominant to both oranges, and the
fourth indicates that orange-2 is dominant to pink. The fifth cross produces four
phenotypes, indicating we are dealing with at least two genes. The presence of two
genes is also suggested by the cross of orange-1 orange-2. If these two traits were
allelic, all the progeny should have been orange. The F1 pink produces progeny
that resemble those from a testcross of a dihybrid. If A–B = wild type, A–bb =
orange-1, aaB– = orange-2, aabb is probably pink. The crosses in question are then
AAbb aaBB
(orange-1)
(orange-2)
AaBb
(wild type)
self
9/16 A–B–
3/16 A–bb
3/16 aaB–
1/16 aabb
wild type
orange-1
orange-2
pink
Thus the F2 ratio is 9 wild type:3 orange-1:3 orange-2:1 pink.
34. Crosses (a) and (b) do not reveal anything about the dominance hierarchy. Each
suggests a mating of a heterozygote with a homozygote. Cross (c) yields a higher
proportion of mallard than dusky offspring, indicating that mallard is dominant to
dusky. Cross (d) yields a higher proportion of restricted than mallard offspring,
indicating that restricted is dominant to mallard. Therefore, the dominance hierarchy
Hyde Chapter 5 – Solutions
29
is MR > M > MD. The crosses are (a) MDMD MMD; (b) MDMD MRMD; (c) MMD
D
D
R D
d
MM ; and (d) MM M M . Note: The dusky allele is actually designated m .
36. This problem can be worked out using the branched-line approach.
1/4 BB
3/4 A–
3/4 C–
9/64 A–BBC–
1/4 cc
3/64 A–BBcc
3/4 C–
18/64 A-BbC–
1/4 cc
6/64 A-Bbcc
3/4 C–
9/64 A–bbC–
1/4 cc
3/64 A–bbcc
3/4 C–
3/64 aaBBC–
1/4 cc
1/64 aaBBcc
3/4 C–
6/64 aaBbC–
1/4 cc
2/64 aaBbcc
3/4 C–
3/64 aabbC–
1/4 cc
1/64 aabbcc
2/4 Bb
1/4 bb
1/4 BB
1/4 aa
2/4 Bb
1/4 bb
Because allele B is recessive for lethality, all individuals with genotype BB will not
survive. So 16/64 (9/64 + 3/64 + 3/64 + 1/64) individuals die before birth. This
leaves 48 surviving offspring. Therefore, the expected phenotypic ratio in the
surviving offspring is 18 A–BbC–:6 A–Bbcc:9 A–bbC–:3 A–bbcc:6 aaBbC–:
2 aaBbcc:3 aabbC–:1 aabbcc.
38. a.
The genotypic frequencies can be obtained from a Punnett square.
Gametes AB
AB
AABB
aB
AaBB
Ab
AABb
AaBb
aB
AaBB
aaBB
ab
AaBb
aaBb
Therefore, 1/8 AABB:1/8 AABb:2/8 AaBB:2/8 AaBb:1/8 aaBB:1/8 aaBb.
The phenotypic frequencies can be derived from the genotypic frequencies.
Therefore, 6/8 horned:2/8 hornless or a 3:1 ratio.
30
Hyde Chapter 5—Solutions
b. Two hornless animals can produce offspring with horns. The animals will have to
be recessive for different genes. For example, if the cross is AAbb aaBB, the
offspring will all be AaBb and will all have horns.
40. Because the two genes are independently assorting, we can consider each trait
separately. First, joy: the offspring are in an approximate ratio of 1 laughing:1
smiling (41:38). Therefore, one of the parents must be laughing (J1J1) and the other
smiling (J1J2). Now, movement: the offspring are in an approximate ratio of
1 runner:2 joggers:1 walker (19:39:21). Thus, both parents must be joggers (M1M2).
Combining the two traits, we can determine that the one parent is a laughing jogger
and the other is a smiling one!
42. a.
To be colorless, individuals must be aabb; the alleles at locus C do not matter.
They could be either aabbC– or aabbcc. Therefore the probability is given by
(1/4 aa)(1/4 bb)(1/1 C– or cc) = 1/16.
b. For red probabilities, determine genotypes that will be red, calculate probability
for each genotype, and add.
A–bbC–
3/4 1/4 3/4 = 9/64
A–bbcc
3/4 1/4 1/4 = 3/64
A–B–cc
3/4 3/4 1/4 = 9/64
aaB–C–
1/4 3/4 3/4 = 9/64
aaB–cc
1/4 3/4 1/4 = 3/64
33/64
Alternatively, calculate the frequency of blacks as 3/4 A– 3/4 B– 3/4 C– =
27/64. We calculated 1/16 or 4/64 colorless, so 1 – (27/64 + 4/64) = 33/64 =
frequency of red.
44. Using the dominance hierarchy, we can determine the potential genotypes for every
phenotype. Thus, a gray rabbit can have one of four genotypes (CC, Ccch, Cch, and Cc); a
chinchilla rabbit, three (cchcch, cchch, and cchc); a Himalayan rabbit, two (chch and chc); and
albino rabbits only one (cc). This order of dominance allows recessive alleles to be
“hidden” in various heterozygotes.
Crosses (a) and (b) suggest that all three gray rabbits are heterozygous. Cross (c),
gray-1 chinchilla-1, resulted in three different genotypes, including Himalayan
rabbits. Both parents must thus be heterozygous, and at least one of them must carry the
Himalayan allele, ch. Therefore, gray-1 = Cch or Cc and chinchilla-1 = cchch or cchc.
(Note: Gray-1 and chinchilla-1 cannot both carry the albino allele at the same time.) We
can now go back and analyze cross (a): gray-1 gray-2. It can be represented as Cch (or
Cc) C–. The offspring include chinchillas, and so the gray-2 rabbit must be
heterozygous for the chinchilla allele, cch. Therefore, gray-2 = Ccch. Let’s move on to
cross (b): gray-2 gray-3. It can be represented as Ccch (or Cc) C–. The offspring
include chinchilla, and so the gray-3 rabbit cannot be homozygous (CC). However, we
cannot tell for sure what the other allele is. Indeed, it can be any of the other three
alleles: cch, ch, or c. And finally, cross (d): chinchilla-2 Himalayan, or cch– ch–. The
Hyde Chapter 5 – Solutions
31
offspring include albinos, and so both rabbits must be heterozygous for albino allele, c.
Therefore, chinchilla-2 = cchc, and Himalayan = chc. The following table summarizes the
results.
Rabbit
Gray-1
Gray-2
Gray-3
Chinchilla-1
Chinchilla-2
Himalayan
Genotype
Cch or Cc (only if chinchilla-1 is cchch)
Ccch
Ccch, Cch, or Cc
cchch or cchc (only if gray-1 is Cch)
cchc
chc
CHAPTER INTEGRATION PROBLEM
a.
The true breeding gray parent is AABBCC. Each of the albino parents must be
homozygous recessive (cc) at the C locus. Because the A, B, and C genes are
independently assorting, each mating can be treated as a series of monohybrid or
dihybrid crosses.
Cross 1: The F2 offspring are in a ratio of 3 gray:1 albino. The presence of no other
color indicates that the F1 rats were homozygous for the A and B genes and have
genotype AABBCc. Otherwise, other colored types would have appeared in the
offspring. Therefore, the albino-1 parent must have been AABBcc.
Cross 2: Without considering the C locus, the F2 offspring are in a ratio of 9 gray:
3 yellow, which can be simplified to a 3:1 ratio. Since yellow offspring have the
genotype A–bb, the preceding ratio indicates that the F1 rats were heterozygous for the
B gene and homozygous for the A gene (genotype would be AABbCc). Therefore, the
albino-2 parent must have been AAbbcc.
Cross 3: Without considering the C locus, the F2 offspring are in a ratio of 9 gray:
3 black, which can be simplified to a 3:1 ratio. Since black offspring have the
genotype aaB–, the preceding ratio indicates that the F1 rats were heterozygous for the
A gene and homozygous for the B gene (genotype would be AaBBCc). Therefore, the
albino-2 parent must have been aaBBcc.
b. Cross 1: The ratio of colored (gray, black, yellow, and cream) to albino rats is
242:83 or approximately 3:1. This indicates a mating between two heterozygotes for
the C locus. The colors are in an approximately 9:3:3:1 ratio, indicating that the
parents must also be heterozygous for the A and B genes.
Cross 2: The ratio of albino to colored (black and cream) rats is 103:99 or
approximately 1:1. This indicates a mating between a heterozygote, Cc, and a
homozygote, cc. The absence of gray and yellow offspring means that the parents
cannot have any A alleles. The approximately 3 black:1 cream ratio means that the
parents were heterozygous at the B locus.
32
Hyde Chapter 5—Solutions
Cross 3: There are no albino rats, meaning that the parents cannot produce the cc
genotype. The colors are in an approximately 9:3:3:1 ratio, indicating that the parents
must be heterozygous for the A and B genes.
Summary:
Parental Phenotypes
Gray gray
Black albino
Gray gray
Cross 1
Cross 2
Cross 3
c.
Parental Genotypes
AaBbCc AaBbCc
aaBbCc aaBbcc
AaBbCC AaBbCC, or
AaBbCC AaBbCc
The parental genotypes can be denoted as A–B–C– (gray) A–bbC– (yellow). The
offspring include an albino (cc), and a cream-colored cat (aabb); therefore, both
parents must be heterozygous for the A and C genes, and the gray parent should also
be heterozygous for the B locus. The cross can be diagrammed as such:
AaBbCc
AabbCc
AbC
Abc
abC
abc
ABC
AABbCC
(gray)
AABbCc
(gray)
AaBbCC
(gray)
AaBbCc
(gray)
ABc
AABbCc
(gray)
AABbcc
(albino)
AaBbCc
(gray)
AaBbcc
(albino)
AbC
AAbbCC
(yellow)
AAbbCc
(yellow)
AabbCC
(yellow)
AabbCc
(yellow)
Abc
AAbbCc
(yellow)
AAbbcc
(albino)
AabbCc
(yellow)
Aabbcc
(albino)
aBC
AaBbCC
(gray)
AaBbCc
(gray)
aaBbCC
(black)
aaBbCc
(black)
aBc
AaBbCc
(gray)
AaBbcc
(albino)
aaBbCc
(black)
aaBbcc
(albino)
abC
AabbCC
(yellow)
AabbCc
(yellow)
aabbCC
(cream)
aabbCc
(cream)
abc
AabbCc
(yellow)
Aabbcc
(albino)
aabbCc
(cream)
Aabbcc
(albino)
The phenotypic ratio is 9/32 gray:9/32 yellow:8/32 albino:3/32 black:3/32 cream.
Hyde Chapter 5 – Solutions
33
d. The polka-dot trait is dominant in males and recessive in females. This is
characteristic of a sex-influenced trait. These traits are controlled by autosomal
genes, so the dotted gene locus would be expected to be found on an autosome.
Therefore, it will assort independently (Mendel’s second law) from X-linked genes,
which have loci on the X chromosome.
e.
Secondary spermatocytes are “future” sperm. Because secondary spermatocytes are
haploid, they would have only one allele per gene. The male has a genotype of
AabbCcDd. However, one has to consider the X and Y chromosomes as well.
Therefore, the genotype can be designated as AabbCcDdXY. The total number of
secondary spermatocytes is 24 = 16 (there are four “heterozygous” positions). The
genotypes can be listed systematically as such: AbCDX, AbCDY, AbCdX, AbCdY,
AbcDX, AbcDY, AbcdX, AbcdY, abCDX, abCDY, abCdX, abCdY, abcDX, abcDY,
abcdX, and abcdY.
f.
Here again, the cross can be treated as a series of monohybrid crosses.
(1) An albino rat has to be cc. The probability is simply 1/4.
(2) The genotype of a yellow-colored rat is A–bbC–. The genotype of a polka-dotted
male rat is D–XY. Therefore, the overall probability is 1/2 1/2 3/4 3/4 1/2 =
9/128.
(3) The genotype of a gray-colored rat is A–B–C–. The dot phenotype depends on the
genotype at the D locus and the sex of the offspring. The genotype of a male rat with
no dots is ddXY, whereas that of a female can be DdXX or ddXX. We can calculate
each of the three possibilities, and then add them all (sum rule) to get to the overall
probability. P(gray male with no dots) = 1/2 1/2 3/4 1/4 1/2 = 3/128.
P(gray female with no dots) = 1/2 1/2 3/4 3/4 1/2 = 9/128. Therefore, the
P(gray rat with no dots) = 3/128 + 9/128 = 12/128 = 3/32.
(4) The dot phenotype depends on the genotype at the D locus and the sex of the
offspring. The genotypes at the A, B, and C loci are irrelevant. A male rat is polkadotted if he has the genotype D–XY, whereas a polka-dotted female has to be
DDXX. We can calculate each of these two probabilities, and then add them up (sum
rule) to get to the overall probability. P(polka-dotted rat) = (3/4 1/2) + (1/4 1/2)
= 4/8 = 1/2.