BIO 184 Laboratory
CSU, Sacramento
Spring 2006
ANSWERS TO SUPPLEMENTAL PROBLEMS
CHAPTER 1
E1.A genetic cross is a mating between two different individuals.
CHAPTER 2
C1. Mendel’s work showed that genetic determinants are inherited in a dominant/recessive manner. This
was readily apparent in many of his crosses. For example, when he crossed two true-breeding plants
for a trait such as height (i.e., tall versus dwarf), all the F1 plants were tall. This is inconsistent with
blending. Perhaps more striking was the result obtained in the F 2 generation: 3/4 of the offspring were
tall and 1/4 were short. In other words, the F2 generation displayed phenotypes that were like the
parental generation. There did not appear to be a blending to create an intermediate phenotype.
Instead, the genetic determinants did not seem to change from one generation to the next.
C2. In the case of plants, cross-fertilization occurs when the pollen and eggs come from different plants
while in self-fertilization they come from the same plant.
C3. The genotype is the type of genes that an individual inherits while the phenotype is the individual’s
observable traits. Tall pea plants, red hair in humans, and vestigial wings in fruit flies are phenotypes.
Homozygous, TT, in pea plants; a heterozygous carrier of the cystic fibrosis allele; and homozygotes for
the cystic fibrosis allele are descriptions of genotypes. It is possible to have different genotypes and the
same phenotype. For example, a pea plant that is TT or Tt would both have a tall phenotype.
C4. A homozygote that has two copies of the same allele.
C5. Conduct a cross in which the unknown individual is mated to an individual that carries only recessive
alleles for the genes in question.
C6. Diploid organisms contain two copies of each type of gene. When they make gametes, only one copy of
each gene is found in a gamete. Two alleles cannot stay together within the same gamete.
C12. Offspring with a nonparental phenotype are consistent with the idea of independent assortment. If
two different traits were always transmitted together as unit, it would not be possible to get nonparental
phenotypic combinations. For example, if a true-breeding parent had two dominant traits and was
crossed to a true-breeding parent having the two recessive traits, the F 2 generation could not have
offspring with one recessive and one dominant phenotype. However, because independent assortment
can occur, it is possible for F2 offspring to have one dominant and one recessive trait.
C13. A. It behaves like a recessive trait because unaffected parents sometimes produce affected
offspring. In other words, we think that the unaffected parents are heterozygous carriers.
B. It behaves like a dominant trait. An affected offspring always has an affected parent. However,
recessive inheritance cannot be ruled out.
C14. A. Barring a new mutation during gamete formation, the chance is 100% because they must be
heterozygotes in order to produce a child with a recessive disorder.
B. Construct a Punnett square. There is a 50% chance of heterozygous children.
C. Use the product rule. The chance of being phenotypically normal is 0.75 (i.e., 75%), so the answer is
D. Use the binomial expansion equation where n = 3, x = 2, p = 0.75, q = 0.25. The answer is 0.422, or
42.2%.
C15. A. 100% because they are genetically identical.
B. Construct a Punnett square. We know the parents are heterozygotes because they produced a blueeyed child. The fraternal twin is not genetically identical, but it has the same parents as its twin. The
answer is 25%.
C. The probability that an offspring inherits the allele is 50% and the probability that this offspring will
pass it on to his/her offspring is also 50%. We use the product rule: (0.5)(0.5) = 0.25, or 25%.
BIO 184 Laboratory
CSU, Sacramento
Spring 2006
D. Barring a new mutation during gamete formation, the chance is 100% because they must be
heterozygotes in order to produce a child with blue eyes.
C20. A. 1/4
B. 1, or 100%
C. (3/4)(3/4)(3/4) = 27/64 = 0.42, or 42%
D. Use the binomial expansion equation where
n = 7, p = 3/4, q = 1/4, x = 3
P = 0.058, or 5.8%
E. The probability that the first plant is tall is 3/4. To calculate the probability that among the next four,
any two will be tall, we use the binomial expansion equation, where n = 4, p = 3/4, q = 1/4, and x = 2.
The probability P equals 0.21.
To calculate the overall probability of these two events:
(3/4)(0.21) = 0.16, or 16%
C21. A. T Y R, T y R, T Y r, T y r
B. T Y r, t Y r
C. T Y R, T Y r, T y R, T y r, t Y R, t Y r, t y R, t y r
D. t Y r, t y
C24. Based on this pedigree, it is likely to be dominant inheritance because an affected child always has
an affected parent. In fact, it is a dominant disorder.
C25. A. 3/16
B. (9/16)(9/16)(9/16) = 729/4096 = 0.18
C. (9/16)(9/16)(3/16)(1/16)(1/16) = 243/1,048,576 = 0.00023, or 0.023%
D. Another way of looking at this is that the probability that it will have round, yellow seeds is 9/16.
Therefore, the probability that it will not is 1 – 9/16 = 7/16.
C29. The drone is sB and the queen is SsBb. According to the laws of segregation and independent
assortment, the male can make only sB gametes, while the queen can make SB, Sb, sB, and sb, in
equal proportions. Therefore, male offspring will be SB, Sb, sB, and sb, and female offspring will be
SsBB, SsBb, ssBB, and ssBb. The phenotypic ratios, assuming an equal number of males and females,
will be:
Males
1 normal wings/black eyes
1 normal wings/white eyes
1 short wings/black eyes
1 short wings/white eyes
Females
2 normal wings, black eyes
2 short wings, black eyes
E4.
According to Mendel’s law of segregation, the genotypic ratio should be 1 homozygote dominant : 2
heterozygotes : 1 homozygote recessive. This data table considers only the plants with a dominant
phenotype. The genotypic ratio should be 1 homozygote dominant : 2 heterozygotes. The homozygote
dominants would be true-breeding while the heterozygotes would not be true-breeding. This 1:2 ratio is very
close to what Mendel observed.
E8. If we construct a Punnett square according to Mendel’s laws, we expect a 9:3:3:1 ratio. Since a total of
556 offspring were observed, the expected number of offspring are
556
BIO 184 Laboratory
CSU, Sacramento
Spring 2006
If we plug the observed and expected values into the chi square equation, we get a value of 0.51. There
are
four categories, so our degrees of freedom equal n – 1, or 3. If we look up our value in the chi square
table
(see Table 2.1), it is well within the range of expected error if the hypothesis is correct. Therefore, we
accept
the hypothesis. In other words, the results are consistent with the law of independent assortment.
E10. A. If we let c+ represent normal wings and c represent curved wings, and e+ represents gray body
and e represents ebony body:
Parental Cross: cce+e+  c+c+ee.
F1 generation is heterozygous c+ce+e.
An F1 offspring crossed to flies with curved wings and ebony bodies is
c+ce+e  ccee.
The F2 offspring would be a 1:1:1:1 ratio of flies.
c+ce+e : c+cee : cce+e : ccee
B. The phenotypic ratio of the F2 flies would be a 1:1:1:1 ratio of flies.
normal wings, gray body : normal wings, ebony bodies : curved wings, gray bodies : curved wings,
ebony bodies
C. From part B, we expect 1/4 of each category. There are a total of 444 offspring. The expected
number of each category is
1/4  444, which equals 111.
(114  111) 2 (105  111) 2 (111  111) 2 (114  111) 2
2 



111
111
111
111
2
  0.49
With 3 degrees of freedom, a value of 0.49 or greater is likely to occur between 95 and 80% of the
time. So we accept our hypothesis.
CHAPTER 3
C2. The term homologue refers to the members of a chromosome pair. Homologues are usually the same
size and carry the same types and order of genes. They may differ in that the genes they carry may be
different alleles.
C3. Sister chromatids are identical copies derived from the replication of a parental chromosome. They
remain attached to each other at the centromere. They are genetically identical, barring rare mutations
and crossing over with homologous chromosomes.
C5. In G1, there should be six linear chromosomes. In G2, there should be 12 chromatids that are attached
to each other in pairs of sister chromatids.
C8. The reduction occurs because there is a single DNA replication event but two cell divisions. Because of
the nature of separation during anaphase I, each cell receives one copy of each type of chromosome.
C15. Bacteria do not need to sort their chromosomes because they only have one type of chromosome.
The attachment of the two copies of the chromosomes to the cell membrane prior to cell division
ensures that each daughter cell receives one copy.
C19. The products of meiosis have only one copy of each type of chromosome. For example, one human
gamete may contain the paternally derived copy of chromosome 11, whereas a different gamete may
contain the maternally derived copy of chromosome 11. These two homologues may carry different
alleles of the same genes and therefore are not identical. In contrast, mitosis produces genetically
identical daughter cells that have both copies of all the pairs of homologous chromosomes.
C29. A. X-linked recessive (unaffected mothers transmit the trait to sons)
BIO 184 Laboratory
CSU, Sacramento
Spring 2006
B. Autosomal recessive (affected daughters and sons are produced from unaffected parents)
C31. Duchenne muscular dystrophy is inherited as an X-linked recessive disorder. Affected individuals are
males. Their mothers are related to males who have the disorder. The mothers of affected males are
heterozygous carriers. Note: This pedigree is also consistent with autosomal recessive inheritance,
although the pattern is more suggestive of X-linked inheritance.
C32. First set up the following Punnett Square:
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 where x = 2, n = 5, p = 1/4, and q = 3/4. The answer is
0.26, or 26%, of the time.
E4.
The basic strategy is to set up a pair of reciprocal crosses. The phenotype of sons is usually the
easiest way to discern the two patterns. If it is Y linked, the trait will only be passed from father to son. If it is
X linked, the trait will be passed from mother to son.
E14.
In X-linked recessive inheritance, it is much more common for males to be affected. In autosomal
recessive inheritance, there is an equal chance of males and females being affected (unless there is a sex
influence as described in Chapter 4). For X-linked dominant inheritance, affected males would produce
100% affected daughters and not transmit the trait to their sons. This would not be true for autosomal
dominant traits, where there is an equal chance of males and females being affected.
CHAPTER 4
C1. Dominance occurs when one allele completely exerts its phenotypic effects over another allele.
Incomplete dominance is a situation in which two alleles in the heterozygote have an intermediate
phenotype. Codominance is when both alleles exert their effects independently in the heterozygote. And
overdominance is a case in which the heterozygote has a phenotype that is superior to either
homozygote.
C4. If the normal allele is dominant, it tells you that one copy of the gene produces a saturating amount of
the protein encoded by the gene. Having twice as much of this protein, as in the normal homozygote,
does not alter the phenotype. If the allele is incompletely dominant, this means that one copy of the
normal allele is not saturating.
C6. There would be a ratio of 1 normal : 2 star-eyed individuals.
C9. If individual 1 is ii, individual 2 could be IAi, IAIA, IBi, IBIB, or IAIB.
If individual 1 is IAi or IAIA, individual 2 could be IBi, IBIB, or IAIB.
If individual 1 is IBi or IBIB, individual 2 could be IAi, IAIA, or IAIB.
Assuming individual 1 is the parent of individual 2:
If individual 1 is ii, individual 2 could be IAi or IBi.
If individual 1 is IAi, individual 2 could be IBi or IAIB.
If individual 1 is IAIA, individual 2 could be IAIB.
If individual 1 is IBi, individual 2 could be IAi or IAIB.
If individual 1 is IBIB, individual 2 could be IAIB.
BIO 184 Laboratory
CSU, Sacramento
Spring 2006
C10. Types O and AB provide an unambiguous genotype. Type O can only be ii, and type AB can only be
IAIB. It is possible for a couple to produce children with all four blood types. The couple would have to
be IAi and IBi. If you construct a Punnett square, you will see that they can produce children with AB, A,
B, and O blood types.
C11. The father could not be IAIB, IBIB, or IAIA. He is contributing the O allele to his offspring.
Genotypically, he could be IAi, IBi, or ii and have type A, B, or O blood, respectively.
C12. A.
1/4
B. 0
C. (1/4)(1/4)(1/4) = 1/64
D. Use the binomial expansion:
n!
p xq(n x)
x !(n  x)!
n  3, p  1/ 4, q  1/ 4, x  2
P  3/ 64  0.047, or 4.7%
P
C28. This is an example of incomplete dominance. The heterozygous horses are palominos. For example,
if C represents chestnut and c represents cremello, the chestnut horses are CC, the cremello horses
are cc, and the palominos are Cc.
E1. Mexican hairless dogs are heterozygous for a dominant allele that is lethal when homozygous. In a
cross between two Mexican hairless dogs, we expect 1/4 to be normal, 1/2 to be hairless, and 1/4 to
die.
CHAPTER 5
C2. An independent assortment hypothesis is used because it enables us to calculate the expected values
based on Mendel’s ratios. Using the observed and expected values, we can calculate whether or not
the deviations between the observed and expected values are too large to occur as a matter of chance.
If the deviations are very large, we reject the hypothesis of independent assortment.
C6. A single crossover produces A B C, A b c, a B C, and a b c.
A. Between 2 and 3, between genes B and C
B. Between 1 and 4, between genes A and B
C. Between 1 and 4, between genes B and C
D. Between 2 and 3, between genes A and B
C8. The likelihood of scoring a basket would be greater if the basket was larger. Similarly, the chances of a
crossover initiating in a region between two genes is proportional to the size of the region between the
two genes. There are a finite number (usually a few) that occur between homologous chromosomes
during meiosis, and the likelihood that a crossover will occur in a region between two genes depends on
how big that region is.
C11. There are four phenotypic categories for the F2 offspring: brown fur, short tails; brown fur, long tails;
white fur, short tails; and white fur, long tails. The recombinants are brown fur, long tails and white fur,
short tails. The F2 offspring will occur in a 1:1:1:1 ratio if the two genes are not linked. In other words,
there will be 25% of each of the four phenotypic categories. If the genes are linked, there will be a lower
percentage of the recombinant offspring.
E5. The rationale behind a testcross is to determine if recombination has occurred during meiosis in the
heterozygous parent. The other parent is usually homozygous recessive, so we cannot tell if crossing
over has occurred in the recessive parent. It is easier to interpret the data if a testcross does use a
completely homozygous recessive parent. However, in the other parent, it is not necessary for all of the
dominant alleles to be on one chromosome and all of the recessive alleles on the other. The parental
generation provides us with information concerning the original linkage pattern between the dominant
and recessive alleles.
E11.
Map distance:
BIO 184 Laboratory
CSU, Sacramento
Spring 2006

64  58
 100
333  64  58  380
 15.1 mu
E12. A.
Since they are 12 mu apart, we expect 12% (or 120) recombinant offspring. This would be
approximately 60 Aabb and 60 aaBb plus 440 AaBb and 440 aabb.
B. We would expect 60 AaBb, 60 aabb, 440 Aabb, and 440 aaBb.
E16. A.
If we hypothesize two genes independently assorting, then the predicted ratio is 1:1:1:1.
There are a total of 390 offspring. The expected number of offspring in each category is about 98.
Plugging the figures into our chi square formula,
(117  98) 2 (115  98) 2 (78  98) 2 (80  98) 2



98
98
98
98
 2  3.68  2.95  4.08  3.31
2 
 2  14.02
Looking up this value in the chi square table under 3 degrees of freedom, we reject our hypothesis,
since the chi square value is above 7.815.
BIO 184 Laboratory
CSU, Sacramento
Spring 2006
B. Map distance:
Map distance =
78  80
117  115  78  80
 40.5 mu
Because the value is relatively close to 50 mu, it is probably a significant underestimate of the true
distance between these two genes.
E20. Let’s use the following symbols: G for green pods, g for yellow pods, S for green seedlings, s for
bluish green seedlings, C for normal plants, c for creepers.
The parental cross is GG SS CC crossed to gg ss cc.
The F1 plants would all be Gg Ss Cc. If the genes are linked, the alleles G, S, and C would be
linked on one chromosome and the alleles g, s, and c would be linked on the homologous chromosome.
The testcross is F1 plants, which are Gg Ss Cc, crossed to ggsscc.
To measure the distances between the genes, we can separate the data into gene pairs.
Pod color, seedling color
2,210 green pods, green seedlings—nonrecombinant
296 green pods, bluish green seedlings—recombinant
2,198 yellow pods, bluish green seedlings—nonrecombinant
293 yellow pods, green seedlings—recombinant
Map distance =
296  293
 100 = 11.8 mu
2,210  296  2,198  293
Pod color, plant stature
2,340 green pods, normal—nonrecombinant
166 green pods, creeper—recombinant
2,323 yellow pods, creeper—nonrecombinant
168 yellow pods, normal—recombinant
Map distance =
166  168
 100 = 6.7 mu
2,340  166  2,323  168
Seedling color, plant stature
2,070 green seedlings, normal—nonrecombinant
433 green seedlings, creeper—recombinant
2,056 bluish green seedlings, creeper—nonrecombinant
438 bluish green seedlings, normal—recombinant
Map distance =
433  438
 100 = 17.4 mu
2,070  433  2,056  438
The order of the genes is seedling color, pod color, plant stature or you could say the opposite order.
Pod color is in the middle. If we use the two shortest distances to construct our map:
S
G
11.8
6.7
C
E21. Let’s use the following symbols: S for normal nose, s for snubnose, p for normal tail, P for pintail, J for
normal gait, j for jerker.
The parental cross is ss PP jj crossed to SS pp JJ.
The F1 offspring would all be Ss Pp Jj. If the genes are linked, the alleles s, P, and j would be
linked on one chromosome and the alleles S, p, and J would be linked on the homologous
chromosome.
The testcross is F1 mice, which are Ss Pp Jj, crossed to ss pp jj mice.
To measure the distances between the genes, we can separate the data into gene pairs.
BIO 184 Laboratory
CSU, Sacramento
Spring 2006
Nose shape, tail length
631 snubnose, pintail—nonrecombinant
111 snubnose, normal tail—recombinant
625 normal nose, normal tail—nonrecombinant
115 normal nose, pintail—recombinant
111  115
 100  15.2 mu
631  111  625  115
Map distance =
Nose shape, walking gait
662 snubnose, jerker—nonrecombinant
80 snubnose, normal—recombinant
652 normal nose, normal—nonrecombinant
88 normal nose, jerker—recombinant
Map distance =
80  88
 100 = 11.3 mu
662  80  652  88
Tail length, walking gait
571 pintail, jerker—nonrecombinant
175 pintail, normal—recombinant
557 normal tail, normal gait—nonrecombinant
179 normal tail, jerker—recombinant
Map distance =
175  179
 100 = 23.9 mu
571  175  557  179
The order of the genes is tail length, nose shape, walking gait or you could say the opposite order. Nose
shape is in the middle.
If we use the two shortest distances to construct our map:
P
15.2
S
11.3
J