1. Genes Are located on chromosomes
Can be visualized using certain techniques

2. The chromosome theory of inheritance states that
Mendelian inheritance has its physical basis in the behavior of chromosomes
The behavior of chromosomes during meiosis was said to account for Mendel’s laws of segregation and independent assortment
The chromosome theory of inheritance states that
Mendelian genes have specific loci on chromosomes that undergo segregation and independent assortment

3. Fertilization among the F1 plants
The chromosomal basis of Mendel’s laws
Yellow-round
seeds (YYRR)
Green-wrinkled
seeds (yyrr)
Meiosis
Fertilization
Gametes
All F1 plants produce
yellow-round seeds (YyRr)
P Generation
F1 Generation
Two equally
probable
arrangements
of chromosomes
at metaphase I
LAW OF SEGREGATION
LAW OF INDEPENDENT ASSORTMENT
Anaphase I
Metaphase II
Fertilization among the F1 plants 9
: 3
: 1 1 4 YR yr yR Y R y r
F2 Generation
Starting with two true-breeding pea plants,
we follow two genes through the F1 and F2
generations. The two genes specify seed
color (allele Y for yellow and allele y for
green) and seed shape (allele R for round
and allele r for wrinkled). These two genes are
on different chromosomes. (Peas have seven
chromosome pairs, but only two pairs are
illustrated here.)
The R and r alleles segregate
at anaphase I, yielding
two types of daughter
cells for this locus.
Each gamete
gets one long
chromosome
with either the
R or r allele. 2
recombines the
R and r alleles
at random. 3
Alleles at both loci segregate
in anaphase I, yielding four
types of daughter cells
depending on the chromosome
arrangement at metaphase I.
Compare the arrangement of
the R and r alleles in the cells on the left and right
Each gamete gets
a long and a short
chromosome in
one of four allele
combinations.
Fertilization results
in the 9:3:3:1
phenotypic ratio in
the F2 generation.

4. Morgan worked with fruit flies
Morgan’s Experiment
Thomas Hunt Morgan
Provided convincing evidence that chromosomes are the location of Mendel’s heritable factors
Morgan worked with fruit flies
Because they breed at a high rate
A new generation can be bred every two weeks
They have only four pairs of chromosomes

5. Morgan first observed and noted
Wild type (normal), phenotypes that were common in the fly populations
Traits alternative to the wild type
Are called mutant phenotypes

6. Correlating Behavior of a Gene’s Alleles with Behavior of a Chromosome Pair
In one experiment Morgan mated male flies with white eyes (mutant) with female flies with red eyes (wild type)
The F1 generation all had red eyes
The F2 generation showed the 3:1 red:white eye ratio, but only males had white eyes

7. Morgan determined
That the white-eye mutant allele must be located on the X chromosome
The F2 generation showed a typical Mendelian
3:1 ratio of red eyes to white eyes. However, no females displayed the
white-eye trait; they all had red eyes. Half the males had white eyes,
and half had red eyes.
Morgan then bred an F1 red-eyed female to an F1 red-eyed male to
produce the F2 generation.
RESULTS P
Generation F1 X F2
Morgan mated a wild-type (red-eyed) female
with a mutant white-eyed male. The F1 offspring all had red eyes.
EXPERIMENT

8. CONCLUSION Since all F1 offspring had red eyes, the mutant
white-eye trait (w) must be recessive to the wild-type red-eye trait (w+).
Since the recessive trait—white eyes—was expressed only in males in
the F2 generation, Morgan hypothesized that the eye-color gene is
located on the X chromosome and that there is no corresponding locus
on the Y chromosome, as diagrammed here. P
Generation F1 F2
Ova
(eggs)
Sperm X Y W W+

9. Each chromosome has hundreds or thousands of genes
Morgan’s discovery that transmission of the X chromosome in fruit flies correlates with inheritance of the eye-color trait
Was the first solid evidence indicating that a specific gene is associated with a specific chromosome
Linked genes tend to be inherited together because they are located near each other on the same chromosome
Each chromosome has hundreds or thousands of genes

10. Recombinant (nonparental-type)
Morgan crossed flies
That differed in traits of two different characters
Double mutant
(black body,
vestigial wings)
Wild type
(gray body,
normal wings)
P Generation
(homozygous)
b+ b+ vg+ vg+ x
b b vg vg
F1 dihybrid
(wild type)
b+ b vg+ vg
TESTCROSS
b+vg+
b vg
b+ vg
b vg+
b+ b vg vg
b b vg+ vg
965
(gray-normal)
944
Black-
vestigial
206
Gray-
185
normal
Sperm
Parental-type
offspring
Recombinant (nonparental-type)
RESULTS
EXPERIMENT
Morgan first mated true-breeding
wild-type flies with black, vestigial-winged flies to produce
heterozygous F1 dihybrids, all of which are wild-type in
appearance. He then mated wild-type F1 dihybrid females with black, vestigial-winged males, producing 2,300 F2 offspring, which he “scored” (classified according to
phenotype).
CONCLUSION
If these two genes were on
different chromosomes, the alleles from the F1 dihybrid
would sort into gametes independently, and we would
expect to see equal numbers of the four types of offspring.
If these two genes were on the same chromosome,
we would expect each allele combination, B+ vg+ and b vg,
to stay together as gametes formed. In this case, only
offspring with parental phenotypes would be produced.
Since most offspring had a parental phenotype, Morgan
concluded that the genes for body color and wing size
are located on the same chromosome. However, the
production of a small number of offspring with
nonparental phenotypes indicated that some mechanism
occasionally breaks the linkage between genes on the
same chromosome.

11. Morgan determined that:
Genes that are close together on the same chromosome are linked and do not assort independently
Unlinked genes are either on separate chromosomes of are far apart on the same chromosome and assort independently
Parents
in testcross
b+ vg+
b vg
b+ vg+
b vg
b vg
Most
offspring X or

12. Recombination of Unlinked Genes: Independent Assortment of Chromosomes
When Mendel followed the inheritance of two characters
He observed that some offspring have combinations of traits that do not match either parent in the P generation
Gametes from green-
wrinkled homozygous
recessive parent (yyrr)
Gametes from yellow-round
heterozygous parent (YyRr)
Parental-
type offspring
Recombinant
offspring
YyRr
yyrr
Yyrr
yyRr YR yr Yr yR

13. Recombinant offspring
Are those that show new combinations of the parental traits
When 50% of all offspring are recombinants
Geneticists say that there is a 50% frequency of recombination

14. Recombination of Linked Genes: Crossing Over
Morgan discovered that genes can be linked
But due to the appearance of recombinant phenotypes, the linkage appeared incomplete
Morgan proposed that
Some process must occasionally break the physical connection between genes on the same chromosome
Crossing over of homologous chromosomes was the mechanism

15. Linked genes Exhibit recombination frequencies less than 50%  =
Testcross
parents
Gray body,
normal wings
(F1 dihybrid) b+
vg+ b vg
Replication of
chromosomes
Meiosis I: Crossing
over between b and vg
loci produces new allele
combinations.
Meiosis II: Segregation
of chromatids produces
recombinant gametes
with the new allele 
Recombinant
chromosome
b+vg+
b   vg
b+ vg
b vg+
b vg
Sperm
Meiosis I and II:
Even if crossing over
occurs, no new allele
combinations are
produced.
Ova
Gametes
offspring
b+  vg+
b   vg
b+   vg
b   vg+
965
Wild type
(gray-normal)
b  vg
b  vg+
944
Black-
vestigial
206
Gray-
185
normal
Recombination
frequency =
391 recombinants
2,300 total offspring
100 = 17%
Parental-type offspring
Recombinant offspring
Black body, vestigial wings (double mutant)

16. Linkage Mapping: Using Recombination Data: Scientific Inquiry
A genetic map
Is an ordered list of the genetic loci along a particular chromosome
Can be developed using recombination frequencies
The farther apart genes are on a chromosome
The more likely they are to be separated during crossing over

17. A linkage map
Is the actual map of a chromosome based on recombination frequencies
Recombination
frequencies 9%
9.5%
17% b cn vg
Chromosome
The b–vg recombination frequency is slightly less than the sum of the b–cn and cn–vg frequencies because double
crossovers are fairly likely to occur between b and vg in matings tracking these two genes. A second crossover
would “cancel out” the first and thus reduce the observed b–vg recombination frequency.
In this example, the observed recombination frequencies between three Drosophila gene pairs
(b–cn 9%, cn–vg 9.5%, and b–vg 17%) best fit a linear order in which cn is positioned about halfway between
the other two genes:
RESULTS
A linkage map shows the relative locations of genes along a chromosome.
APPLICATION
TECHNIQUE
A linkage map is based on the assumption that the probability of a crossover between two genetic loci is proportional to the distance separating the loci. The recombination frequencies used to construct a linkage map for a particular chromosome are obtained from experimental crosses, such as the cross depicted
in Figure The distances between genes are expressed as map units (centimorgans), with one map unit
equivalent to a 1% recombination frequency. Genes are arranged on the chromosome in the order that best fits the data.

18. Many fruit fly genes
Were mapped initially using recombination frequencies
Mutant phenotypes
Short
aristae
Black
body
Cinnabar
eyes
Vestigial
wings
Brown
Long aristae
(appendages
on head)
Gray
Red
Normal
Wild-type phenotypes II Y I X IV
III
48.5
57.5
67.0
104.5

19. The Chromosomal Basis of Sex
An organism’s sex
Is an inherited phenotypic character determined by the presence or absence of certain chromosomes
In humans and other mammals
There are two varieties of sex chromosomes, X and Y

20. (d) The haplo-diploid system
Different systems of sex determination
Are found in other organisms
22 + XX X
76 + ZZ ZW 16
(Haploid)
(Diploid)
(b) The X–0 system
(c) The Z–W system
(d) The haplo-diploid system

21. Inheritance of Sex-Linked Genes
The sex chromosomes
Have genes for many characters unrelated to sex
A gene located on either sex chromosome
Is called a sex-linked gene
Some recessive alleles found on the X chromosome in humans cause certain types of disorders
Color blindness, Duchenne muscular dystrophy and hemophilia

22. Sex-linked genes Follow specific patterns of inheritance XAXA XaY Xa Y
Ova
Sperm
XaXA 
XaXa
A father with the disorder will transmit the mutant allele to all daughters but to no sons. When the mother is a dominant homozygote, the daughters will have the normal phenotype but will be carriers of the mutation.
If a carrier mates with a male of normal phenotype, there is a 50% chance that each daughter will be a carrier like her mother, and a 50% chance that each son will have the disorder.
If a carrier mates with a male who has the disorder, there is a 50% chance that each child born to them will have the disorder, regardless of sex. Daughters who do not have the disorder will be carriers, where as males without the disorder will be completely free of the recessive allele.
(a)
(b)
(c)

23. X inactivation in Female Mammals
In mammalian females
One of the two X chromosomes in each cell is randomly inactivated during embryonic development

24. If a female is heterozygous for a particular gene located on the X chromosome
She will be a mosaic for that character
Two cell populations
in adult cat:
Active X
Orange
fur
Inactive X
Early embryo:
X chromosomes
Allele for
black fur
Cell division
and X
chromosome
inactivation
Black
Figure 15.11
Allele for
orange fur

25. Large-scale chromosomal alterations
Alterations of chromosome number or structure cause some genetic disorders
Large-scale chromosomal alterations
Often lead to spontaneous abortions or cause a variety of developmental disorders
Aneuploidy
Results from the fertilization of gametes in which nondisjunction occurred
Is a condition in which offspring have an abnormal number of a particular chromosome

26. Abnormal Chromosome Number
When nondisjunction occurs
Pairs of homologous chromosomes do not separate normally during meiosis
Gametes contain two copies or no copies of a particular chromosome
Meiosis I
Nondisjunction
Meiosis II
Gametes
n + 1
n  1
n – 1
n –1 n
Number of chromosomes
Nondisjunction of homologous
chromosomes in meiosis I
Nondisjunction of sister
chromatids in meiosis II
(a)
(b)

27. If a zygote is monosomic
If a zygote is trisomic
It has three copies of a particular chromosome
If a zygote is monosomic
It has only one copy of a particular chromosome

28. Polyploidy
Is a condition in which there are more than two complete sets of chromosomes in an organism

29. Alterations of Chromosome StructureB C D E F G H
Deletion
Duplication M N O P Q R
Inversion
Reciprocal
translocation
(a) A deletion removes a chromosomal
segment.
(b) A duplication repeats a segment.
(c) An inversion reverses a segment within
a chromosome.
(d) A translocation moves a segment from one chromosome to another,
nonhomologous one. In a reciprocal
  translocation, the most common type,
nonhomologous chromosomes exchange
fragments. Nonreciprocal translocations
also occur, in which a chromosome
transfers a fragment without receiving a
fragment in return.

30. Human Disorders Due to Chromosomal Alterations
Alterations of chromosome number and structure
Are associated with a number of serious human disorders

31. Down Syndrome Down syndrome
Is usually the result of an extra chromosome 21, trisomy 21

32. Aneuploidy of Sex Chromosomes
Nondisjunction of sex chromosomes
Produces a variety of aneuploid conditions
Klinefelter syndrome
Is the result of an extra chromosome in a male, producing XXY individuals
Turner syndrome
Is the result of monosomy X, producing an X0 karyotype

33. Two normal exceptions to Mendelian genetics include
Some inheritance patterns are exceptions to the standard chromosome theory
Two normal exceptions to Mendelian genetics include
Genes located in the nucleus
Genes located outside the nucleus

34. Genomic Imprinting In mammals
The phenotypic effects of certain genes depend on which allele is inherited from the mother and which is inherited from the father

35. (a) A wild-type mouse is homozygous for the normal igf2 allele.
Genomic imprinting
Involves the silencing of certain genes that are “stamped” with an imprint during gamete production
(a) A wild-type mouse is homozygous for the normal igf2 allele.
Normal Igf2 allele
(expressed)
with imprint
(not expressed)
Paternal
chromosome
Maternal
Wild-type mouse
(normal size)
Mutant lgf2 allele
Dwarf mouse
Normal size mouse
(b) When a normal Igf2 allele is inherited from the father, heterozygous mice grow to normal size But when a mutant allele is inherited from the father, heterozygous mice have the dwarf phenotype.

36. Inheritance of Organelle Genes
The inheritance of traits controlled by genes present in the chloroplasts or mitochondria
Some diseases affecting the muscular and nervous systems
Are caused by defects in mitochondrial genes that prevent cells from making enough ATP
Depends solely on the maternal parent because the zygote’s cytoplasm comes from the egg

37. Animations and Videos
Independent Assortment and Gamete Diversity
Alleles That Do Not Assort Independently
Meiosis - Nondisjunction – 1
Meiosis - Nondisjunction – 2
Karyotype Animation
Bozeman - Chi-squared Test
Speciation by Ploidy
Changes in Chromosome Structure

38. Animations and Videos
Chapter Quiz Questions – 1
Chapter Quiz Questions - 2

39. Why did the improvement of microscopy techniques in the late 1800s set the stage for the emergence of modern genetics?
It revealed new and unanticipated features of Mendel’s pea plant varieties.
It allowed the study of meiosis and mitosis, revealing parallels between behaviors of the Mendelian concept of the gene and the movement/pairing of chromosomes.
It allowed scientists to see the nucleotide sequence of DNA.
It led to the discovery of mitochondria.
It showed genes functioning to direct the formation of enzymes.
Answer: B 39

40. Why did the improvement of microscopy techniques in the late 1800s set the stage for the emergence of modern genetics?
It revealed new and unanticipated features of Mendel’s pea plant varieties.
It allowed the study of meiosis and mitosis, revealing parallels between behaviors of the Mendelian concept of the gene and the movement/pairing of chromosomes.
It allowed scientists to see the nucleotide sequence of DNA.
It led to the discovery of mitochondria.
It showed genes functioning to direct the formation of enzymes. 40

41. Morgan and his colleagues worked out a set of symbols to represent fly genotypes. Which of the following is representative?
AaBb  AaBb
46 or 46w
w or w on X
2  3
Answer: C 41

42. Morgan and his colleagues worked out a set of symbols to represent fly genotypes. Which of the following is representative?
AaBb  AaBb
46 or 46w
w or w on X
2  3 42

43. on the O chromosome of a male on the X chromosome of a male
Imagine that Morgan had used a grasshopper (2n  24, and sex is determined as follows: male has X, and female has XX) to study sex linkage. Predict where the first mutant would have been discovered.
on the O chromosome of a male
on the X chromosome of a male
on the X chromosome of a female
on the Y chromosome of a male
Answer: B
This question is designed to show students that there are a variety of sex determination mechanisms and to help students understand the significance of Morgan’s work. Answer A is incorrect because there is no O chromosome—the X chromosome has no pairing partner. Answer C is incorrect because a mutant is likely recessive and is likely to be masked by a dominant on the other X chromosome in a female. Answer D is incorrect because grasshopper males do not have a Y chromosome. Answer B is correct because a mutant on the male’s single X chromosome would not be masked by a normal allele on a second X chromosome. 43

44. on the O chromosome of a male on the X chromosome of a male
Imagine that Morgan had used a grasshopper (2n  24, and sex is determined as follows: male has X, and female has XX) to study sex linkage. Predict where the first mutant would have been discovered.
on the O chromosome of a male
on the X chromosome of a male
on the X chromosome of a female
on the Y chromosome of a male 44

45. Think about bees, which have no X and Y sex chromosomes
Think about bees, which have no X and Y sex chromosomes. Males are haploid, whereas fertilization results in females, as diploid cells become females. Which of the following are accurate statements about bee males when they are compared to species in which males are XY and diploid for the autosomes?
Bee males have half the DNA of bee females, whereas human males have nearly the same amount of DNA as human females.
Considered across the genome, harmful (deleterious) recessives will negatively affect bee males more than Drosophila males.
Human and Drosophila males have sons, but bee males do not.
Inheritance in bees is like inheritance of sex-linked characteristics in humans.
none of the above
Answer: A, B, or C
The point of this question is that there are a variety of sex determination mechanisms among different species. Answers A, B, and C are correct. Answer A is correct because bee males are haploid, but the only difference in the amount of DNA in human males and females is that the Y chromosome is slightly smaller than the X chromosome. Answer B is correct because deleterious alleles from the whole genome will affect male bees, but deleterious recessives on the autosomes in Drosophila males can be masked by a dominant allele. Answer C is correct because, in humans and Drosophila, half of a male’s sperm carry a Y chromosome and cause the production of male offspring, but in bees and ants males are haploid and any egg the sperm fertilizes develops into a female. Answer D is incorrect because male bees have no sons. 45

46. Think about bees, which have no X and Y sex chromosomes
Think about bees, which have no X and Y sex chromosomes. Males are haploid, whereas fertilization results in females, as diploid cells become females. Which of the following are accurate statements about bee males when they are compared to species in which males are XY and diploid for the autosomes?
Bee males have half the DNA of bee females, whereas human males have nearly the same amount of DNA as human females.
Considered across the genome, harmful (deleterious) recessives will negatively affect bee males more than Drosophila males.
Human and Drosophila males have sons, but bee males do not.
Inheritance in bees is like inheritance of sex-linked characteristics in humans.
none of the above 46

47. This mutation occurs in all offspring of a male with the mutation.
Determination of sex in Drosophila is similar to that in humans. In some species of Drosophila, there are genes on the Y chromosome that do not occur on the X chromosome. Imagine that a mutation of one gene on the Y chromosome reduces the size by half of individuals with the mutation. Which of the following statements is accurate with regard to this situation?
This mutation occurs in all offspring of a male with the mutation.
This mutation occurs in all male but no female offspring of a male with the mutation.
This mutation occurs in all offspring of a female with the mutation.
This mutation occurs in all male but no female offspring of a female with the mutation.
This mutation occurs in all offspring of both males and females with the mutation.
Answer: B
The point of this question is to help students understand sex linkage by looking at the effect of genes on the Y chromosome. Answer A is incorrect because half of the sperm of the mutant male will contain an X chromosome and produce normal size daughters. Answer C is incorrect because females do not have a Y chromosome and cannot carry the mutation. Answer D is incorrect because, again, females do not have a Y chromosome and thus cannot carry the mutation. Answer E is incorrect because females cannot carry the mutation (no Y chromosome) and only the half of the eggs that are fertilized by a Y-bearing sperm will receive the mutation (i.e., only half of the male offspring). Answer B is correct because all the sons of the mutant male receive his Y chromosome. 47

48. This mutation occurs in all offspring of a male with the mutation.
Determination of sex in Drosophila is similar to that in humans. In some species of Drosophila, there are genes on the Y chromosome that do not occur on the X chromosome. Imagine that a mutation of one gene on the Y chromosome reduces the size by half of individuals with the mutation. Which of the following statements is accurate with regard to this situation?
This mutation occurs in all offspring of a male with the mutation.
This mutation occurs in all male but no female offspring of a male with the mutation.
This mutation occurs in all offspring of a female with the mutation.
This mutation occurs in all male but no female offspring of a female with the mutation.
This mutation occurs in all offspring of both males and females with the mutation. 48

49. In cats, a sex-linked gene affects coat color
In cats, a sex-linked gene affects coat color. The O allele produces an enzyme that converts eumelanin, a black or brown pigment, into phaeomelanin, an orange pigment. The o allele is recessive to O and produces a defective enzyme, one that does not convert eumelanin into phaeomelanin. Which of the following statements is/are accurate?
The phenotype of o-Y males is black/brown because the nonfunctional allele o does not convert eumelanin into phaeomelanin.
The phenotype of OO and Oo males is orange because the functional allele O converts eumelanin into phaeomelanin.
The phenotype of Oo males is mixed orange and black/brown because the functional allele O converts eumelanin into phaeomelanin in some cell groups (orange) and because in other cell groups the nonfunctional allele o does not convert eumelanin into phaeomelanin.
The phenotype of O-Y males is orange because the nonfunctional allele O does not convert eumelanin into phaeomelanin, while the phenotype of o-Y males is black/brown because the functional allele o converts eumelanin into phaeomelanin.
Answer: A
This focuses on the color of males and the action of the enzyme that converts eumelanin (brown/black pigment) to phaeomelanin (orange pigment). Male genotypes will be either O-Y or o-Y, with phenotypes of either orange or black/brown, respectively. In O-Y males, the eumelanin is converted to phaeomelanin, and in o-Y males, the eumelanin is unchanged. To answer this question, a student must know that males have only one copy of the gene and must understand that a functional allele produces an enzyme that catalyzes the chemical reaction. 49

50. In cats, a sex-linked gene affects coat color
In cats, a sex-linked gene affects coat color. The O allele produces an enzyme that converts eumelanin, a black or brown pigment, into phaeomelanin, an orange pigment. The o allele is recessive to O and produces a defective enzyme, one that does not convert eumelanin into phaeomelanin. Which of the following statements is/are accurate?
The phenotype of o-Y males is black/brown because the nonfunctional allele o does not convert eumelanin into phaeomelanin.
The phenotype of OO and Oo males is orange because the functional allele O converts eumelanin into phaeomelanin.
The phenotype of Oo males is mixed orange and black/brown because the functional allele O converts eumelanin into phaeomelanin in some cell groups (orange) and because in other cell groups the nonfunctional allele o does not convert eumelanin into phaeomelanin.
The phenotype of O-Y males is orange because the nonfunctional allele O does not convert eumelanin into phaeomelanin, while the phenotype of o-Y males is black/brown because the functional allele o converts eumelanin into phaeomelanin. 50

51. Imagine two species of cats that differ in the timing of Barr body formation during development. Both species have genes that determine coat color, O for the dominant orange fur and o for the recessive black/brown fur, on the X chromosome. In species A, the Barr body forms during week 1 of a 6-month pregnancy, whereas in species B, the Barr body forms during week 3 of a 5-month pregnancy. What would you predict about the coloration of heterozygous females (Oo) in the two species?
Both species will have similar sized patches of orange and black/brown fur.
Species A will have smaller patches of orange or black/brown fur than will species B.
The females of both species will show the dominant fur color, orange.
Answer: B
This question relates genetics, embryonic development, and Barr body formation. The sizes of the patches should be related to the timing of Barr body formation in development. Early formation of the Barr body would mean that larger clusters of cells are affected by which X chromosome is inactivated compared to later Barr body formation. 51

52. Imagine two species of cats that differ in the timing of Barr body formation during development. Both species have genes that determine coat color, O for the dominant orange fur and o for the recessive black/brown fur, on the X chromosome. In species A, the Barr body forms during week 1 of a 6-month pregnancy, whereas in species B, the Barr body forms during week 3 of a 5-month pregnancy. What would you predict about the coloration of heterozygous females (Oo) in the two species?
Both species will have similar sized patches of orange and black/brown fur.
Species A will have smaller patches of orange or black/brown fur than will species B.
The females of both species will show the dominant fur color, orange. 52

53. Imagine a species with three loci thought to be on the same chromosome
Imagine a species with three loci thought to be on the same chromosome. The recombination rate between locus A and locus B is 35%, and the recombination rate between locus B and locus C is 33%. Predict the recombination rate between A and C.
The recombination rate between locus A and locus C is either 2% or 68%.
The recombination rate between locus A and locus C is probably 2%.
The recombination rate between locus A and locus C is either 2% or 50%.
The recombination rate between locus A and locus C is either 2% or 39%.
The recombination rate between locus A and locus C cannot be predicted.
Answer: C
The recombination rate between loci A and B is 35%. Locus C can be either between A and B or on the opposite side of B from A. If locus C is in between locus A and locus B (ACB), the distance between locus A and C would be 2%. Locus C cannot be on the other side of A from B (CAB) because the recombination rate would have to be higher than 35%. Thus far, answers A, B, C, and D could be correct. If locus C is on the other side of locus B from locus A (ABC), adding the two recombination rates gives a prediction of 68%, but the maximum recombination rate between two loci is 50%, the same frequency of recombinants that are on different chromosomes. Using this information, answer A cannot be correct because a recombination rate cannot be greater than 50%. Answer B could be right, but it is not the only possible placement and so answer B is incomplete. Answer C is the best answer (2% if C is between A and B, and 50% if C is on the other side of B than A). 53

54. Imagine a species with three loci thought to be on the same chromosome
Imagine a species with three loci thought to be on the same chromosome. The recombination rate between locus A and locus B is 35%, and the recombination rate between locus B and locus C is 33%. Predict the recombination rate between A and C.
The recombination rate between locus A and locus C is either 2% or 68%.
The recombination rate between locus A and locus C is probably 2%.
The recombination rate between locus A and locus C is either 2% or 50%.
The recombination rate between locus A and locus C is either 2% or 39%.
The recombination rate between locus A and locus C cannot be predicted. 54

55. Triploid species are usually sterile (unable to reproduce), whereas tetraploids are often fertile. Which of the following is likely a good explanation of these facts? (Hint: Synapsis.)
In mitosis, some chromosomes in triploids have no partner at synapsis, but chromosomes in tetraploids do have partners.
In meiosis, some chromosomes in triploids have no partner at synapsis, but chromosomes in tetraploids do have partners.
In mitosis, some chromosomes in tetraploids have no partner at synapsis, but chromosomes in triploids do have partners.
In meiosis, some chromosomes in tetraploids have no partner at synapsis, but chromosomes in triploids do have partners.
Answer: B
The point of this question is to make students think about mitosis and meiosis in relation to polyploids. To answer this question, students should draw chromosomes of a triploid and a tetraploid as they go through mitosis and meiosis. Answers A and C are incorrect because chromosomes do not synapse during mitosis. Answer D is incorrect because tetraploids do have partners at synapsis but triploids do not. Answer B is correct—one-third of the chromosomes do not have a partner. 55

56. Triploid species are usually sterile (unable to reproduce), whereas tetraploids are often fertile. Which of the following is likely a good explanation of these facts? (Hint: Synapsis.)
In mitosis, some chromosomes in triploids have no partner at synapsis, but chromosomes in tetraploids do have partners.
In meiosis, some chromosomes in triploids have no partner at synapsis, but chromosomes in tetraploids do have partners.
In mitosis, some chromosomes in tetraploids have no partner at synapsis, but chromosomes in triploids do have partners.
In meiosis, some chromosomes in tetraploids have no partner at synapsis, but chromosomes in triploids do have partners. 56

57. Chromosomal rearrangements can occur after chromosomes break
Chromosomal rearrangements can occur after chromosomes break. Which of the following statements is most accurate with respect to alterations in chromosome structure?
Chromosomal rearrangements are more likely to occur in mammals than in other vertebrates.
Translocations and inversions are not deleterious because no genes are lost in the organism.
Chromosomal rearrangements are more likely to occur during mitosis than during meiosis.
An individual that is homozygous for a deletion of a certain gene is likely to be more damaged than one that is homozygous for a duplication of that same gene because loss of a function can be lethal.
Answer: D
Chromosomal rearrangements are important in evolution. For example, duplications provide raw material on which natural selection can act (e.g., the globin genes are thought to have arisen via gene duplication). This question will make students think about the consequences of chromosomal rearrangements. Answer A is not correct and should be nixed by students because there is no information in the chapter that would lead to this conclusion. Answers B and C directly contradict material in the text and are therefore incorrect. 57

58. Chromosomal rearrangements can occur after chromosomes break
Chromosomal rearrangements can occur after chromosomes break. Which of the following statements is most accurate with respect to alterations in chromosome structure?
Chromosomal rearrangements are more likely to occur in mammals than in other vertebrates.
Translocations and inversions are not deleterious because no genes are lost in the organism.
Chromosomal rearrangements are more likely to occur during mitosis than during meiosis.
An individual that is homozygous for a deletion of a certain gene is likely to be more damaged than one that is homozygous for a duplication of that same gene because loss of a function can be lethal. 58

59. Which of the following statements about crossing over is false?
It accounts for the recombination of linked genes.
It occurs while replicated homologs are paired during prophase I of meiosis.
Portions of sister chromatids change places.
It breaks the physical connection between specific alleles on the same chromosome.
Recombinants may result.
Answer: C

60. Which of the following statements about crossing over is false?
It accounts for the recombination of linked genes.
It occurs while replicated homologs are paired during prophase I of meiosis.
Portions of sister chromatids change places.
It breaks the physical connection between specific alleles on the same chromosome.
Recombinants may result.

61. Which of the following is a type of chromosomal alteration that differ from all of the others?
aneuploidy
polyploidy
triploidy
tetraploidy
octaploidy
Answer: A

62. Which of the following is a type of chromosomal alteration that differs from all of the others?
aneuploidy
polyploidy
triploidy
tetraploidy
octaploidy
Answer: A