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Last Updated 4/17/06  

Gene Mutations
1) The figure below shows the mRNA made by a very small wild type gene without introns and four mutant alleles of this same gene (the changes are underlined). Explain the phenotypic effects of each mutation below. The start codon is in bold type and the mutation is underlined.

wild type mRNA . . . . . . . . ...AUGGCGAACUUGAAUUUGUCGUGA...
mutant A mRNA . . . . . . . . ...AUGGCAAACUUGAAUUUGUCGUGA...
mutant B mRNA . . . . . . . . ...AUGGCGAACUAGAAUUUGUCGUGA...
mutant C mRNA . . . . . . . . ...AUGG_GAACUUGAAUUUGUCGUGA...
mutant D mRNA . . . . . . . . ...AUGGCGAACUUGAAUUUGAGUUCGUGA...

2) The Ames test is used to screen for... A) potential mutagens and carcinogens. B) the presence of enol forms of thymine and guanine. C) frameshift mutations. D) spontaneous mutations. E) oncogenes.

3) Which of the following are true?
A) Mutations are caused by genetic recombination. B) Mutations are changes in genetic information C) Mutations are caused by faulty transcription of the genetic code. D) Mutations are usually but not always beneficial to the development of individuals in which they occur. E) A nonsense mutation changes a codon for an amino acid to another codon for the same amino acid. F) A mutation that results in the addition or deletion of a bp within the coding region of a gene is a frameshift mutation. G) When guanine is mutated to adenine this is a transition mutation. H) A neutral mutation causes an amino acid substitution that does not alter the function of the translated protein. I) Reverse mutations turn mutant into wild type genes. J) A point mutation always results in a phenotypic change. K) Mutation rate and mutation frequency are the same things. L) Somatic mutations are carried to subsequent generations through the germ line.

4) Ultraviolet light usually causes mutations by a mechanism involving... A) induction of thymine dimers and their persistence or imperfect repair. B) inversion of DNA segments. C) one-strand breakage in DNA. D) light-induced change of thymine to alkylated guanine. E) deletion of DNA segments.

Cytogenetics
1) Answer the following about the karyotype below (assume there are no homeologus chromosomes).

A) For this species n = _________________________ B) What is the ploidy level of this species? _________________________ C) For this species X = _________________________

2) (Fill in the blank) A diploid apple tree and a tetraploid apple tree cross and produce seeds which grow to become mature apple trees. The progeny are most likely ____X.

3) If a wild tetraploid strawberry has x = 8, how many chromosomes does the plant have? ________
n = ____________ in this strawberry plant.

4) Circle all true statements below.
A) 2n + 1 is a euploid. B) A 5x plant has a greater chance of being fertile than a 6x plant. C) Chromatin is Scattered throughout the nucleus in non-dividing cells. D) Supercoiled DNA is only found in non-dividing cells. E) Chromosomes are always composed of two chromatids. F) Aneuploids never form bivalents. G) Bivalents are pairs of chromosomes. H) Univalents are unpaired chromosomes. I) Allopolyploids are the result of crosses between 2 different species

5) An insertion... A) removes a piece of chromosome. B) adds a piece of chromosome to a chromosome. C) changes the direction of a piece of chromosome. D) exchanges pieces between chromosomes.

6) If a hexaploid plant has x = 21, how many chromosomes does the plant have?____________________
If a hexaploid plant has x = 21, n = ____________________ .

7) Nondisjunction can result in... A) auto- & allopolyploidy B) aneuploidy C) aneuploidy & allopolyploidy D) allopolyploidy D) autopolyploidy F) aneuploidy & autopolyploidy

Population Genetics
1) Name the assumptions made for a population in Hardy-Weinberg equilibrium?

2) In a population exhibiting Hardy-Weinberg equilibrium, 23% of the individuals are homozygous for a recessive character. What will the genotypic, phenotypic and allelic frequencies be for the next generation?

3) A population has twelve times as many heterozygotes as homozygous recessives. What is the frequency of the recessive gene?

4) Which of the populations in the table below are in Hardy-Weinburg equilibrium?
Population [Genotypic frequencies for AA, AB, AC, BB, BC and CC, respectively]
a [ 0.300 , 0.173 , 0.321 , 0.025 , 0.093 , 0.086 ]
b [ 0.775 , 0.155 , 0.051 , 0.007 , 0.015 , 0.001 ]
c [ 0.253 , 0.353 , 0.146 , 0.023 , 0.102 , 0.121 ]
d [ 0.020 , 0.000 , 0.243 , 0.000 , 0.000 , 0.736 ]
e [ 0.561 , 0.142 , 0.232 , 0.009 , 0.029 , 0.024 ]

5) A population of lizards in New Mexico have allele nif for night foraging which masks the expression of the allele daf for day foraging. Lizards which forage during the day have a reproductive fitness of 0.4, in comparison to those that forage at night (W=1). Out of 1000 lizards, 134 were found to be homozygous for daf while 743 were found to be homozygous for nif. After one generation what will the allelic, genotypic, and phenotypic frequencies be for this population of lizards if natural selection is the only disturbing factor?

6) A population of violets (The NJ state flower) has an allele B for blue flowers which is dominant to the allele b for white flowers. BB genotypes have a reproductive fitness of 1, Bb genotypes have a reproductive fitness of 1, and bb genotypes have a reproductive fitness of 0.4. The allelic frequency for B is 0.3. After one generation what will the allelic and genotypic frequencies be for this population of violets if natural selection is the only disturbing factor?

7) In a population exhibiting Hardy-Weinberg equilibrium, 81% of the individuals are homozygous for a recessive character. What % of the population is homozygous dominant and what % is heterozygous?

Quantitative Genetics and other questions
1) Name any three likely multifactorial traits.

2) Variance is...A) a measure of how a set of data is spread out about the mean. A) a measure of the accuracy of a measurement. C) the mean of a set of measurements. D) the square root of the average of a set of measurements. E) the range of a set of measurements.

3) Heritability is... A) the fraction of a population's phenotype which is strictly do to genetic factors. B) the fraction of a population's phenotypic variation that is strictly do to genetic factors. C) the degree to which related members of a population resemble each another. D) the degree to which a continuously varying trait is controlled by genotype. E) all phenotypic plasticity.

4) The narrow-sense heritability of the fruit color in strawberries is very close to 1. Which of the following is true? A) Fruit color will be difficult to select for. B) Most fruit color variance is additive genetic variance. C) Most fruit color variance has an environmental cause. D) Both A and B. E) Both A and C.

5) Match the description on the left with BEST answer on the right by filling in the blanks.

____1) Additive genetic variance. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A) Artificial selection
____2) The individuals in a population declines to a very low number. . . . .B) M-cytotype
____3) Phenotypic variations do not overlap for each genotype. . . . . . . . . .C) Multifactorial trait
____4) Change in allelic frequencies. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . D) Correlation
____5) Non-additive genetic variance. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .E) Natural selection
____6) All the alleles of all the loci in a population. . . . . . . . . . . . . . . . . . . G) Evolution
____7) Characters affected by more than one gene. . . . . . . . . . . . . . . . . . . H) Bottleneck
. . . . . . .locus and/or the environment.
____8) Allelic frequencies stay the same. . . . . . . . . . . . . . . . . . . . . . . . . . . I) Heterosis
____9) A mutation transmitted to subsequent generations. . . . . . . . . . . . . . L) Discrete traits
____10) Enzyme coded by autonomous transposable elements. . . . . . . . . .M) Germ line
____11) The proportion of a population's phenotypic variation. . . . . . . . . .O) Somatic
. . . . . . . .attributable to genetic factors.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .P) Gene pool
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .R) Hybrid dysgenesis
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .S) Transposase
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .T) Narrow sense heritability
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .U) Broad sense heritability
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .G) Hardy-Weinberg equilibrium

6) Circle only the TRUE statements about the figure shown below (2 points each).

A) The heterozygosity of population A is greater than that of population C. B) The h2 of the trait is close to 1. C) Most of the phenotypic variance results from environmental variance. D) The mean value of the trait has increased. E) The variance of population A is less than population C. F) The trait is easy to select for.

7) Circle only the TRUE statements about the figure shown below.

A) The narrow-sense heritability of the trait is close to 1. B) Most of the phenotypic variance results from environmental variance. C) The heterozygosity of population A is greater than that of population B. D) The variance of population A is less than population B. E) The mean value of the trait of has increased.

ADDITIONAL QUESTIONS
1) Circle all that reduce polymorphism?
A) Mutation B) Genetic drift C) Inbreeding D) Bottlenecks E) Selfing F) Interspecific crosses.

2) Circle only the TRUE statements below.
A) Autonomous transposons do not have a transposase gene. B) Transposable elements can cause chromosome rearrangements C) Natural selection is the ultimate source of all genetic variation D) Variance is the estimate of the accuracy of a measurement. E) Bottlenecks increase genetic variation. F) Heterosis is hybrid vigor. G) Populations in Hardy-Weinberg equilibrium are free of migration and sexual selection. H) Blood type is a quantitative trait I) Homeologous chromosomes rarely cross over. J) 2n -2 is an allopolyploid. K) A 3x plant has a greater chance of being fertile than a 2x plant. L) Transposases can inactivate genes. M) Dr. Barbara McClintock was the first to describe transposable elements. N) Transposons can cause deletions or insertions. O) All transposons have a transposase gene. P) Transposons end in inverted repeats. Q) The first transposable elements described were those in maize (corn). R) The haploid number (n) is the number of chromosomes found in gametes. S) Trisomy 21 causes Down syndrome. T) Mutation rate and mutation frequency are the same things. U) Somatic mutations are carried to subsequent generations through the germ line. V) Reverse mutations turn mutant into wild type genes. W) A point mutation always results in a phenotypic change. X) Recombination often occurs at regions of DNA homology. Y) Branch migration is the movement of a crossover point between DNA complexes. Z) M cytotype is lethal in Drosophila. aa) Branch migration is the movement of a crossover point between DNA complexes. bb) Recombination rarely occurs at regions of DNA homology. cc) Gene duplications caused by unequal crossing-over have played an important role in gene evolution. dd) P elements can be used to transform Drosophila. ee) Ac and Ds are plant transposable elements.

3) Match the description on the left with BEST answer on the right by filling in the blanks.
___1) Homologous DNA molecules are restricted and religated during crossing over . . A) Site directed mutagenesis
___2) A mutation which is not passed on to subsequent generations . . . . . . . . . . . . . . . B) Transposon mutagenesis
___3) A kind of Drosophila transposable element . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .C) Holliday model
___4) Genetic engineering using homologous recombination . . . . . . . . . . . . . . . . . . . . . D) Ac-Ds
___5) Allows transposable elements to move from one location to another . . . . . . . . . . . E) Germ line
___6) Genetic rearrangement in Drosophila due to transposable elements . . . . . . . . . . . .F) Somatic
___7) Often forms pyrimidine dimers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . G) PCR
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . H) Repressor
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . I) M-cytotype
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . J) UV mutagenesis
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . K) P-element
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . L) Hybrid dysgenesis
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . M) Transposase

4) _____________ requires homologous recombination.
A) DNA sequencing . . B) PCR . . C) Restriction length polymorphisms . . D) Site directed mutagenesis . . E) Transposon mutagenesis.

5) In house flies, two genes, A and R, affect behavior. Neither is completely dominant, and the two interact on each other to produce seven different behaviors:

AA RR = High circling . . . . . Aa RR = Low circling . . . . . AA Rr = High hovering . . . . . Aa Rr = Low hovering . . . . . AA rr = High non hovering . . . . . Aa rr = Low non hovering . . . . . aa RR, aa Rr, and aa rr = walking

In a cross of a high circling fly with a walking one of genotype (a/a r/r), what will be the appearances of the F₁, the F₂, and the offspring of the F₁ backcrossed to the high circling parent?

6) In Tiger Lilies, flower color is determined by three loci. Dr. Pete Moss worked out the inheritance of flower color in this species and developed the table below which shows the genotypes and their associated phenotypes. The genes are not linked.

AA BB ee = white . . . . . AA BB E__ = white with dots . . . . . Aa BB ee = yellow . . . . . Aa BB E__ = yellow with dots . . . . . AA Bb ee = orange . . . . . AA Bb E__ = orange with dots . . . . . Aa Bb ee = red . . . . . Aa Bb E__ = red with dots . . . . . aa B__ ee, A__ bb ee, and aa bb ee = rose . . . . . aa B__ E__, A__ bb E__, and aa bb E__ = rose with dots

Dr. Moss selfed Aa Bb Ee Lilies. What is the expected phenotypic ratio of the progeny?

7) Circle only the TRUE statements about the figure shown below.

A) The narrow-sense heritability of the trait is close to 1. B) Most of the phenotypic variance results from environmental variance. C) The heterozygosity of population A is greater than that of population B. D) The variance of population A is less than population B. E) The mean value of the trait of has increased. F) The trait is easy to select for. G) The broad-sense heritability of the trait is close to 0.

8) Match the description on a mutation on the left with the BEST answer on the right by filling in the blanks.
___1) A purine becomes another purine . . . . . . . . . . . . . . . . . . .A) Duplication
___2) Produces a termination codon . . . . . . . . . . . . . . . . . . . . . B) Point
___3) A second mutation corrects the first . . . . . . . . . . . . . . . . .C) Deletion
___4) A purine becomes a pyrimidine . . . . . . . . . . . . . . . . . . . . D) Inversion
___5) Changes a polypeptides amino acid sequence . . . . . . . . . E) Reverse
___6) Changes most amino acids downstream of a mutation . . .F) Nonsense
___7) Mutation within a single gene . . . . . . . . . . . . . . . . . . . . . G) Missense
___8) Loss of nucleotides . . . . . . . . . . . . . . . . . . . . . . . . . . . . . H) Transversion
___9) Gain of nucleotides. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . I) Transition
___10) A DNA sequence is reversed . . . . . . . . . . . . . . . . . . . . . J) Frameshift

Gene Mutations
1) The phenotypes are given in terms of the polypeptide changes resulting from the mutations.

wild type polypeptide...Met Ala Asn Leu Asn Leu Ser
mutant A polypeptide...Met Ala Asn Leu Asn Leu Ser... Silent mutation
mutant B polypeptide...Met Ala Asn... Nonsense mutation truncates polypeptide
mutant C polypeptide...Met Gly Thr... Frameshift mutation changes and truncates polypeptide
mutant D polypeptide...Met Ala Asn Leu Asn Leu Ser Ser... A 3 pase pair insertion mutant results in an additional serine on the C-terminus

2) A) potential mutagens and carcinogens.

3) The true statements are below.
B) Mutations are changes in genetic information. F) A mutation that results in the addition or deletion of a bp within the coding region of a gene is a frameshift mutation. G) When guanine is mutated to adenine this is a transition mutation. H) A neutral mutation causes an amino acid substitution that does not alter the function of the translated protein. I) Reverse mutations turn mutant into wild type genes.

4) Ultraviolet light usually causes mutations by a mechanism involving... A) induction of thymine dimers and their persistence or imperfect repair.

Cytogenetics

1) A) n = 10 B) Tetraploid or 4X C) X = 5

2) 3X or triploid.

3) 32 chromosomes, n = 16

4) The true statements are below.
C) Chromatin is scattered throughout the nucleus in non-dividing cells. G) Bivalents are pairs of chromosomes. H) Univalents are unpaired chromosomes. I) Allopolyploids are the result of crosses between 2 different species

5) An insertion... B) adds a piece of chromosome to a chromosome.

6) 126 chromosomes, n = 63

7) Nondisjunction can result in... B) aneuploidy

Population Genetics

1) No Mutation, No Migration, No Natural Selection, No Sexual Selection, Infinite Population Size, No Gametic Drive.

2) Recessive allele frequency = 0.4796 ... Dominant allele frequency = 0.5204
Homozygous recessive = 0.2300 ... Heterozygous = 0.4992 ... Homozygous Dominant = 0.2708
Recessive trait = 0.2300 ... Dominant trait = 0.7700

3) 2pq = 12(q²) ........... p = 6q .......... p = 6(1-p) .......... p = 6 - 6p .......... 7p = 6 .......... p = 0.8571 .......... q = 0.1428
Frequency of the recessive allele = 0.1428

4)
Population a Allelic frequencies A = 0.547, B = 0.158, C = 0.293
Genotypic frequencies should be AA = 0.299, BB = 0.025, CC = 0.085, AB = 0.173, AC = 0.320, BC = 0.092 which match the observed values, therefore pop. a is in H.-W. equilibrium.
Population b Allelic frequencies A = 0.878, B = 0.092, C = 0.034
Genotypic frequencies should be AA = 0.771, BB = 0.008, CC = 0.001, AB = 0.162, AC = 0.060, BC = 0.006 which are similar to the observed values, but BC is 150 % greater than it should be and AC is 15% less than it should be and AB is 4% less than it should be, therefore pop. b is NOT in H.-W. equilibrium.
Population c Allelic frequencies A = 0.502, B = 0.250, C = 0.245
Genotypic frequencies should be AA = 0.252, BB = 0.062, CC = 0.060, AB = 0.251, AC = 0.246, BC = 0.122 which are very different from the observed values, therefore pop. c is NOT in H.-W. equilibrium.
Population d Allelic frequencies A = 0.142, B = 0.000, C = 0.858
Genotypic frequencies should be AA = 0.020, BB = 0.000, CC = 0.736, AB = 0.000, AC = 0.244, BC = 0.000 which match the observed values, therefore pop. d is in H.-W. equilibrium.
Population e Allelic frequencies A = 0.748, B = 0.094, C = 0.154
Genotypic frequencies should be AA = 0.560, BB = 0.009, CC = 0.024, AB = 0.141, AC = 0.230, BC = 0.029 which match the observed values, therefore pop. e is in H.-W. equilibrium.

5) W_{nif __} = 1.0 .......... W_{daf daf} = 0.4

daf daf = 0.134 .......... nif nif = 0.743 .......... nif daf = 0.123
Alleles before selection .......... daf = 0.196 .......... nif = 0.804

Genotypic frequencies after selection
daf daf = 0.015 .......... nif nif = 0.662 .......... nif daf = 0.323

Phenotypic frequencies after selection
Day foraging = 0.015 .......... Night foraging = 0.985

Alleles after selection
daf = 0.176 .......... nif = 0.824

6) W_{B __} = 1.0 .......... W_bb = 0.4

Alleles before selection .......... B = 0.3 .......... b = 0.7

Genotypic frequencies after selection
BB = 0.127 .......... Bb = 0.595 .......... bb = 0.278

Alleles after selection
B = 0.424 .......... b = 0.576

7) Homozygous dominant = 1 % .......... Heterozygous = 18%

Quantitative Genetics and other questions
1) Height, number of leaves, % body fat, mycelial weight, number of eggs, % yield, lung cancer, intelligence, wood quality, etc.

2) Variance is...A) a measure of how a set of data is spread out about the mean.

3) Heritability is... B) The fraction of a population's phenotypic variation that is strictly do to genetic factors.

4) The narrow-sense heritability of the fruit color in strawberries is very close to 1. Which of the following is true? B) Most fruit color variance is additive genetic variance.

5 -- 1) T) Narrow sense heritability 2) H) Bottleneck 3) L) Discrete traits 4) G) Evolution 5) I) Heterosis 6) P) Gene pool 7) C) Multifactorial trait 8) G) Hardy-Weinberg equilibrium 9) M) Germ line 10) S) Transposase 11) U) Broad sense heritability

6) The true statements are listed below.
C) Most of the phenotypic variance results from environmental variance. E) The variance of population A is less than population C.

7) The true statements are listed below.
A) The narrow-sense heritability of the trait is close to 1. C) The heterozygosity of population A is greater than that of population B. E) The mean value of the trait of has increased.

ADDITIONAL QUESTIONS
1) B) Genetic drift C) Inbreeding D) Bottlenecks E) Selfing

2) True statements below.
B) Transposable elements can cause chromosome rearrangements. F) Heterosis is hybrid vigor. G) Populations in Hardy-Weinberg equilibrium are free of migration and sexual selection. I) Homeologous chromosomes rarely cross over. M) Dr. Barbara McClintock was the first to describe transposable elements. N) Transposons can cause deletions or insertions. P) Transposons end in inverted repeats. Q) The first transposable elements described were those in maize (corn). R) The haploid number (n) is the number of chromosomes found in gametes. S) Trisomy 21 causes Down syndrome. V) Reverse mutations turn mutant into wild type genes. X) Recombination often occurs at regions of DNA homology. Y) Branch migration is the movement of a crossover point between DNA complexes. aa) Branch migration is the movement of a crossover point between DNA complexes. cc) Gene duplications caused by unequal crossing-over have played an important role in gene evolution. dd) P elements can be used to transform Drosophila. ee) Ac and Ds are plant transposable elements.

3) Match the description on the left with BEST answer on the right by filling in the blanks.
1) C) Holliday model 2) F) Somatic 3) K) P-element 4) A) Site directed mutagenesis 5) M) Transposase 6) L) Hybrid dysgenesis 7) J) UV mutagenesis

4) D) Site directed mutagenesis

5) AA RR X aa rr
F₁ = All Aa Rr = All Low hovering

Aa Rr X Aa Rr
F₂ . . . 1/16 AA RR : 1/8 AA Rr : 1/16 AA rr : 1/8 Aa RR : 1/4 Aa Rr : 1/8 Aa rr : 1/16 aa RR : 1/8 aa Rr : 1/16 aa rr
F₂ = 1/16 High circling : 1/8 High hovering : 1/16 High non hovering : 1/8 Low circling : 1/4 Low hovering : 1/8 Low non hovering : 1/4 walking

Aa Rr X AA RR . . . . . . . . . . 1/2 AA : 1/2 Aa and 1/2 RR : 1/2 Rr
1/4 AA RR : 1/4 AA Rr : 1/4 Aa RR : 1/4 Aa Rr

Backcrossed F₁ = 1/4 High circling : 1/4 High hovering : 1/4 Low circling : 1/4 Low hovering

6) Aa Bb Ee X Aa Bb Ee

1/4 AA : 1/2 Aa : 1/4 aa ...and...1/4 BB : 1/2 Bb : 1/4 bb ...and...1/4 EE : 1/2 Ee : 1/4 ee

AA BB ee = White = 1/4 X 1/4 X 1/4 = 1/64
AA BB E__ = White with dots = 1/4 X 1/4 X 3/4 = 3/64
Aa BB ee = Yellow = 1/2 X 1/4 X 1/4 = 1/32
Aa BB E__ = Yellow with dots = 1/2 X 1/4 X 3/4 = 3/32
AA Bb ee = Orange = 1/4 X 1/2 X 1/4 = 1/32
AA Bb E__ = Orange with dots = 1/4 X 1/2 X 3/4 = 3/32
Aa Bb ee = Red = 1/2 X 1/2 X 1/4 = 1/16
Aa Bb E__ = Red with dots = 1/2 X 1/2 X 3/4 = 3/16
aa B__ ee + A__ bb ee + aa bb ee = Rose = (1/4 X 3/4 X 1/4) + (3/4 X 1/4 X 1/4) + (1/4 X 1/4 X 1/4) = 3/64 + 3/64 + 1/64 = 7/64
aa B__ E__ + A__ bb E__ + aa bb E__ = Rose with dots = (1/4 X 3/4 X 3/4) + (3/4 X 1/4 X 3/4) + (1/4 X 1/4 X 3/4) = 9/64 + 9/64 + 3/64 = 21/64

7) B) Most of the phenotypic variance results from environmental variance. C) The heterozygosity of population A is greater than that of population B. G) The broad-sense heritability of the trait is close to 0.

8) 1) I) Transition 2) F) Nonsense 3) E) Reverse 4) H) Transversion 5) G) Missense 6) J) Frameshift 7) B) Point 8) C) Deletion 9) A) Duplication 10) D) Inversion