Download Final Genetic Problems for IBO 2014 PART I In Drosophila

Final Genetic Problems for IBO 2014
PART I
1. In Drosophila melanogaster, mutant alleles b (black body), sc (scarlet eyes) and vg (vestigial
wings) are all recessive compared to wild type. A cross was made between a wild type fly with a
fly that have all three mutations (black body, scarlet eyes and vestigial wings). F1 produced from
this cross were all normal (wild type). F1 from this cross were subsequently crossed with flies that
have black body, scarlet eyes and vestigial wings. F2 produced from this cross is shown in the table
below.
Phenotype
Percentage
Normal (wild type)
20.5
Scarlet eyes
20.5
Vestigial wings
4.5
Black body
4.5
Scarlet eyes, black body
4.5
Vestigial wings, black body
20.5
Scarlet eyes, vestigial wings
4.5
Black body, scarlet eyes, vestigial wings
20.5
Indicate whether this statements is TRUE or FALSE based on the result of the cross
1. Gene for vestigial wings, scarlet eyes and black body are all located on the same chromosome.
2. On same chromosome, the distance between b locus and vg locus is farther than the distance
between b locus and sc locus
3. Drosophila with b+ b sc+ sc genotype is able to produce four different gametes from the
combination of b and sc alleles. Those four gametes are produced in the same number.
4. When Drosophila with genotype vg+ vg sc+ sc are crossed with each other, 18.75% of the progeny
produced from this cross will have vestigial wings.
2. In a mouse population, there are three different coat colors: Yellow (Y), Agouti (A) and Black (B).
To understand the inheritance of the coat colors, several crosses were made, and the results are
shown in the following table:
No
Crosses (Parental Phenotypes)
Ratio of F1 Phenotypes
1
YxY
2Y : 1A
2
YxY
2Y : 1B
3
AxA
All A
4
AxA
3A : 1B
5
BxB
All B
6
Ax B
All A
7
Ax B
1A : 1B
8
Y (F1 from cross 1) x A (F1 from cross 3)
1Y : 1A
9
Y (F1 from cross 1) x B (F1 from cross 5)
1Y : 1A
10
Y (F1 from cross 2) x A (F1 from cross 3)
1Y : 1A
11
Y (F1 from cross 2) x B (F1 from cross 5)
1Y:1B
Indicate whether the following statements are TRUE or FALSE based on the results
of the crosses.
1. Coat color in mouse is coded by one gene with multiple alleles. The dominance of the
alleles is yellow > agouti > black
2. You will never find a pure breed yellow mouse in the population.
3. If you cross a yellow mouse from cross 1 with a yellow mouse from cross 2, then the
ratio of the F2 mice will be 3:1 for yellow: agouti
4. If you cross a F1 yellow mouse from cross 2 with a F1 agouti mouse from cross 7, a
ratio of 1:2:1 (agouti: yellow: black) will be observed in F2.
3. In a particular plant species, height is the result of polygenic inheritance. Many genes contribute to
the plant’s height and the dominant allele of each gene has similar contributions in increasing the
plant height. For example, if aabb plant’s height is 5 cm and every dominant allele will contribute
2 cm increment in height, then the height of plants with genotypes Aabb and aaBb will be 7 cm.
A researcher found that the height of a particular plant species varies between 6 to 36 cm. When 6
cm and 36 cm plants were crossed, all plants were 21 cm. In the F2 population, a continuous range
of heights was observed. Most of the plants were 21 cm, and 1/64 of the plants have a height of 6
cm.
Indicate whether these statements are TRUE or FALSE based on the result of the above
experiment
1. Three genes are involved in determining the height of the plants
2. Six different phenotypes was observed in F2
3. There are seven possible genotypes for plants with a height of 21 cm
4. In F2, the number of 11 cm plants was similar to the number of 26 cm plants
4. Below is shown a pedigree of a family where some of its members has a genetic disease
(shown in black). The frequency of this disease in the population is 9% (assume the
population is in Hardy Weinberg equilibrium). Study the pedigree carefully:
1
2
8
7
4
3
5
6
?
109
Indicate whether these statements are TRUE or FALSE based on the mode
inheritance of this disease
1. The disease is caused by autosomal recessive allele.
2. Probability of 5 being heterozygous is 50%.
3. Probability of 3 being homozygous is 54%.
4. There is a possibility that 9 is homozygote.
5. Neurospora crassa has two mating types, a and α. Under certain environmental conditions,
these two mating types can be induced to mate (fuse into diploid cells). After fusion, they
undergo meiosis and four haploid spores are produced.
Mating can be induced by inoculating the medium with one strain at 25oC, and after three-four
days, the opposite strain is added. When the second strain is added into the medium, the first first
strain has already grown into larger cells. The first strain is larger and contribute all of its
cytoplasm and nucleus to the zygote, whereas the second strain contributes only its nucleus.
Neurospora is an aerobic microorganisme, it means that the energy is produced in mitochondria
by oxidative phosphorylation. Researchers have isolated mutants of Neurospora that have
mitochondria malfunction. Because of this mutation, the growth rate of the mutant Neurospora
are slower than normal (slow growing). Two of the mutants are poky and cyt. Mating between
poky, cyt with wildtype (wt) has different outcomes in their progenies (spores produced after
meiosis). The table below shows the result of the crosses.
No of
First
Second
Progeny
Cross
Parent
Parent
Normal
Growing
Slow
Growing
1
cyt
wt
200
200
2
wt
cyt
200
200
3
poky
wt
0
400
4
wt
poky
400
0
Indicate whether this statement is TRUE or FALSE based on the result of the experiments
6. Poky mutation is a mutation located on the gene in nuclear DNA
7. Cyt mutation is a mutation located on the gene in mitochondrial DNA
8. If you mate between cyt and poky, with cyt as first parent and poky as the second parent, 50%
of progenies are normal-growing and 50% are slow growing.
9. If you mate between cyt and poky, with poky as first parent and cyt as the second parent, all of
progeny will have slow-growing phenotype
11. In humans, alleles for MN blood type and hair texture phenotypes are co-dominant. A person
with SS genotype has straight hair, Ss genotype has wavy hair and a person will have curly hair
if the genotype is ss. A survey was conducted to find out the distribution of MN blood type and
hair phenotypes among a population in an island. The distribution of the phenotypes is shown
below:
No
1
2
3
4
5
6
7
8
9
Phenotypes
M straight hair
M wavy hair
M curly hair
MN straight hair
MN wavy hair
MN curly hair
N straight hair
N wavy hair
N curly hair
Number of
people in the
Population
25
50
25
12
25
13
13
25
12
Indicate whether the statements are TRUE or FALSE based on the result of the survey
1. Both gene (MN blood type and hair texture) were not in Hardy Weinberg Equilibrium
2. If both loci are in Hardy Weinberg Equilibrium, 64 % of population will have N blood type
and wavy hair
3. If both loci are in Hardy Weinberg Equilibrium, 47 people in the population will have MN
blood type and wavy hair
4. People in the population did not have a tendency to marry people with a certain hair texture.
12. The following is a list of mutational changes. For each of the specific mutations described
below, indicate whether the mechanism is TRUE or FALSE as possible cause of the
mutation
mutations
1
2
3
4
An AT base pair in the wildtype gene is changed to a
GC pair
The sequence AACGTCACACACACATCG is
changed to AACGTCACATCG
The gene map in a given chromosome is changed from
bog-rad-fox1-fox2-try-duf to bog-rad-fox1-mel-qui-txusqm
The sequence AAGCTTATCG is changed to
AAGCTTTATCG
possible cause
Deamination
Intercalator
Unequal crossing-over
Intercalator
13. Life at high altitudes requires special adaptations of the body. In a genome wide study Yi et al.
identified the Endothelial PAS domain-containing protein 1 (EPAS1) as a promising candidate
gene for such adaptations. For this study, Yi et al. only used genetic data.
Indicate whether the following conclusions can be obtained from Yi et al.’s study
1. EPAS1 increases the capability to take up oxygen at low partial pressure of oxygen.
2. Most of the genetic difference between people living at a low altitude and those living at a
high altitude do not have any measurable impact on fitness.
3. EPAS1 is under natural selection.
4. Tibetans (who live at altitudes higher than 4000MASL) have a different form of EPAS1 than
the Chinese living in lowland areas.
14. dN / dS is the ratio of the number of non-synonymous substitutions to the number of synonymous
substitutions sites in protein coding genes. Synonymous substitution means that the nucleotide
substitution will change the sequence of the nucleotide but won’t change the sequence of the
amino acid in the protein. Non-synonymous substitution means that the change of nucleotide
sequence also will change the amino acid in the protein.
This ratio is often used as an indicator of selective pressure on the protein-coding gene. There are
three types of selective pressure, positive, negative and neutral selection. Positive selection is
selection that favors the substitution in the specific environment (advantageous, superior than
wild type) and will be maintained in the population. On the other hand, negative selection is
selection when the substitutions will have a deleterious effect so that it will be eliminated from
the population.
Indicate this statement is TRUE or FALSE based on the usage of dN/dS ratio to determine
selective pressure on a protein coding gene
1. dN / dS ratio for genes that are under positive selection will be higher than 1
2. An example of a gene that undergoes negative selection is a gene that is responsible for
antibiotic resistance through a population of bacteria that are being exposed to that antibiotic
3. A conserved protein among species will have dN / dS below 1 and under negative selection
4. For all ratio of dN/dS, the selection process will always decrease genetic variation of the
population
15. There are two competing theories about how Homo sapiens arise. First theory was multiregional
hypothesis. This theory declared that Homo sapiens independently evolved in each region of the
world from H erectus that has migrated out of Africa. This hypothesis considers H. erectus to be
an early version of H. sapiens, and not a different species. The second hypothesis was Out of
Africa, this hypothesis declared that All H. sapiens evolved in Africa, and migrated out of Africa
about 100,000 years ago. .
Genetic analysis has been done using mitochondrial DNA to resolve this puzzle. The data from
mtDNA can be summarized into two main points below:
1. African people has greater mitochondrial genetic diversity compared with people from other
continent
2. mtDNA genetic variations in modern human populations are low.
Indicate whether the following statements are TRUE or FALSE based on the genetic data
to resolve the two competing theories about how modern human arises.
1. Mitochondrial DNA was chosen for analysis rather than nuclear DNA because it has more
conserved sequences and the gene didn’t recombine when it inherited across generations.
2. Out of Africa theory can explain the low mtDNA genetic variation in modern human
population because of small founding population of Homo sapiens.
3. Genetic flow between the populations can explain the low mtDNA genetic variation in
modern human population in the Multiregional theory.
4. The greater mtDNA genetic diversity among African people support Out of Africa theories.
PART II
1. A test cross between a wild type trihybrid corn plant and a pure-breeding corn plant that has recessive
phenotypes (plant-color booster, liguleless, and silkless) produces progeny as follows:
where B, L, and S denote dominant alleles for plant-color booster, ligule and silk, respectively, and b,
l, and s denote corresponding recessive alleles.
Each of these three characteristics is controlled by a single gene that has two possible alleles, and all
three genes are on the same chromosome.
Indicate the following statements as TRUE or FALSE based on the result of the cross.
1.
2.
3.
4.
The genotype of the wild type parent plant was BSL / bsl (“ / ” indicate different chromosome)
Recombinant frequency between B locus and L locus was 37.18 %
S locus is located between L locus and B locus
S locus is closer to the L locus than B locus
2. There are two flower colors in plant Y, red and white. Three different crosses between pure
breeds with white flowers were performed. F1 produced from these crosses were all red flowers.
When F1 was crossed with each other, the ratio in F2 was 9 red: 7 white. Summarized results
was tabulated on the table below
No of Cross
Cross
F1
F2 (F1 x F1)
1
White 1 x White 2
All Red
9 Red : 7 White
2
White 2 x White 3
All Red
9 Red : 7 White
3
White 1 x White 3
All Red
9 Red : 7 White
Indicate this statement is TRUE or FALSE based on the mode of inheritance of flower color
in plant Y
1. From result of the cross, it can be concluded that there are two genes which control flower
color in plant Y.
2. F1 plant (red plant) was heterozygous on all the flower color genes.
3. There are 8 possible genotypes for plant with red flower color
4. If F1 from cross 1 is crossed with F1 from cross 3, then the ratio of the progeny will be 13 : 3
for white : red
3. Five strains of Escherichia coli mutants (mutants 1 – 5) have one of the following mutations
affecting the lac operon were isolated. Mutant 1 has a mutation in the lacY gene, but the
mutations in the other mutants are unknown. The mutations in mutants 2 – 5 could be:
1.
2.
3.
4.
Nonsense mutation in lacZ : nonfunctional β-galactosidase is produced.
lacOc mutation: the repressor cannot bind to the operator.
Promoter mutation: RNA polymerase cannot bind to the promoter.
Super-represor mutation: lactose cannot bind and inactivate the repressor
The lac operons from each mutants were inserted into a wildtype and mutant E. coli, creating
merodiploid cells (the cells contained two copies of the lac operon). The ability of the
merodiploid cells to grow on lactose-containing medium was assayed. + indicated the ability
to grow and – indicated that no growth was observed. Growth requires functional LacZ and
LacY.
Host cell
Inserted
lac
operon
1
2
3
4
1
2
3
4
5
Wildtype
-
+
+
-
-
+
+
+
+
+
+
-
-
-
+
-
-
-
-
5
+
Indicate this statement is TRUE or FALSE based on the type of mutation in mutant 2 - 5
1. Mutation in mutant 2 was LacOc mutation
2. Mutation in mutant 3 was Promoter mutation
3. Mutation in mutant 4 was Super-repressor mutation
4. Mutation in mutant 5 was LacZ-Nonsense mutation
4. A geneticist crossed two highly inbreed strains of a plant (P1 x P2) to get F1. Then the F1 was
crossed to each other to produce F2. From each generation, he calculated the average of the stem
height and its variance. The stem height phenotype is determined both by polygenic inheritance and
environmental factor. He maintained P1, P2, F1 and F2 on the same location. Some data missing
from his notes are shown by A and B.
Generation
Stem Height
Stem Height
(average)
(Variance)
P1
60.5 cm
4.3 cm
P2
85 cm
4.2 cm
F1 (P1 x P2)
A (data lost)
5.1 cm
F2 (F1 x F1)
B (data lost)
25.4 cm
Indicate this statement is TRUE or FALSE based on the result of the experiment
1. Values in A and B are approximately similar
2. Variances in F1 generation is caused both by genetic and environmental influences
3. In F2, environmental factor has greater effect on stem length variation compared with genetic
factor
4. It is likely that the location where the plant maintained was in a greenhouse (stable and
homogenous habitat) than in a hill (diverse habitat)
5. Pathways for amino acid synthesis in bacteria were worked out in part by using cross-feeding
experiments among mutants that were each defective for a single step in the pathway. There were three
mutants of tryptophan biosynthesis pathway, TrpB-, TrpD- and TrpE-. Each mutant has a mutation in
one step of the tryptophan biosynthetic pathway (the pathway was shown below). These three mutants
were streaked on specific regions of the petri dish and allowed to grow briefly in the presence of a very
small concentration of tryptophan, forming pale streaks. After tryptophan was used up from the media,
the mutants must synthetize tryptophan by themselves. Some regions of the streak were thicker (heavier
growth) than others (See figure below). This heavier growth indicated that one mutant could cross-feed
(supply metabolic intermediate) to another mutant..
TrpE
-
TrpD
-
-
Diagram of Tryptophan biosynthetic pathway
Chorismate
Anthranilate
Indole
Tryptophan
Indicate whether these statements are TRUE or FALSE based on the nature of the
mutations in TrpB, TrpD and TrpE:
1.
2.
3.
4.
The experiment can be done if all of the metabolic intermediate accumulated inside the cell
TrpD- has a mutation in the enzyme that catalyzes the conversion of Indole into Tryptophan
TrpE- has capability to synthetize tryptophan if the media contains Anthranilate or Indole
Indole will be accumulated in media if TrpB synthetize tryptophan using chorismate as
starting material
6. The pedigree below shows the inheritance of a rare genetic disease (inherited by autosomal
dominant allele) in a family. Locus for this disease is unknown. A genetic consultant is curious
whether the inheritance of this disease is linked with a certain genetic marker (microsatellite).
This microsatellite has been known to be located on chromosome 5 in humans and has two main
allele variants: allele 1 and allele 2. The pedigree shows the inheritance of the disease (dark means
the person has the disease) and the microsatellite marker associated with it in each person (1/2)
Indicate whether these statements are TRUE or FALSE based on the inheritance of the
disease and the microsatellite marker
3.
4.
5.
6.
It is possible that the gene for the disease is also located on chromosome no 5
If a person has the disease, he/she would always have microsatellite allele 1 variant
In the 4th generation, only one child obtained a recombinant chromosome from his/her father
From the 3rd and 4th generation, it could be concluded that the recombination frequency
between the gene of the disease and the microsatellite was 3/1
7. The locations of six deletions have been mapped to the Drosophila chromosome as shown in
the following diagram.
Recessive mutations a, b, c, d, e and f are known to be located in the same regions as the
deletions, but the order of the mutations on the chromosome is not known. When flies
homozygous for the recessive mutations are crossed with flies homozygous for the deletions,
the following results are obtained, where the letter “m” represents a mutant phenotype and a
plus sign (+) represents the wild type.
Mutations
Deletion
a
b
c
d
e
f
1
m
+
+
m
+
m
2
+
+
+
m
+
m
3
+
m
+
m
+
+
4
+
m
m
+
m
+
5
+
m
m
m
+
+
6
+
+
m
+
m
+
Indicate this statement is TRUE or FALSE based on possible relative orders of the six
mutant genes on the chromosome.
1. Location of the deletion in mutation d is in the overlapping area between deletion 4 and
deletion 6.
2. Location of the deletion in mutation b can be deduced using information of overlapping area
between deletion 3, 4 and 5.
3. Among the six mutations, mutation d has the shortest deletion.
4. The relative order of the six mutations on the chromosome is a, f, d, b, c, e.
8. Suppose that the horse’s color is determined by one gene which has two alleles B and b. B allele
is dominant over b and expresses brown color, while b allele is recessive and codes for black
color. There were two populations of horses in two separate locations. In population 1, the
allele frequency of B was 0.5, while in population 2 it was 0.2. The size of population 1 was
five times larger than population 2. Initially, both populations were in Hardy Weinberg
Equilibrium. Then, the two populations joined into one unified population. After both
populations were joined, the unified population remained stable. 10 years after that, 1000 baby
horses were born from this unified population.
Indicate whether the following statements are TRUE or FALSE based the information
above
1. The phenomenon is example of genetic drift
2. 10 years after migration, the allele B frequency is higher than frequency allele b.
3. In the unified population, the probability that mating between brown horses will produce
black horses is 12,6 %
4. Among those 1000 newborn horse-babies, 698 of them were brown horses.
9. Figures (A, B, and C) illustrate the results of a series of computer simulations of changes in
allele frequencies in a group of populations due to chance alone (genetic drift). The simulations
were done in different population sizes. X axis indicates the number of generation, Y axis
indicates allele frequency and different colors indicate the repetition of simulations.
Simulation on Population I
Simulation on Population II
Simulation on Population III
Indicate whether the following statements are TRUE or FALSE based on the result of genetic
drift simulation:
1. Among three simulations, simulation III was done on the smallest population size while
simulation I was done on the largest population size.
2. Time for fixation of a new allele in population is shorter in population I than in population III.
3. Probability that new mutation will be lost in a population will be higher in population III than
in population I.
4. Inbreeding rate will be higher in population III compared to population I.
10. Robertsonian races are races within a species which have differing chrosome numbers due to the
so-called Robertsonian fusion of chromosomes. This combines two acrocentric chromosomes by
their centromeres to a single metacentric chromosome. Thus Robertsonian fusions reduce the
number of chromosomes in the karyotype. In the picture, house mice chromosome 1 and 15 and
combines to form a new chromosome, called Rb(1.15).
The Robertsonian fusion might cause problems during meiosis. As the fusion heterozygotes have
high chances of producing non-viable offspring (see figure below), their fitness is normally lower
than either of the homozygotes. Normally these forms are found only in the hybrid zones of
different Robetsonian races.
Indicate whether these following statements are TRUE or FALSE based on the
information of Robertsonian races:
1. Robertsonian fusion could lead to lower fitness by causing new gene linkages.
2. New races evolving by the Robertsonian fusions must have higher fitness for the new
karyotype to be fixed.
3. The broader the hybrid zone, the higher is the difference in the fitness between homozygotes
and heterozygotes.
4. Two Robertsonian races would be more strongly reproductively isolated than Robertsonian
and wild-type race.
Backup Problems
1. Speciation is the process of the formation of new species. Speciation involves several stages. Initially,
individuals in a population can mate with each other, then a barrier appears that separate the population
(it can be geographic or non-geographic barrier) that prevents the separated population to mate with one
another. Along with time, each population will accumulate different genetic makeup, so that when the
separated populations are mixed again, they cannot mate or produce fertile offsprings. The following
picture shows four possible speciation mechanism:
Mechanism 1
Mechanism 2
Mechanism 3
Mechanism 4
Indicate this statement is TRUE or FALSE based on different mode of speciation:
1. Only 1 mechanism from the above four mechanisms of speciation involve geographic barrier.
2. A small population of fruit fly move into an island. This population brings rare allele from the original
population and get fixated in the island because of genetic drift. No gene flow occurs during allele
fixation. After sometime, the island population and original population, when mixed together, cannot
produce fertile offsprings. This case is an example of “Mechanism 2” speciation.
3. A particular plant was distributed uniformly near mines. The soil near mines has been contaminated by
heavy metal. Because of this contamination, plants near the mines evolved different flowering time
compared to plants located further away from the mines, which were not exposed to the contamination.
This case leads to “Mechanism 3” speciation.
4. Polyploidy can only be caused by “Mechanism 4” speciation
Answer & Explanations
1.
2.
3.
4.
False : Mechanism 1 & 2 involve geographical barrier
True
True
False
Reference:
Reece et.al. 2009. Campbell Biology 9th edition. Pearson. Chapter 24 Concept 24.2 Page 493-498.
2. A man with a certain disease marries a normal woman. They have eight children (four boys and four
girls); all of the girls have their father’s disease, but none of the boys do.
Indicate whether the following statements is the possible mechanism inheritance of the disease:
1. The disease appears to be autosomal, because the trait is inherited in a sex-specific way.
2. It is Y linked, because females are affected.
3. The disease is X-linked dominant. All daughters and none of the sons will be affected if the father has
an X-linked dominant trait.
4. The disease is X-linked recessive. If the mother is a carrier, she would pass the gene to all her daughters
and none of her sons.
Answer
The disease is X linked dominant since the mate between diseased male with normal female resulted in all
daughter has disease but none of the son has the disease.
1.
2.
3.
4.
False
False
True
False
References:
Reece et.al. 2009. Campbell Biology 9th edition. Pearson. Chapter 14 Concept 14.4 Page 275-276.
3. Humans, rats, and mice female individuals, which are heterozygous for the mutant gene Tfm (Testicular
feminization locus), have normal phenotype. However, the half of the individuals with male genotype is
sterile and has female phenotype. Females homozygous for the mutant gene are normal and fertile. The
disorder is caused by the lack of a specific protein that serves as a testosterone receptor.
Indicate if each of the following statements is true or false
A. Normal sexual growth of female can occur without the product of this gene- testosterone
B. Mutant males are not able to synthesize the testosterone receptor
C. Mutant males are not able to synthesize the testosterone
D. The gene Tfm (Testicular feminization locus) is inherited as sex-linked gene.
Answer & Explanations:
1. True
2. True
3. False
4. True
References: Proposed by Armenia, No reference
5. A fish population had been separated to form two populations in isolated freshwater ponds.
One population of fishes lived in a predator-rich pond and the fishes could swim in short, fast
bursts. The other population lived in a predator-poor pond and the fishes could swim
continuously for a long time. When fishes from both populations were placed together in the
same body of water, the females from each population exhibited exclusive breeding
preferences.
Indicate whether the following statements are TRUE or FALSE based on the above information:
1. Habitat selection was the reason why the different population of fishes could not mate with one another,
even after the fishes from both populations were placed in the same body of water.
2. Sexual selection could increase the divergence between both fish populations.
3. To maintain the capability of interbreeding between both populations, a canal linking the ponds of both
populations must be build, where the fish can pass through the canals, but not the predators.
4. The capability of interbreeding between the populations could be maintained if females from one
population were regularly moved into the other populations and vice versa (reprocical transfer)
Answer & Explanations:
1. False : The causes was behavioral selection
2. True
3. True
4. False
References:
Reece et.al. 2009. Campbell Biology 9th edition. Pearson. Chapter 24 Concept 24.1 Page 488-492.
6. There are two dominant alleles, TH and TV and one recessive allele t from one single gene.
Indicate whether the following statements are TRUE or FALSE based on the information
above:
1. There can be 4 diffferent phenotypes
2. There can be 6 possible genotypes
3. There can be 3 different phenotypes
4. There can be 4 possible genotypes
Answer & Explanations
1. True : There are 4 possible genotype, TH, THTV, TV and t
2. True : There are 6 possible genotype: THTH, THTV, THt, TVTV, TYt, tt
3. False
4. False
Reference:
Reece et.al. 2009. Campbell Biology 9th edition. Pearson. Chapter 14 Concept 14.3 Page 272-273.