Mendelian Genetics & Inheritance Patterns Mendelian Genetics

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Mendelian Genetics & Inheritance Patterns
Mendelian Genetics
Classwork
1. Sexual reproduction takes a very significant toll on those species that utilize this
process. What is the benefit that sexual reproduction offers for a species?
2. What is the relationship between traits and heredity?
3. Explain the concept of ‘Blending Inheritance.’ Was this theory supported or
rejected by Mendel and his work?
4. Explain, in terms of the theory of ‘Inherited Characteristics,’ how an anteater has
developed a long nose.
5. Identify the characteristics that made pea plants ideal for Mendel to use as a
model organism.
6. Explain the process by which Mendel was able to control his breeding between
pea plants.
7. What is a phenotype?
8. Pick three traits Mendel chose to observe and identify two potential phenotypes
for each trait.
9. What is ‘true breeding?’
10. Identify the ratio of offspring for the F1 and F2 generations in a monohybrid cross
involving parents with white and purple flowers.
11. Identify the four rules that Mendel established following his studies of the
monohybrid cross process.
12. What is it called when an organism possesses two different alleles for a trait?
13. Explain how the ratio of the F2 generation in Mendel’s monohybrid crosses
helped him to determine the existence of dominant and recessive alleles.
14. What is the ‘Law of Segregation?’
15. Identify the relationship between meiosis and Mendel’s patterns of inheritance.
16. Suppose one of Mendel’s pea plants had experienced a non-lethal form of
trisomy for one of the characteristics he was studying. Explain how this could
have complicated Mendel’s observations if this plant were part of a parent
generation in one of his monohybrid crosses.
17. Suppose Mendel had made his observations a few decades later when there as
more widespread knowledge of chromosomes throughout the scientific
community. How do you suppose this may have impacted the development of
his theory?
Homework
18. ______________________ is widely considered to be the ‘father of modern
genetics’ because of the role that he played in the development of this field.
19. What is the concept of ‘Inheritance of Acquired Characteristics?’ Explain this
theory and identify the individual who developed the theory.
20. What is a ‘model organism?’ Identify Mendel’s model organism.
21. Why was it essential that Mendel was able to control breeding between his pea
plants?
22. Identify the details regarding the phenotypes investigated by Mendel that
contributed to the ideal nature of his model organism.
23. What is a ‘monohybrid cross?’ Explain the steps involved in this process and
identify the names of the generations involved.
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24. Compare the phenotypes of the F1 generation with the phenotypes of the P
generation in a monohybrid cross.
25. What did Mendel conclude regarding inheritance of alleles from his monohybrid
cross experiments?
26. What is it called when an organism possesses two copies of the same allele for a
given trait?
27. Why did Mendel never discuss chromosomes in the development of his theories
on genetics?
28. What is the relationship between homologous chromosomes and allele pairs
described in Mendel’s theories of inheritance?
29. What is a ‘heritable factor?’
30. What is the difference between a homozygous and heterozygous allele
combination?
Punnett Squares
Classwork
31. Explain the purpose of a Punnett Square.
32. What is the difference between a genotype and a phenotype?
33. Identify the genotypes of the P generation involved in Mendel’s monohybrid
crosses.
34. Suppose large ears (L) are dominant over small ears in elephants. Cross an
elephant with small ears with a heterozygote elephant. What are the genotypic
and phenotypic ratios?
35. Suppose broad stripes (B) are dominant over skinny stripes in tigers. If two
heterozygote tigers produce a cub, what is the percent chance that the cub will
have skinny stripes?
36. Suppose the allele for blue color (B) is dominant to the allele for yellow in a
flower species. What combination of parent genotypes could potentially yield
yellow offspring?
37. Are offspring genotypic and phenotypic ratios always the same? Explain your
answer.
38. What is the purpose of a test-cross?
39. In Mendel’s pea plants, tall was the dominant allele for plant height. Perform a
test cross to determine the unknown genotype of a tall pea plant. Provide both
possible outcomes.
40. Why does it make sense that test-crosses are more accurate and reliable
involving species that reproduce with multiple offspring at one time?
41. What is the difference between a monohybrid and dihybrid cross?
42. Suppose in eagles, short wings (S) are dominant over long wings (s) and sharp
talons (T) is dominant over dull talons (t). Cross an eagle heterozygous for both
traits with an eagle homozygous recessive for both traits. Provide genotypic and
phenotypic ratios for offspring.
43. Suppose in lions, long tail (L) is dominant to short tail and solid coat (S) is
dominant to mottled coat. Cross a lion homozygous recessive for tail length and
heterozygous for coat with a lion heterozygous for tail length and homozygous
dominant for coat. Provide genotypic and phenotypic ratios.
44. Explain, in terms of probability, why the chances of producing two organisms that
are genetically identical decreases as the number of genes in a genome
increase.
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45. Use the rule of multiplication to determine the following: Suppose you cross an
organism with the genotype AaBbCc with an organism with the genotype aabbcc.
What is the probability that the offspring will have the genotype aabbcc?
46. Use the rule of addition to determine the following: Suppose you cross two pea
plants heterozygous for flower color (remember purple is dominant to white).
What is the probability the offspring will be a heterozygote?
47. Use the rule of addition to determine the following: Suppose you cross two pea
plants heterozygous for flower color (remember purple is dominant to white).
What is the probability the offspring will be purple?
48. Use the rules of multiplication and addition to determine the following: Suppose
you cross two individuals with the following genotypes: AABbCC x aaBBCc.
What is the probability that you will produce an offspring with genotype AaBBCC
or AaBbCc?
Homework
49. In developing a Punnett Square, what do the letters actually represent?
50. In Mendel’s monohybrid cross, which allele is expressed in a heterozygous
genotype?
51. In Mendelian traits, is it always possible to determine the genotype of an
organism with a dominant phenotype? Explain your answer.
52. Suppose in pine trees long needles (L) are dominant to short needles. Cross a
homozygous dominant tree with a short-needled tree. Provide genotypic and
phenotypic ratios.
53. Suppose in great white sharks that a tall dorsal fin (T) is dominant to a short
dorsal fin. Cross a homozygous dominant shark with a heterozygous shark.
Provide genotypic and phenotypic ratios.
54. Suppose in panda bears small ears (S) are dominant over big ears. You work at
a zoo, and need to determine the genotype of your small eared panda. Explain
how you could potentially determine the unknown genotype.
55. Explain how dihybrid crosses helped Mendel to develop his theory of
independent assortment.
56. Suppose in largemouth bass, green skin color (G) is dominant over olive (g) and
broad stripe (B) is dominant over thin (b). Cross two bass that are heterozygous
for both traits, and provide phenotypic ratios.
57. Suppose in spruce trees, long needles (L) are dominant over short needles (l)
and thick bark (T) is dominant over thin (t). Cross a spruce tree heterozygous for
needle length and homozygous dominant for bark thickness with a tree
homozygous dominant for needle length and homozygous recessive for bark
thickness. Provide genotypic and phenotypic ratios for offspring.
58. Use the rule of multiplication to determine the following: Suppose you cross two
organisms with the genotype AaBbCc. What is the probability the offspring will
have the same genotype?
59. Use the rule of addition to determine the following: Suppose you cross two pea
plants heterozygous for flower color (remember purple is dominant to white).
What is the probability the offspring will be a homozygote?
60. Use the rules of multiplication and addition to determine the following: Suppose
you cross two individuals with the following genotypes: AaBbCc x AaBbCc. What
is the probability that you will produce an offspring with genotype AaBbCc or
AAbbCc?
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Non-Mendelian Inheritance
Classwork
61. Provide one example in which the patterns of inheritance observed by Mendel
would not apply
62. How many phenotypes may exist for a trait that possesses two allele options?
63. Suppose Mendel’s pea plants demonstrated incomplete dominance, with pink
being the intermediate phenotype. What would the genotypic and phenotypic
ratios have been in Mendel’s F1 generation?
64. Suppose in crocodiles, the tooth length trait exhibits incomplete dominance.
Cross a long toothed (L) crocodile with a short toothed crocodile. Provide ratios.
65. In horses, red hair (R) is codominant with white hair (W). The codominant
phenotype is a roan coat. Cross a red stallion with a roan mare. Provide ratios.
66. What is pleiotropy?
67. Explain the principle of polygenic inheritance.
68. Why is height in humans considered to be a polygenic trait?
69. In humans there are three alleles for blood type: A, B and O. A and B are
codominant, and O is recessive. Cross a type AB parent with an AO parent,
provide potential genotypic and phenotypic ratios for offspring.
70. Refer to the question above. What is the probability that the parents would
produce a child with type AB blood?
Homework
71. Explain the principle of incomplete dominance.
72. In snapdragons, incomplete dominance results in a pink flower. Cross a red
flower (R) with a pink flower. Provide ratios.
73. Explain the principle of codominance.
74. Identify and explain the difference between codominance and incomplete
dominance.
75. Explain why sickle cell anemia would be considered to be a pleiotropic trait.
76. Why is skin color in humans considered to be a polygenic trait?
77. Consider a genetic mutation in which a single gene creates multiple effects in an
afflicted individual. Would this be considered a pleiotropic or polygenic situation?
78. Suppose parents with type A blood had a child with type O blood. Explain how
this situation could potentially occur.
79. In rabbits, coat color is a trait that demonstrates complete dominance, but has
multiple alleles. Brown coat is dominant to all other alleles, Black coat is
dominant to Gray coat. Cross a homozygous Brown rabbit with a heterozygous
black rabbit. Provide ratios.
80. Explain, using examples of non-Mendelian inheritance, why Gregor Mendel may
have had a more difficult time developing his rules for inheritance if his pea
plants were not as straightforward in their hereditary patterns.
Inherited Disease
Classwork
81. What genotype would an individual who suffers from a recessive inherited
disease have to exhibit? Explain.
82. Would you be able to identify a carrier of a genetic disease by examining
phenotype? Why or why not?
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83. Suppose two parents who were carriers of a recessive genetic disorder on a
single allele decided to have children. What is the percent chance that their child
could be born with the disorder?
84. Explain why consanguineous marriages may result in a higher rate of rare
inherited disorders.
85. Can an individual be a carrier for a dominant genetic disorder? Explain your
answer.
86. What is a sex-linked disorder?
87. Suppose a female is a carrier for a sex-linked disorder that her husband does not
have. What is the percent chance that she will have a child with the disorder?
Does the gender of the child matter in this situation?
88. Suppose a mother with an X-linked dominant genetic disorder has a child with a
male who does not exhibit the disorder. What is the likelihood that their child will
have the disorder? Does the gender of the child matter in this situation?
89. What is the role of a genetic counselor in the process of family planning?
90. Briefly explain the process of amniocentesis.
91. Explain, in terms of genotype versus phenotype, how ultrasounds and
amniocentesis are used to predict the occurrence of genetic disorders.
Homework
92. Define carrier, in terms of inherited diseases.
93. Suppose parents who are carriers of a recessive genetic disorder on a single
allele decided to have a child with a parent who exhibits the same disorder.
What is the percent chance that their child could be born with the disorder?
Does this percentage change if they had a second child? Explain your answer.
94. Huntington’s disease is a dominant genetic disorder. Would an individual
heterozygous for the Huntington’s allele demonstrate the disease? Explain your
answer.
95. Suppose a father, who is heterozygous for Huntington’s disease, has a child with
a mother who does not have the disease. What is the percent chance that the
child will have the disease?
96. Identify the combination of sex chromosomes exhibited by a male and a female.
97. Why do males typically exhibit x-linked disorders more often than females?
98. Suppose a father with an X-linked recessive disorder has a child with a woman
who does not have the disorder. What is the percent chance that the child will
have the disorder? Does gender matter in this situation?
99. Explain why, in relationship to sex chromosomes, most of the genes that are
critical for survival are found on the X chromosome instead of the Y
chromosome.
100.
Briefly explain the process of chorionic villi sampling. What may be the
advantage to this process over amniocentesis?
101.
What methods can be used to mitigate the symptoms of a child born with
phenylketonuria?
102.
Why are amniocentesis, CVS and fetoscopy generally only performed if
the risk of genetic disease is deemed to be relatively high?
Pedigrees
Classwork
103.
Why are pedigrees useful in studying genetic disorders?
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104.
If two individuals in a pedigree do not have an autosomal recessive
disorder, but one of their four children does have the selected disorder, what are
the genotypes of the parents? How do you know?
105.
Suppose you are using a pedigree to study a sex-linked recessive genetic
disorder. If a female demonstrates this disorder, what is her genotype? What
can you infer about her father’s genotype?
106.
Refer to the pedigree. Is the studied disorder a recessive or dominant
disorder? How do you know?
107.
Refer to the pedigree. What can you infer about the genotype of
individual II-1? How do you know?
108.
Refer to the pedigree. Suppose this a sex-linked disorder. What can you
infer about the genotype of individual I-1? Explain your answer.
Homework
109.
When studying a pedigree, how can you tell if a trait is autosomal
dominant or autosomal recessive?
110.
Suppose you are studying a pedigree for a genetic disorder that only
shows up in male individuals. What details may you be able to infer about this
disorder?
111.
Suppose you are using a pedigree to study a sex-linked disorder. You
come across a situation where a female has the disorder, but her father did not.
Explain how this could occur.
112.
Refer to the pedigree. Could this be an autosomal dominant disorder?
Explain your answer.
113.
Refer to the pedigree. If this is an autosomal dominant disorder, what is
the genotype of II-4?
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114.
Refer to the pedigree. If this is an autosomal recessive disorder, what is
the genotype individual II-1?
Free Response
1. The results of Mendel’s experiments allowed him to develop four (4) hypotheses
regarding inheritance. Although we have added greatly to this knowledge, his
understandings still hold true. Explain each hypothesis in 1-2 sentences.
a. Hypothesis #1: Alleles.
b. Hypothesis #2: Two alleles.
c. Hypothesis #3: Dominant and recessive
d. Hypothesis #4: Law of Segregation
2. Two mice heterozygous for the long tail gene are crossed (P generation). Let T =
long tail and t = short tail. These mice have 18 offspring, some which have long
tails and some that have short tails.
a. Draw a Punnett Square illustrating the cross of the parental generation)
b. We want to select out the mice with long tails, but only the ones with the
genotype of TT. Which mice will you cross?
c. What possible results will you get from your cross? (show your Punnett
square(s)).
d. What result do we get if we have a true breeding long tailed mouse?
e. What do we call this type of experimental crossing?
3. Two dihybrid plants are crossed. Let R = allele for red flowers; r = allele for white
flowers; T = allele for tall plants; t = allele for short plants. The inheritance of
these alleles exhibits complete dominance.
a. What is the probability for offspring that are RRTT: (show all work)
b. What is the probability for offspring that are RrTt: (show all work)
4. Two trihybrid plants are crossed. Let R = allele for red flowers; r = allele for white
flowers; T = allele for tall plants; t = allele for short plants; B = big flowers; b =
small flowers. The inheritance of these alleles exhibits complete dominance.
a. What is the probability for offspring that are RRTTBb: (show all work)
b. What is the probability for offspring that are RrTTBB: (show all work)
5. Male patterned baldness cannot be determined by observing the hair pattern of a
man’s father. This is because this trait is a recessive sex-linked trait. Using the
letter B to represent the not bald allele and b to represent baldness, allele
respond to the prompts below
a. Because this is a sex-linked trait, what will be the symbol you will use for
a female who is a carrier for male patterned baldness?
b. Because this is a sex-linked trait, what will be the symbols you will use for
a male who is not bald?
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c. Show the cross between such a female and a male, using a Punnett
square.
d. What is the percent probability that this couple will have a child with male
patterned baldness?
e. What is the percent probability that, given they have a male child, the
child will have male patterned baldness.
f. What is the percent probability that, given they have a female child, the
child will have male patterned baldness.
6. A pedigree is shown below. Refer to this pedigree to respond to the prompts
below.
https://migrc.org/Library/Sexlinked
a. Is the X-linked trait indicated by the darkened circles and squares, dominant or
recessive?
b. For each individual within the pedigree (1-15) write in the line below the genotype
(don’t forget the Xs). Use the letters B and b to indicate the allele. (there are two
possibilities for individuals 10 and 12)
c. For the man labeled as #11 above, if he has children with a woman who does
not exhibit the x-linked trait, what is the percent probability they will have a child
who will exhibit the trait?
d. For the man labeled as #11 above, if he has children with a woman who does
exhibit the x-linked trait, what is the percent probability they will have a child who
will exhibit the trait?
Mendelian Genetics Answer Key
1. Sexual reproduction offers genetic diversity and variability, which allows species
and individuals to respond to environmental changes with greater success and
flexibility.
2. Traits are characteristics possessed by individuals, heredity is the passing of
these traits from one generation to another.
3. ‘Blending Inheritance’ was the idea that the traits of offspring were a direct result
of the ‘blending’ of traits exhibited by their parents. Mendel disproved this theory.
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4. According to ‘Inherited Acquired Characteristics,’ anteaters stretch their nose to
access termite mounds and anthills, and in doing so, grow a bigger nose. The
genes for having a longer nose will now be passed to the offspring of these
anteaters, as a result of the physical stretching used to access food.
5. There are many different varieties of easily observable traits. There are two
alleles for each trait. Self-pollination is possible. Cross-pollination is possible.
Control mating, easy to grow, and fast generation time.
6. Mendel removed the stamen (male reproductive organs) from the plant to be
pollinated- then he would add desired pollen to the carpels (female reproductive
organs).
7. Phenotype is the physical characteristics exhibited by an organism
8. Flower color- purple or white; flower position- terminal or axial; seed color- yellow
or green; seed shape- round or wrinkled; pod shape- inflated or wrinkled; pod
color- yellow or green; plant height- tall or short
9. True breeding is allowing a plant to self-pollinate for a few generations in order to
insure that they produce offspring identical to themselves
10. F1- 1:0 or 100% purple; F2- 3:1 or 75%-25% purple:white
11. A. Alternative versions exist for each gene; B. All organisms have two alleles for
each gene; C. There are dominant and recessive alleles; D. Each gamete carries
only one allele for each gene.
12. Heterozygous
13. Since Mendel saw the recessive phenotype show up in the F2 generation, he
realized that this allele must have been present in the F1 generation, but was
masked by the other allele present, the dominant allele.
14. The Law of Segregation is that alleles separate during meiosis, resulting in
gametes that only carry a single allele for each gene.
15. Most of Mendel’s patterns of inheritance are a result of the process of meiosis.
16. This could have made it difficult for Mendel to refine his hypothesis because the
plant would carry three alleles for a particular trait, which could potentially alter
his offspring ratios. However, because the pea plants demonstrated complete
dominance, the third allele would not make as much of a difference as it may in
other species with more alleles and variable patterns of inheritance.
17. Mendel’s ideas would have been even more complete than they already were.
His accuracy was quite remarkable without this knowledge, but if Mendel had a
better understanding of what his ‘heritable characteristics’ were, his hypotheses
would have been even further developed.
18. Gregor Mendel
19. This is the idea that the physical characteristics that an individual acquires over
its lifetime may be passed to its offspring. Jean-Baptiste Lamarck pioneered the
ultimately incorrect theory.
20. An individual/species that is ideal for experimentation because they possess
certain characteristics: short-lived, inexpensive, short generation time, many
offspring, etc. Mendel used pea plants.
21. If breeding were random, Mendel would not know which pea plants were
fertilizing each other, and he would not have been able to narrow down his ratios
and results of certain breeding pairs, which was the overall success of his
experimentation.
22. The pea plants exhibited seven easily observable traits, for which two versions of
each trait existed.
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23. A cross in which two individuals which exhibit different alleles for a specific trait
are involved.
24. The F1 generation all showed the phenotype of just one of the P generation
individuals.
25. Each plant inherited two alleles for each trait, one of which was dominant, one of
which was recessive
26. Homozygous
27. The existence and role of chromosomes was not known to Mendel.
28. Homologous chromosomes contain the different alleles for various genes
29. A gene
30. Homozygous means two of the same allele; heterozygous contains two different
alleles.
31. A Punnett Square allows one to predict the potential outcomes of a genetic
cross.
32. A genotype is the combination of alleles possessed by an individual, a phenotype
are the physical characteristics displayed as a result of the genotype.
33. Genotypes were homozygous dominant and homozygous recessive
34. Genotype: Ll1:ll1 or 50%Ll:50%ll; Phenotype: Long 1:Small 1
35. 25%
36. bb-homozygous recessive
37. No, because in situations that exhibit complete dominance, homozygous
dominant and heterozygous genotypes will produce a dominant phenotype
38. A test-cross allows you to determine the unknown genotype of an individual that
has a dominant phenotype
39. T? TT x tt – all offspring will be tall; Tt x tt – some offspring will be short, proving
that the unknown genotype had to be heterozygous, the only way that short
offspring could be produced.
40. The more offspring, the more reliable the results will be, because numbers
increase accuracy. If there was only one offspring produced in the test-cross,
you could not reliably conclude the unknown genotype of the parent. If multiple
offspring were produced, and none displayed a recessive phenotype, you could
reliably conclude that the unknown genotype was homozygous dominant.
41. A monohybrid cross only involves one trait, whereas a dihybrid cross involves
crossing individuals that differ in two traits.
42. SsTt1: Sstt1: ssTt1: sstt1; Short Sharp, Short dull, Tall Sharp, Tall dull
43. LlSS1: LlSs1: llSS1: llSs1; Long Solid 1: Short Solid 1
44. The more genes introduced into a genome, the more variety that can exist in an
individual’s gene combinations. For example, humans have somewhere around
21,000 genes. With the exception of twins, the chances of two humans having
the exact same combination of genes is nearly impossible
45. ½ x ½ x ½ = 1/8
46. ½
47. ¾
48. ½
49. The letters represent the alleles that could be carried by the gametes of the
parent individuals.
50. Dominant allele
51. Not without performing a test cross, because the dominant phenotype will be
expressed in a heterozygote or homozygous dominant genotype
52. Ll 100%; Long Needle 100%
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53. TT 1: Tt 1; 100% Tall
54. You would have to cross your small eared panda with a big eared panda. If any
of the offspring have big ears, you know that your panda was heterozygous, if all
of the offspring have small ears, you can assume homozygous dominant, but you
would need multiple offspring to achieve reliable results.
55. Dihybrid crosses allowed Mendel to determine whether or not individual genes
have an impact on other genes, in other words: are purple flowers always tall,
etc. A monohybrid cross cannot achieve this because it only involves one trait
56. 9 Green Broad, 3 Green Thin, 3 Olive Broad, 1 Olive Thin
57. LLTt1: LlTt1; Long Thick 100%
58. 1/8
59. ½
60. 5/32
61. Incomplete dominance, codominance, multiple alleles, polygenic traits
62. Three - dominant, recessive, and (in cases of incomplete dominance or
codominance) the blended phenotype or both phenotypes at the same time.
63. Genotype: 100% heterozygous; Phenotype: 100% Pink
64. 100% medium teeth- because of incomplete dominance, long teeth must be
homozygous dominant, short teeth must be homozygous recessive, therefore,
the offspring will all be heterozygous which will all have medium teeth.
65. RR x RW- Genotype: 50% RR 50% RW; Phenotype: 50% Red, 50% Roan
66. One gene influences multiple traits.
67. Polygenic means that multiple genes affect a single trait.
68. Many genes contribute to the weight of an individual, rather than just one gene
determining the weight.
69. AB x AO- Genotype: AA, AO, AB, BO- 25% for each; Phenotype: Type A- 50%,
AB 25%, B 25%
70. 25%
71. Incomplete dominance is when a heterozygote has a phenotype that is an
intermediate of the two alleles.
72. Genotype: 50% homozygous dominant, 50% heterozygous; Phenotype: 50%
Red, 50% Pink
73. Codominance is when neither allele is dominant or recessive, and both are
expressed equally when possessed.
74. Codominance is when neither allele is dominant and both alleles are expressed
equally, incomplete dominance results in an intermediate phenotype between the
two alleles when heterozygote genotype exists.
75. The gene that causes sickle cell anemia influences many traits.
76. Many genes contribute to skin color.
77. Pleiotropic
78. Since the A allele is completely dominant over O, both parents could be AO,
which would produce A blood, but could both produce gametes with the O allele,
potentially creating a child with type O blood.
79. Genotype: 50% heterozygous Brown/Black, 50% heterozygouse Brown/Gray;
Phenotype: 100% Brown
80. If pea plants exhibited multiple alleles, codominance, incomplete dominance,
polygenic inheritance, pleiotropic genes, Mendel would not have obtained the
ratios that he did in his offspring generations, and it would have been more
difficult for him to develop his theories without the clean ratios that allowed him to
figure out the genetic qualities of the plants.
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81. Homozygous recessive- if the disease is a recessive disease, the individual
would need to possess two alleles for the disease in order to have it.
82. No, a carrier is heterozygous for a recessive allele, and will show a normal
phenotype.
83. 25%
84. These marriages lack genetic diversity, and therefore the disease allele has a
greater chance of remaining in the gene pool. The more you introduce foreign
alleles into the gene pool, the less of a chance the disease allele has of
remaining and increasing in occurrence
85. No. A carrier only applies for a recessive disorder, for a dominant disorder, if an
individual is heterozygous, he/she will have the disorder
86. A disorder that is carried on the X or Y chromosome
87. 25%, yes, there is no chance that a female produced from this marriage could
have the disorder
88. 50%, No, because the disorder is dominant and every child has at least 1 X
chromosome, a male or female could have the disorder.
89. A genetic counselor can provide a couple with guidance related to the possibility
of a genetic disorder or other specific traits occurring in their offspring.
90. A needle is used to remove amniotic fluid from a pregnant woman, which can be
used to test for the presence of genetic diseases in the baby.
91. Amniocentesis focuses on the genotype of the fetus by analyzing the genetic
structure, ultrasounds focus on the phenotype by analyzing the physical features
of the fetus.
92. A carrier is an individual that has an allele for a disease, but who does not have
the disease, this only occurs in recessive disorders where the individuals would
be heterozygous.
93. 50%- No, the chance is the same for each new pregnancy
94. Yes. Since it is a dominant disorder, you only need one allele to have the
disorder.
95. 50%
96. Male- XY; Female- XX
97. Because males only have one X chromosome, so if their x chromosome carries
the disorder, they do not have a second x chromosome to mask the affected
allele
98. 0%- If the pairing produces a female, she will have a normal x chromosome from
the mother which will dominate over the affected x chromosome from the father.
A male from this pairing would only have a normal x chromosome from the
mother.
99. Both genders have an x chromosome, females do not have a y chromosome, so
it would be evolutionarily detrimental if essential genes were found on the y
chromosome, as no women would have them
100.
A sample of the placenta is removed and genetically tested; it can be
performed earlier in the pregnancy and the results can be obtained more quickly.
101.
A modified diet
102.
They pose some risk of complication, which is preferable to avoid if
possible.
103.
Pedigrees trace the occurrence of a genetic disorder through a family
history.
104.
Heterozygous- they are both carriers
105.
Homozygous recessive. Her father has the disorder
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106.
Recessive. There is a situation where neither parent has the disorder but
the offspring does.
107.
Heterozygous. The individual does not have the disorder but one of her
offspring does.
108.
The individual has the gene for the disease on the X chromosome
109.
If the trait is recessive, you could have parents that do not have the
disease, but offspring that do. If it is dominant, it is more likely to be exhibited in
every generation, if you have the allele, you have the disease.
110.
It is most likely sex-linked.
111.
It could be an x-linked dominant disorder that the mother had and passed
on.
112.
Yes it could. There is no situation where the offspring have the disorder
but neither parent did.
113.
Homozygous recessive
114.
Heterozygous
1. Mendel
a. There are alternate forms of genes, called alleles
b. All organisms inherit two alleles for each trait, one from each parent
c. If alleles are heterozygous and one allele determines the appearance of
the trait, this allele is dominant. If the second allele has no noticeable
affect, it is recessive.
d. Each gamete carries only one allele for each trait – the pairs separate (or
segregate) from each other during meiosis
2. Mice
T
t
a. Punnett Square
T
TT
Tt
t
Tt
tt
b. We will take one long tailed mouse and one short tailed mouse and cross
them.
c. If the two mice chosen are TT x tt , all of the offspring will be long tailed.
If the long tailed mouse we choose is heterozygous, then we will get short
tailed offspring
T
t
t
Tt
tt
t
Tt
tt
T
T
t
Tt
Tt
t
Tt
Tt
d. This is a test cross
3. Dihybrid Cross
a. RR = .25 TT = .25 probability 0.0625
b. Rr = .50 Tt = .50
probability = .25
4. Trihybrid Cross
a. RR = .25
TT = .25 Bb = .5 probability = 0.03125
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b. Rr = .5 TT = .25 BB =
.25 probability = 0.03125
5. Baldness
B
b
X
X
a. XB Xb
B B
XBXb
b. XB y
XB X X
c. Punnett
Square
d. 25%
B
b
X y
y X y
e. 50%
f. 0%
6. Pedigree
a. Recessive
b. 1: XbXb; 2: XBY; 3: XBY; 4: XBXb; 5: XbY; 6: XBXb; 7: XBY; 8: XbXb; 9: XBY;
10: XBXb or XBXB; 11: XBY; 12: XBXb or XBXB; 13: XBY; 14: XbXb; 15: XbY
c. 0%
d. 50%
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