Chapter 11: Mendelian Patterns of Inheritance€¦ · Chapter 11: Mendelian Patterns of Inheritance . AP Curriculum Alignment. Without variation within a population, it is impossible

Mader, Biology, 12th Edition, Chapter 11 164

Chapter 11: Mendelian Patterns of Inheritance

AP Curriculum Alignment

Without variation within a population, it is impossible for evolution to occur. The fact that some variations can increase or decrease the fitness of an organism is explained in the genetic diseases that are profiled in Chapter 11, such as sickle cell anemia. These concepts draw on Big Idea 1.

Genetic information makes up a large portion of the type of information that is essential to life processes. It is no surprise then that a large portion of Big Idea 3 is examined in Chapter 11. The work of Gregor Mendel is fully explained. Today we know that Mendel’s Law of Segregation applies to genes that are carried on different chromosomes. Mendel’s explanation of dominant and recessive traits is explained as being part of the Law of Segregation. Both Mendelian genetics (monohybrid and dihybrid crosses) and non-Mendelian Genetics (sex-linked genes, codominance, incomplete dominance, and genes linked on the same homologous chromosome) are fully explained in chapter 11.

The rules of probability, the product rule (also known as the Rule of Multiplication) and the Sum Rule, (also known as the Rule of Addition) are presented in this chapter and should be noticed. Both of these probability rules appear on the AP Biology Formula Sheet and may appear on the AP Exam.

Big Idea 4 is in the spotlight during the discussion of heterozygous advantage, particularly in relation to sickle cell anemia. It should be noted that epistasis and pleiotropy are not part of the AP Biology curriculum.

ALIGNMENT OF CONTENT TO THE CURRICULUM FRAMEWORK Big Idea 1: The process of evolution drives the diversity and unity of life. Enduring understanding (EU) 1.A: Change in the genetic makeup of a population over time is evolution. Essential knowledge (EK) 1.A.2: Natural selection acts on phenotypic variations in populations.

b. Phenotypic variations are not directed by the environment but occur through random changes in the DNA and through new gene combinations. c. Some phenotypic variations significantly increase or decrease fitness of the organism and the population.

To foster student understanding of this concept, instructors can choose an illustrative example such as: • Sickle cell anemia • Peppered moth • DDT resistance in insects

Big Idea 3: Living systems store, retrieve, transmit, and respond to information essential to life processes.

Mader, Biology, 12th Edition Chapter 11 165

Enduring understanding (EU) 3.A: Heritable information provides for continuity of life. Essential knowledge (EK) 3.A.1: DNA, and in some cases RNA, is the primary source of heritable information.

a. Genetic information is transmitted from one generation to the next through DNA or RNA.

Evidence of student learning is a demonstrated understanding of each of the following: 1. Genetic information is stored in and passed to subsequent generations through DNA molecules and, in some cases, RNA molecules.

Essential knowledge 3.A.3: The chromosomal basis of inheritance provides an understanding of the pattern of passage (transmission) of genes from parent to offspring.

a. Rules of probability can be applied to analyze passage of single gene traits from parent to offspring. b. Segregation and independent assortment of chromosomes result in genetic variation.

Evidence of student learning is a demonstrated understanding of each of the following: 1. Segregation and independent assortment can be applied to genes that are on different chromosomes. 2. Genes that are adjacent and close to each other on the same chromosome tend to move as a unit; the probability that they will segregate as a unit is a function of the distance between them. 3. The pattern of inheritance (monohybrid, dihybrid, sex-linked, and genes linked on the same homologous chromosome) can often be predicted from data that gives the parent genotype/phenotype and/or the offspring phenotypes/genotypes.

c. Certain human genetic disorders can be attributed to the inheritance of single gene traits or specific chromosomal changes, such as nondisjunction.

To foster student understanding of this concept, instructors can choose an illustrative example such as: • Sickle cell anemia • Tay-Sachs disease • Huntington’s disease • X-linked color blindness • Trisomy 21/Down syndrome • Klinefelter’s syndrome

d. Many ethical, social and medical issues surround human genetic disorders. To foster student understanding of this concept, instructors can choose an illustrative example such as: • Reproduction issues • Civic issues such as ownership of genetic information, privacy, historical contexts, etc.


Essential knowledge (EK) 3.A.4: The inheritance pattern of many traits cannot be explained by simple Mendelian genetics.

a. Many traits are the product of multiple genes and/or physiological processes. Evidence of student learning is a demonstrated understanding of the following: 1. Patterns of inheritance of many traits do not follow ratios predicted by Mendel’s laws and can be identified by quantitative analysis, where observed phenotypic ratios statistically differ from the predicted ratios.

b. Some traits are determined by genes on sex chromosomes. To foster student understanding of this concept, instructors can choose an illustrative example such as: • Sex-linked genes reside on sex chromosomes (X in humans). • In mammals and flies, the Y chromosome is very small and carries few genes. • In mammals and flies, females are XX and males are XY; as such, X-linked recessive traits are always expressed in males. • Some traits are sex limited, and expression depends on the sex of the individual, such as milk production in female mammals and pattern baldness in males.

c. Some traits result from nonnuclear inheritance. Evidence of student learning is a demonstrated understanding of each of the following: 1. Chloroplasts and mitochondria are randomly assorted to gametes and daughter cells; thus, traits determined by chloroplast and mitochondrial DNA do not follow simple Mendelian rules. 2. In animals, mitochondrial DNA is transmitted by the egg and not by sperm; as such, mitochondrial-determined traits are maternally inherited. ✘✘ Epistasis and pleiotropy are beyond the scope of the course and the AP Exam.

Big Idea 4: Biological systems interact, and these systems and their interactions possess complex properties. Enduring understanding (EU) 4.C: Naturally occurring diversity among and between components within biological systems affects interactions with the environment. Essential knowledge 4.C.1: Variation in molecular units provides cells with a wider range of functions.

b. Multiple copies of alleles or genes (gene duplication) may provide new phenotypes.

Evidence of student learning is a demonstrated understanding of each of the following: 1. A heterozygote may be a more advantageous genotype than a homozygote under particular conditions, since with two different alleles, the organism has two forms of proteins that may provide functional resilience in response to environmental stresses. 2. Gene duplication creates a situation in which one copy of the gene maintains its original function, while the duplicate may evolve a new function.


To foster student understanding of this concept, instructors can choose an illustrative example such as: • The antifreeze gene in fish

Concepts covered in Chapter 11 also align to the learning objectives that provide a foundation for the course, an inquiry-based laboratory experience, class activities, and AP exam questions. Each learning objective (LO) merges required content with one or more of the seven science practices (SP), and one activity or lab can encompass several learning objectives. The learning objectives and science practices from the Curriculum Framework that pertain to Mendelian Patterns of Inheritance are shown in the table below. Note that other learning objectives may apply as well.

LO 3.9 The student is able to construct an explanation, using visual representations or narratives, as to how DNA in chromosomes is transmitted to the next generation via mitosis, or meiosis followed by fertilization. LO 3.10 The student is able to represent the connection between meiosis and increased genetic diversity necessary for evolution. LO 3.11 The student is able to evaluate evidence provided by data sets to support the claim that heritable information is passed from one generation to another generation through LO 3.12 The student is able to construct a representation that connects the process of meiosis to the passage of traits from parent to offspring. LO 3.13 The student is able to pose questions about ethical, social or medical issues surrounding human genetic disorders. LO 3.14 The student is able to apply mathematical routines to determine Mendelian patterns of inheritance provided by data sets. LO 3.15 The student is able to explain deviations from Mendel’s model of the inheritance of traits. LO 3.16 The student is able to explain how the inheritance patterns of many traits cannot be accounted for by Mendelian genetics. LO 3.17 The student is able to describe representations of an appropriate example of inheritance patterns that cannot be explained by Mendel’s model of the inheritance of traits. LO 4.22 The student is able to construct explanations based on evidence of how variation in molecular units provides cells with a wider range of functions.


Key Concepts Summary Mendel and the science of genetics

• Science of genetics explains the stability of inheritance as well as variations between offspring from one generation to the next.

• The “father of genetics” is Gregor Mendel. • He established rules of inheritance without knowing about the process of

meiosis. • Mendel’s famous work involved crossing pea plants.

o He created pure-breeding plants (P generation) through self-pollination. o Mendel began by crossing P generation plants that occurred in two distinct

but alternative forms to produce the F1 generation. o All offspring in this F1 generation expressed only one trait such as purple

flower color. One of the physical types had disappeared. o He crossed F1 plants with other F1 plants to produce the F2 generation. o The F2 generation contained both of the original phenotypes. o This type of cross is called a monohybrid cross.

• Mendel used the terms recessive to describe the trait that had disappeared in the F1 generation and dominant to the trait that had not.

• Mendel also conducted dihybrid crosses, in which the F1 individuals showed dominant characteristics for two traits, but there were four phenotypes in a 9:3:3:1 ratio among the F2 offspring.

• Mendel deduced his law of independent assortment, which states that the members of one pair of alleles separate independently of those of another pair. This applies to genes that are carried on different chromosomes.

Genetics and probability

• Each gene has a specific location, or locus, on a chromosome, and dominant alleles mask the expression of recessive alleles.

• The sum rule of probability tells us that when the same event can occur in more than one way, we can add the results to determine the likelihood of a phenotype.

• The product rule of probability tells us that we have to multiply the chances of independent events to determine how likely is it that an offspring will inherit a specific set of two alleles, one from each parent.

• Alleles on the X chromosome are called X-linked alleles; the Y chromosome does not carry those alleles. The Y chromosome does not carry these genes, and, because males inherit only one copy of an allele for an X-linked trait, they cannot be heterozygous for the traits.


Key Terms

alleles autosome carriers codominance dihybrid cross genotype hemizygous heterozygous

homozygous incomplete penetrance law of independent assortment law of segregation monohybrid cross multifactorial traits multiple alleles

phenotype pleiotropy polygenic inheritance Punnett square X-linked

Teaching Strategies

If you have the classroom space, methods for caring for plants over missed school days, and time, then College Board’s Investigation 1: Artificial Selection should be conducted. You should try to set this lab up early in the unit or during the Mitosis unit. An alternative lab is to use characteristics that germinating fast plant seed display (Activity 1 below) or genetic corn. Both are available from science supply houses.

Class time: Six 45-minute class periods

Day 1: Lecture on Mendelian genetics including monohybrid and dihybrid crosses – 25 minutes

Activity 1: genetics with Wisconsin fast plants – 20 minutes

Day 2: Lecture on non-Mendelian genetics – 25 minutes

Practice problems – 5 minutes

Tend to plants from Activity 1 – 5 minutes

Day 3: Complete Day 3 of Activity 1 – 15 minutes

Lecture on genetic diseases – 15 minutes

Begin Activity 2 – 15 minutes

Day 4: Complete Day 4 of Activity 1 – 30 minutes

Continue Activity 2 – 15 minutes

Day 5: Lecture on rules of probability – 15 minutes

Activity 2 presentations – 30 minutes

Day 6: Summative assessment


Suggested Approaches

This unit will mix lab activities, lectures and group projects. It will be helpful to recap the work at the beginning and end of each class. The students can be called on to recap as well. Student Misconceptions and Pitfalls

Students get the concepts of homologous chromosomes and alleles confused and may write about “homologous alleles”. The pool noodle activity that students completed during the meiosis unit will help them keep these two concepts separated. You may want to show students these models again.

Suggested Activities

Activity 1: Genetics of Organisms Using Wisconsin Fast Plants

This investigation was designed for students to develop understanding of the following biological concepts and skills:

• Mendel’s law of segregation and law of independent assortment • Inheritance of 2 traits • How genotypes influence phenotypes • Scientific inquiry, including interpretation of evidence • Practice with Chi-Square statistical analyses

Full instructions for students are included on the following worksheet. Activity 2: Genetic Disorder Project

Students will research and present on genetic disorders and their patterns of inheritance. Full instructions for students follow on a separate worksheet.


Genetics of Organisms Using Wisconsin Fast Plants

Introduction

Fast Plants®, scientific name: Brassica rapa, are called rapid-cycling Brassicas because of their short life cycle. They are members of the crucifer family of plants, closely related to cabbage, turnips, broccoli and other cruciferous vegetables. They make ideal organisms for genetic studies to the ease of caring for them in the lab, the large number of progeny and the short life cycle.

We will be studying two traits that can be observed in seedlings that are just 3- 4 days old. You will be given seeds of one of the P generation, the F1 generation, and the F2 generation of a dihybrid cross. After observing the P and F1 generations, you will form your hypothesis for the outcome of the F2 generation.

Procedures

Day 1

• Obtain two Petri dishes. • Cut a filter to fit each dish. • Mark Dish 1 as P and F1. • Mark Dish 2 F2. • Label your dishes with your period and group name. • Obtain 10 each of the appropriate seeds. • Moisten the filter paper. • Lay the seeds in a straight line on the moistened paper. • Place paper in Petri dish. • Place both dishes sideways in the plastic container with water and under grow

light.

Day 2

• Water your seeds

Day 3

• Water your seeds. • If seeds have germinated, take Dish 1 for analysis. • Observe at P and F1 plants • Determine the dominant traits


Day 4 • Obtain your F2 Petri dish. • Then count the seedlings and categorize them by phenotype. • Record your observed phenotypes in the table below. • Compute the Chi-square for your data after completing the statistical section

below. • State whether or not you would accept your hypothesis based on this statistical

analysis and why. Analysis Questions to be completed after Day 3

1. What can these two generations of plants tell us?

2. How can you determine the P2 (the other parent plant)?

3. What are the dominant traits? How do you know this?

Analysis Questions so be completed after Day 4

1. How are you going to formulate your hypothesis?

2. What can you use to determine what your hypothesis should be?

3. In terms of statistical analysis, will your hypothesis be your expected numbers or your observed numbers?


4. Determine your expected numbers.

5. Fill in the first two columns below.

6. State whether or not you would accept your hypothesis based on this statistical analysis and why.

Traits

Expected

(e)

Observed

(o)

Expected-Observed

(o – e)

(o - e)2

(o - e)2 / e

Σ (o - e)2 / e

Computed Chi Square


Activity 2: Genetic Disorder Project

Background: How do genetic disorders or conditions occur? Does a single mutation, a mutation in one gene, or even a whole chromosome cause the disorder? Genetic disorders can affect many different aspects of human development: mental or intellectual development such as height, the nervous system, and even gender. Sometimes during meiosis, there is an error called nondisjunction, in which the chromosomes fail to separate properly. Nondisjunction can result in other combinations of an individual's chromosomes, including the sex chromosomes. If the autosomes are affected (pairs #1-22) disorders such as Down syndrome may result. When the sex chromosomes are involved, the new combinations can affect gender development in a variety of ways. Purpose: In this activity, you will be exploring the process of how genetic disorders are inherited and how they develop. You will focus on a specific genetic condition and research how that particular condition shapes an individual. Vocabulary: Autosomal: refers to a trait whose gene is located on an autosome (non-sex chromosome) Meiosis: cell division that produces egg cells and sperm cells Mutation: a chemical change in a gene, resulting in a new allele; or, a change in the portion of a chromosome that regulates the gene (controls when a gene should make its protein) Nondisjunction: failure of chromosomes to separate properly during meiosis Sex-linked or X-linked: refers to a trait whose gene is located on the X chromosome Syndrome: set of symptoms that typically occur together Procedure: A. You will be assigned one of the following genetic disorders: 1) XO female (Turner's syndrome) 2) XXY male (Klinefelter's syndrome) 3) Xeroderma Pigmentosum 4) Down Syndrome 5) Achondroplasia (dwarfism) 6) Sickle-cell anemia 7) Cystic Fibrosis 8) Tay Sachs 9) Osteogenesis Imperfecta 10) Phenylketonuria (PKU) 11) Huntington's disease 12) Tourette Syndrome 13) Marfan's Syndrome 14) Hemophilia


15) Neurofibromatosis 16) Thalassemia 17) Gaucher Disease 18) Fragile X Syndrome B. Research your particular genetic disorder Use your Biology textbook or the Internet (a list of helpful websites is provided below)

C. Design a trifold brochure about your selected disorder, addressing the following questions:

1) How would you recognize this condition in a family member (e.g. what physical & psychological characteristics are associated with this genetic condition)? 2) What is the cause of this disorder? Is it sex-linked, autosomal, recessive, dominant, mutation, chromosomal abnormality, etc.? 3) Is there any treatment for this condition? Is there a way to prevent this condition? 4) Is there a way to screen individuals for this condition? 5) What percentage of the population is thought to have this condition? Is it more common in certain populations (ethnicity, region of the world)? 6) How has society interpreted these variant genetic conditions? (Are individuals with this condition accepted by society, shunned, institutionalized, teased, etc.) D. Include a Bibliography using APA format.


Student Edition Chapter Review Answers

Answers to Assess Questions

1. b; 2. d; 3. d; 4. a; 5. b; 6. b; 7. a; 8. b; 9. c; 10. c; 11. b; 12. a; 13. d; 14. b Answers to Applying the Big Ideas Questions 1. Inheritance of genes within a population is a cornerstone of species’ ability to change

over time. a) Describe TWO kinds of data that could be collected by scientists to provide

a direct answer to the question, how can scientists investigate the role of natural selection in evolution?

b) Explain how the data you suggested in part (a) would provide a direct answer to the question.

Essential Knowledge

1.A.1: Natural selection is a major mechanism of evolution.

Science Practice

5.3: The student can evaluate the evidence provided by data sets in relation to a particular scientific question.

Learning Objective

1.2: The student is able to evaluate evidence provided by data to qualitatively and quantitatively investigate the role of natural selection in evolution.

4 points maximum. Mathematical routines used to describe the natural phenomena may include:

Descriptions of kind of data (1 point each)

Explanations (1 point each)

Data that supports Mendel’s particulate theory of inheritance would support the ideas of natural selection’s role in evolution. In this theory, inheritance involves the reshuffling of the same genes from generation to generation.

If parents of contrasting appearance always produced offspring of intermediate appearance, then no individuals would have a selective advantage because over time variation would decrease as individuals became more and more alike (natural selection would have nothing to act upon).

Genetic variation and mutation play roles in natural selection and can play a part in populations changing over time.

Individuals with genes that when expressed confer favorable variations are more likely to survive and produce more offspring. Individuals that have experienced harmful mutations may not survive or reproduce more


offspring.

A diverse gene pool is important for the survival of a species in a changing environment.

An adaptation is a genetic variation present in a population that is favored by selection and is manifested as a trait that provides an advantage, allowing a population to continue even in the face of change.

2. Methemoglobinemia is an autosomal recessive disorder. If a man and a woman are both carriers of the recessive allele, what is the probability of each of the following (show all work):

a) All three of their offspring will be of normal phenotype? a) All three of their offspring will have methemoglobinemia?

Essential Knowledge

3.A.3: The chromosomal basis of inheritance provides an understanding of the pattern of passage (transmission) of genes from parent to offspring.

Science Practice

2.2: The student can apply mathematical routines to quantities that describe natural phenomena.

Learning Objective

3.14: The student is able to apply mathematical routines to determine Mendelian patterns of inheritance provided by data sets.

4 points maximum. Mathematical routines used to describe the natural phenomena may include: Work shown (1 point each) Answer (1 point each) Normal = MM; Carrier = Mm; Methemoglobinemia = mm

If both parents are carriers, their genotypes should reflect this: Mother = Mm; Father = Mm

According to Mendelian patterns of inheritance, each genotype may occur in offspring. Student may construct a Punnett square.

The probability for each of the genotypes to occur in offspring are: MM = ¼; Mm = ½; and mm = ¼.

(a) This is equal to the odds of not producing an affected child (notice this is phenotype, not genotype, so MM and Mm are added together). (¾) x (¾) x (¾) = (27/64) or (0.75)3

Probability that all three offspring will be of normal phenotype: (27/64) or 0.42 or 42%.


(b) This is equal to the odds of producing a child who has the disease (mm). (¼) x (¼ x (¼) = (1/64) or (0.25)3

Probability that all three offspring will have methemoglobinemia: (1/64) or 0.016 or 1-2%.

3. In a paragraph, describe at least THREE situations that illustrate the influence of environmental factors on the phenotype of an organism. Essential Knowledge

4.C.2: Environmental factors influence the expression of the genotype in an organism.

Science Practice

6.2: The student can construct explanations of phenomena based on evidence produced through scientific practices.

Learning Objective

4.23: The student is able to construct explanations of the influence of environmental factors on the phenotype of an organism.

3 points maximum. Examples of the influence of environmental factors on phenotype may include (1 point each): • Multifactorial traits are those controlled by polygenes subject to environmental

influences. If each gene has several alleles, and each allele responds slightly differently to environmental factors, then the phenotype can vary considerably. Many genetic disorders such as diabetes, schizophrenia and allergies are probably multifactorial.

• One environmental factor could be diet, and that can impact height and weight in humans, as well as certain health issues.

• Effect of increased UV on melanin in animals: Skin color is a polygenic trait controlled by many pairs of alleles, which result in a range of phenotypes. Additionally, UV rays form the sun can contribute to the darkening of the skin.

• Flower color on based on temperature: Primroses have white flowers when grown above 32°C but red flowers when grown at 24°C.

• Darker fur in cooler regions of the body in certain mammal species such as Siamese cats and Himalayan rabbits. Experimental evidence suggests that the enzyme encoded by a gene involved in the production of melanin is active only at a low temperature, so dark coloring occurs only at the extremities where body heat is lost to the environment.


Answers to Applying the Science Practices Questions Think Critically 1. While not the case for all plants, for the purpose of this prompt, the students should

consider this unknown plant to have genes that would behave in the way that Mendel’s pea plant genes did with discrete traits. Student responses are expected to reflect the work of Mendel. Answers may include:

• Ensure that the plants will be true-breeding (offspring exactly like the parent plants – so plants with serrated leaf edges only have plants with serrated leaf edges, while smooth-edged plants only have offspring with smooth leaves) by only allowing self-pollination over several generations.

• For best, most apparent results, test for only one trait at a time, controlling all other variables. So, first we will look at the trait of leaf edges.

• Perform reciprocal crosses: dust the pollen of true-breeding (P generation) serrated-leaf plants onto the stigmas of smooth-leaf plants (P generation), and dust the pollen of smooth-leaf plants onto the stigmas of serrated-leaf plants.

• The first generation (F1) should all display the same phenotype and one could infer that this will be the dominant trait, if Mendelian genetics is followed.

• To further confirm the idea that the traits being display are the dominant traits, allow the F1 plants to self-pollinate, and if a 3:1 ratio is achieved in the F2 generation, where three out of four individuals display the trait shown by the F1 generation, then that is the dominant trait.

• To confirm that the F1 plants of the first cross are, in fact, heterozygous for the trait being examined, cross F1 generation plants with true-breeding plants of what you understand to be the recessive trait to perform a testcross. If the resulting offspring show a 1:1 phenotypic ratio, then the correct dominant and recessive traits have been identified properly.

• Perform the same steps to test the next trait, being thoroughly conscientious in controlling which plants are able to pollinate with which plants.

2. This question requires us to do a dihybrid cross (or remember the 9:3:3:1 ratio). We can represent the gene for leaf edges with the symbol “L” for dominant serrated margins and “l” for recessive smooth margins. Likewise, for flower color, we can use “F” for white flowers and “f” for yellow flowers. The problem states that the birds being crossed are heterozygous for each trait, implying that the genotype for both birds would be LlFf. There are 16 total offspring. The genotypes of the offspring would then be 1 LLFF, 3 Llff, 3 llFf, 8 LlFf, 1 llff. Therefore, 12 of the offspring carry the dominant L allele, giving them the serrated leaf edge phenotype (1 LLFF, 3 Llff, 8LlFf). Of those 12, 9 carry the dominant F allele for white flowers (1 LLFF, 8 LlFf). This means 9


out of the 16 offspring will express both the serrated edges and the white flowers phenotypes.

Additional Questions for AP Practice

1. Construct a flow chart of genetic material passing from parents to offspring.

2. Describe two ethical issues that surround genetic disease conditions.

3. In the Sail-fin Molly fish, gold color (g) is recessive to normal color (G). When a gold colored fish was crossed with a normal color fish, 55 of the offspring were normal color and 45 were gold color.

One hypothesis is that the normal parent was heterozygous. The Chi-squared value is 1. The table below shows the probability values.

Probability

Degrees of freedom 0.99 0.950 0.05 0.01

1 0.000 0.004 3.84 6.64

2 0.020 0.103 5.99 9.21

3 0.115 0.352 7.82 11.35

Which of the following is the correct response? A) Accept the hypothesis because the Chi-squared value is less than 3.84. B) Reject the hypothesis because the probability is less than 0.05. C) Accept the hypothesis because the Chi-squared value is less than 5.99. D) Reject the hypothesis because there is not enough evidence.

4. In a cross between AaBbCc x AaBBCC, what is the probability that the offspring will be AaBbCC?

5. A man with hemophilia (a recessive, sex-linked condition) has a daughter of normal phenotype. She marries a man who is normal for the trait. What is the probability that: A) a daughter of this mating will be a hemophiliac?

B) a son will be a hemophiliac?


C) if the couple has four sons, all four will be born with hemophilia?

Grid-In Questions

1. Poison dart frogs carry alleles for producing either a red (R) pigment or a yellow (r) pigment. The allele for red pigment is dominant over the allele for yellow. If two heterozygous frogs mate, what percentage of their offspring will inherit two recessive alleles and be yellow in color?

2. If two pigs breed which are both heterozygous for having curly tails Cc, what percentage of their offspring will have curly tails?

3. Balsam firs have particularly aromatic needles if they carry two dominant A alleles and grow quickly if they have at least one dominant H allele. If a tree farmer cross-pollinates two trees with the alleles AaHh, what percentage of offspring will be both highly aromatic and fast growing trees?

4. Two parents are both carriers for an autosomal recessive disease, what is the

probability they will have a child that is also a carrier?


Answers to Additional Questions for AP Practice

1. Possible answer:

2. Answers could include: Notification of condition to insurance company. The dilemmas of whether or not genetic testing should be chosen by offspring and whether or not to choose prenatal testing.

3. Answer is A.

4. This is an “and” question – all events happening at the same time so multiply.

½ x ½ x ½ = 1/8

father mother Genetic material 2n

Gametes n Gametes n

Fertilization

Zygote 2n


5. Genotypes: A man with hemophilia is XhY where h = hemophilia gene and H = the normal gene. Any daughter with normal phenotype whose father has hemophilia will be a carrier. Her genotype must be: XhXH and NOT XHXH We can use a Punnett square to show the probability of a daughter or son having hemophilia.

daughter x normal man XhXH x XHY

XH Y

XH XHXH

1

XHY

2

Xh XhXH

3

XhY

4

A. If the daughter marries a normal male the probability of a daughter having hemophilia is zero. B. About 50% of male children would have hemophilia (Boxes 2 and 4 above) C. The probability that all 4 sons have inherited hemophilia would be: 1/2 x 1/2 x 1/2 x 1/2 or 1/16.


Answers to Grid-In Questions

1. Chapter: 11 Mendelian Patterns of Inheritance Answer: 25% Rr X Rr = RR = ¼ Rr = ¼ rR = ¼ rr = ¼

2. Chapter: 11 Mendelian Patterns of Inheritance Answer: 75% Use the rule of sums: CC = 1/ 4 Cc= 1/ 4 cC =1 /4

3. Chapter: 11 Mendelian Patterns of Inheritance

Answer: 18.7% Good smelling, fast growing trees would have the alleles: AAHH or AAHh. Only 1/4th of the population will carry two dominant A alleles and out of these four, only three have either HH or Hh. The percentage of AAHH and AAHh therefore would be 3/16.

AH Ah aH ah

AH AAHH AAHh AaHH AaHh

Ah AAHh AAhh AaHh Aahh

aH AaHH AaHh aaHH aaHh

ah AaHh Aahh aaHh aahh

4. Chapter: 11 Mendelian Patterns of Inheritance Answer: 50% Assuming Aa is a carrier: Aa + Aa Aa = 1/4 aA = 1/4 AA = 1/4 aa = 1/4

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Chapter 11: Mendelian Patterns of Inheritance€¦ · Chapter 11: Mendelian Patterns of Inheritance . AP Curriculum Alignment. Without variation within a population, it is impossible