Student Sheet
Name:
Date:
Instructor’s Name:
Assignment: SCIE207 Phase 3 Lab Report
Title: Lab to Determine the Outcome of Heredity
Instructions: You will fill out the Punnett squares and answer a set of questions for each exercise.
When your lab report is complete, submit this document to your instructor in your assignment box.
Using what you learned on the lab animation (and the images below), you will fill in the following Punnett squares and answer the questions that follow:
Exercise 1: Color Blindness
A. Choose the parental gametes, and align these in the correct positions around the Punnett square. (Type in the correct gametes. Pink are the female gametes, and blue are the male gametes. XC is the gene for normal vision, and Xc is the gene for color blindness.)
C
B. Choose the correct genotype of the progeny to fill in the Punnett square.
(Type the correct genotypes in the boxes.)
______
______
______
______
Question
Answer
What is the genotype of the mother?
What is the genotype of the father?
What are the possible phenotypes of their children?
What is the probability of the color blind trait in female children and the probability of the color blind trait in male children?
X and Y genes code for male and female gender. XX is female and XY is male. Why is XcY color blind, but XcXC not color blind? Can an XcXC female pass the color blind trait to her children?
Exercise 2: Freckles
C. Choose the parental gametes and align these in the correct positions around the Punnett square. (Type in the correct gametes. The pink are female gametes, and the blue are male gametes.)
A. Choose the correct genotype of the progeny to fill in the Punnett square.
(Type in the genotypes.)
______
______
______
______
Question
Answer
What is the genotype of the mother?
What is the genotype of the father?
What are the possible phenotypes of the children?
What is the probability of freckles in their children?
Freckles in humans are inherited by which pattern: dominant/recessive, incomplete dominance, or codominance? Why does the mother have freckles, even though she has a gene for no freckles? Why does the father have no freckles?
Exercise 3: Blood Type
A male has Type A blood, and a female has Type B blood. Could they have a child with Type O blood? Demonstrate how this is genetically possible by filling out the Punnett square.
Choose the correct parental gametes from all of the potential blood type gametes below, and align these in the correct positions around the Punnett square. (Type in the correct gametes. The pink are female gametes, and the blue are male gametes.)
A. Choose the correct genotypes of the progeny, and fill in the Punnett square.
(Type the correct genotypes into the boxes from the possible genotypes shown below).
______
______
______
______
Question
Answer
What is the genotype of the mother?
What is the genotype of the father?
What are possible phenotypes of their children?
What is t.
Student Sheet
Name:
Date:
Instructor’s Name:
Assignment: SCIE207 Phase 3 Lab Report
Title: Lab to Determine the Outcome of Heredity
Instructions: You will fill out the Punnett squares and answer a set of questions for each exercise.
When your lab report is complete, submit this document to your instructor in your assignment box.
Using what you learned on the lab animation (and the images below), you will fill in the following Punnett squares and answer the questions that follow:
Exercise 1: Color Blindness
A. Choose the parental gametes, and align these in the correct positions around the Punnett square. (Type in the correct gametes. Pink are the female gametes, and blue are the male gametes. XC is the gene for normal vision, and Xc is the gene for color blindness.)
C
B. Choose the correct genotype of the progeny to fill in the Punnett square.
(Type the correct genotypes in the boxes.)
______
______
______
______
Question
Answer
What is the genotype of the mother?
What is the genotype of the father?
What are the possible phenotypes of their children?
What is the probability of the color blind trait in female children and the probability of the color blind trait in male children?
X and Y genes code for male and female gender. XX is female and XY is male. Why is XcY color blind, but XcXC not color blind? Can an XcXC female pass the color blind trait to her children?
Exercise 2: Freckles
C. Choose the parental gametes and align these in the correct positions around the Punnett square. (Type in the correct gametes. The pink are female gametes, and the blue are male gametes.)
A. Choose the correct genotype of the progeny to fill in the Punnett square.
(Type in the genotypes.)
______
______
______
______
Question
Answer
What is the genotype of the mother?
What is the genotype of the father?
What are the possible phenotypes of the children?
What is the probability of freckles in their children?
Freckles in humans are inherited by which pattern: dominant/recessive, incomplete dominance, or codominance? Why does the mother have freckles, even though she has a gene for no freckles? Why does the father have no freckles?
Exercise 3: Blood Type
A male has Type A blood, and a female has Type B blood. Could they have a child with Type O blood? Demonstrate how this is genetically possible by filling out the Punnett square.
Choose the correct parental gametes from all of the potential blood type gametes below, and align these in the correct positions around the Punnett square. (Type in the correct gametes. The pink are female gametes, and the blue are male gametes.)
A. Choose the correct genotypes of the progeny, and fill in the Punnett square.
(Type the correct genotypes into the boxes from the possible genotypes shown below).
______
______
______
______
Question
Answer
What is the genotype of the mother?
What is the genotype of the father?
What are possible phenotypes of their children?
What is t.
Study Guide Chapter 15 -Chromosomal Basis of Inheritance-Answers.docxhanneloremccaffery
Study Guide Chapter 15 -Chromosomal Basis of Inheritance-Answers
15.1 Mendelian Inheritance and chromosome theory______________________________
1. Thomas Hunt Morgan identified the first solid evidence associating a specific___gene_____ on a specific chromosome.
2. Why did Morgan choose Drosophila as his experimental organism? (List 3 reasons)
They reproduce quickly, a new generation of adults forms every two weeks!
Prolific (single matting = hundreds of offspring)
Only 4 chromosome pairs = simple genetics
3A. The normal phenotype for a character (phenotype most common in nature) is called___wild type_____.
3B. Provide at least two examples this phenotype in Drosophila.
Red eyes
gray body
4A. An alternative phenotype for a character (phenotype not common in nature) is called____mutant____.
4B. Provide at least two examples of this phenotype in Drosophila.
White eyes
Black body
5. Morgan and his students invented a notation to symbolize alleles in Drosophila that differed from the notation Mendel used to represent alleles. Describe how Morgan’s Drosophila characters are named and the symbol used for the allele type.
You can think of Morgan’s wild-type allele as equivalent to the dominant allele in Mendel’s naming system. And you can think of Morgan’s mutant allele as equivalent to Mendel’s recessive allele. So whenever you have one wild-type allele and one mutant allele, that organism will have the wild-type phenotype.
Mendel
Used the first letter of the dominant character name to represent the dominant allele. This letter was capitalized for the dominant allele, lower case for the recessive allele
Ex: purple flower allele dominant to white flower allele: P= dominant allele, p=recessive allele
Morgan
-Used the first letter of the mutant character name to represent the wild-type allele. The letter for the wild-type and mutant alleles are both lower case. The wild-type allele gets a + sign and the mutant does not.
Ex: red eye allele wild-type to white eye mutant: w+ = wild-type allele, w= recessive allele.
6. Morgan and his students invented a notation to symbolize alleles in Drosophila. Which of the following genotypes would produce a fly that is wild-type for eye color (red vs. white eyes)?
w+ w+
w+ w
w w
w+w+ and w+w
7. Morgan performed an experiment that yielded a 3:1 ratio of offspring in the F2 generation; however, only the males of this F2 generation had white eyes.
White- eye females DO EXIST, so why were only white-eye males observed in this cross?
Complete the following chart and describe why Morgan observed these results and how it allowed him to conclusively determine that the gene for eye color was located on the X chromosome.
Morgan’s experiment (see Figure 15.4) Use notation developed by Morgan and his students
If the gene for eye color is located on the X chromosome….
P generation
Red eye Female genotype for eye color = w+w+
White eye Male genotype for eye color= w
F1 generatio.
Dragon Genetics Hands on LabIntroductionThere are many patterns o.docxmadlynplamondon
Dragon Genetics Hands on LabIntroduction:
There are many patterns of inheritance. This lab will explore Mendelian traits where alleles are dominant or recessive. You will also have to understand genes that express incomplete dominance, codominance and are sex-linked. We will utilize this lab to explore those terms and to use our knowledge of gene combinations to create an organism.
Utilize this link to learn more about Mendelian traits, incomplete dominance and codominance.(Opens in new window) This link helps you to learn more about sex-linked genes. (Opens in new window)Part 1 Procedure: Checking for Understanding (Worth up to 15pts)
A. For each genotype, tell me if it’s homozygous or heterozygous. If it is homozygous also tell me if it is dominant or recessive.
1. BB
2. jj
3. Pp
4. RR
5. ff
B. Dimples (D) is dominant to (d) no dimples. Brown Eyes (B) are dominant to (b) blue eyes. For each genotype indicate the phenotype.
1. Dd
2. DD
3. dd
4. BB
5. bb
6. Bb
C. For each phenotype list the genotype(s). Trait: Red flower (R); White flower (r)
1. Red flower (homozygous)
2. Red flower (heterozygous)
3. White flowerPart 2: Creating the Parents (Worth up to 48 pts)
1. Flip a coin to determine the genotype for each trait and record it in the data table.
a. Heads= allele 1, Tails = allele 2 ( Example: if you flipped heads twice, your dragon will have two copies of allele 1 for her genotype)
2. Determine the phenotype resulting from the allele pair for each trait.
3. Repeat steps 1-2 for each trait and complete the Mama Dragon’s Table 1.
a. Pay attention to the notes under the trait; it means it does not fully follow Mendelian rules.
Example:
For the trait Eye Size. I flipped my coin 2 times. Flip 1= heads (E); Flip 2= heads (E). So I would put EE in the Genotype box and Two Small Eyes in the Phenotype box for that trait.
Table 1: Genotypes and Phenotypes for Mama Dragon
Trait
Allele 1
Allele 2
Genotype
Phenotype
Eye Size & Number
Two Small (E)
One large (e)
Eye Color
(incomplete Dom)
Red (R)
White (r)
Tail Shape
Curly (T)
Straight (t)
Tail Color
Purple (P)
Orange (p)
Teeth
Sharp (S)
Round (s)
Feet
Four toes (F)
Two toes (f)
Horn Color
(Codominant)
Purple (P)
White (W)
Ear Shape
Pointy (Y)
Round (y)
Claws
Long (L)
Short (l)
Skin Color
(Codominant)
Blue (B)
Yellow (Y)
Mane Texture
(incomplete Dom)
Curly (M)
Straight (m)
Dragon Hemophilia
(Sex- Linked)
Hemophilia (XH)
Normal Clotting (Xh)
Table 2: Genotype and Phenotypes for Daddy Dragon
Trait
Allele 1
Allele 2
Genotype
Phenotype
Eye Size & Number
Two Small (E)
One large (e)
Eye Color
(incomplete Dom)
Red (R)
White (r)
Tail Shape
Curly (T)
Straight (t)
Tail Color
Purple (P)
Orange (p)
Teeth
Sharp (S)
Round (s)
Feet
Four toes (F)
Two toes (f)
Horn Color
(Codominant)
Purple (P)
White (W)
Ear Shape
Pointy (Y)
Round (y)
Claws
Long (L)
Short (l)
Skin Color
(Codominant)
Blue (B)
Yellow (Y)
Mane Texture
(incomplete Dom)
Curly (M)
Straigh ...
Human body cells contain 46 chromosomes- The first 22 pairs are called.pdfkrishnac481
Human body cells contain 46 chromosomes. The first 22 pairs are called autosomes, and they
contain numerous genes that affect the traits of the individual. The last pair, number 23, are the
sex chromosomes. The sex chromosomes determine gender (i.e., either male or female), but there
are other genes on this pair of chromosomes as well. Males sex chromosomes are XY, while
female sex chromosomes are XX . The gametes in a human (either egg cells or sperm cells)
contain only 23 chromosomes. Fertilization, the fusion of an egg and a sperm, restores the total
of 46 chromosomes in a human zygote. Non-gamete cells are called somatic cells, and they have
all 46 chromosomes in them. 1. Your sex chromosomes: XX Just as in Mendel's pea plant
experiments, genes in humans can be dominant or recessive, and the results of "crosses" can be
predicted using Punnett squares. A phenotype is the physical expression of a gene (made up of a
pair of alleles). The genotype is the actual genetic makeup of the allele pair. An individual
having two identical alleles for a gene is said to be homozygous. There can be homozygous
dominant or homozygous recessive combinations. Dominant traits are represented by capital
letters; recessive by lower case letters. An individual having non-identical alleles for a gene is
said to be heterozygous. Note that the phenotype of a heterozygous individual is determined by
the dominant gene. Dominant alleles tend to cover up the presence of any recessive alleles. In a
case of alleles that show simple dominance / recessiveness, it is not possible to know if an
individual who possesses a dominant trait has the homozygous dominant or the hatarnzunniic
nanntune haced an nhanntuns tha onlv nne wo knnwe for rertain is the Accessibility: Investigate
Example: What phenotypes and genotypes could one expect from a cross between two pea
plants, one true-breeding for yellow seeds and the other true-breeding for green seeds? Yellow
seeds are dominant to green. Complete the Punnett square below. Y The true-breeding The true-
breeding green seed plant yellow seed plant can only contribute can only contribute a recessive
allele. a dominant allele. Many human traits are controlled by a single pair of alleles and through
simple dominant and recessive rules. Example: Tongue rolling - If you can roll your tongue
lengthwise, you have the trait controlled by the dominant allele. Let " R " represent the dominant
allele in your genotype and r represent the recessive allele. If you have the dominant phenotype,
how do you know if you are homozygous dominant or heterozygous. That depends upon
knowing if one of your parents couldn't roll their tongue. For example, my Dad cannot roll his
tongue but I can. So, my genotype is Rr for this trait. If you do know know about your parents,
then you have to put both possible genotypes for yourself, i.e. RR or Rr . 2. What is your
phenotype (roller or non-roller)? 3. What is your genotype? A. If you and your parents can both
ro.
Gregor Mendel used pea plants to study heredity in a series of exper.docxisaachwrensch
Gregor Mendel used pea plants to study heredity in a series of experiments. Mendel worked by carefully observing and recording traits in successive generations of plants. Knowledge about DNA and chromosomes came later.
This lab will apply genetic laws to human inheritance using Punnett squares.
Recall that DNA is wound tightly into chromosomes. Cells with only one set of chromosomes, such as sex cells, are
haploid
. When two haploid cells fuse during fertilization, a diploid zygote with two full sets of chromosomes is formed. Most cells of a mature individual are diploid.
Homologous chromosomes
have the same genes, but they might have different versions (
alleles
) of those genes.
Diploid
cells have two alleles for each gene. These alleles might be identical (gene A) or different (gene B). Each gene’s
locus
is its location on a chromosome.
Human traits come through dominant or recessive inheritance. For example, the cystic fibrosis traits carried by a dominant allele are always expressed, even if the recessive gene is present (FF or Ff). The recessive is only expressed when two copies of the recessive allele are present (ff).
Mother: Healthy carrier
F
f
Father: Healthy carrier
F
FF
Healthy non-carrier
Ff
Healthy carrier
f
Ff
Healthy carrier
FF
Affected
Human gender is carried on the X and Y chromosomes. Females are XX and males are XY. Heredity traits such as color blindness, which is the inability to distinguish among some colors, are carried on the X chromosome (X
c
). The presence of one normal X
C
will allow normal vision.
In this next set of exercises, you will determine the genotypes of the parents by considering the inheritance patterns of traits in their children. The following is a table of the phenotypes of the family members:
Phenotype
Alleles
Parents
Mother
Not color blind
Freckles
Type B blood
X
c
X
C
Ff
I
B
i
Father
Color blind
No freckles
Type A blood
X
c
Y
Ff
I
A
i
Children
Abby
Color blind
Freckles
X
c
X
c
Ff or ff
Brady
Not color blind
No freckles
X
C
Y
ff
Carly
Not color blind
No freckles
X
c
X
C
ff
Dennis
Color blind
Freckles
X
c
Y
Ff or ff
Exercise 1: Color Blindness
Using the alleles XC (not color blind) and Xc (color blind), distribute the gametes from each parent to the outside of the Punnett square. Drag and drop the child with the correct phenotype to the box within the Punnett square that has the corresponding genotype that would occur from the fusion of egg and sperm as indicated by your placement of the gametes.
Exercise 2: Freckles
Freckles are groups of cells on the skin that produce the pigment melanin, often in response to exposure to ultraviolet (UV) light. The gene for freckles is inherited in a dominant/recessive pattern. A person carrying even a single copy of the dominant allele (F) will have freckles. A person who is homozygous recessive (ff) will have no freckles.
Using the alleles F (freckles) and f (no freckles), distribute the gametes from each parent to the outside of the Punne.
Biology 106 EpistasisSex linked TraitsAnswer each question in.docxhartrobert670
Biology 106 Epistasis/Sex linked Traits
Answer each question in the space provided (in your own words of course). There is more than ample space for the answers. Don’t worry if your answer doesn’t take up the entire space!! Remember to use your name and the assignment name for the file name.
1. Marfan syndrome is a dominant disorder. In the cross Mm x mm, what percentage of the children are expected to inherit the disorder? Explain your answer.
2. How does non-disjunction cause Down syndrome? What is specifically happening in which process to cause this disorder?
3a. Labrador retrievers coat color is controlled by two genes. For the cross bbEe x BbEe, list the gametes produced by each parent.
3b. What color puppies will this cross produce?
3c. List the percentage for each color from this cross.
4a. Why are there more males with red/green colorblindness than females?
4b. If Xc denotes the colorblindness allele and X denotes the normal vision allele; From the cross XcX x XY What percentage of boys will be colorblind? What percentage of girls will be colorblind?
5. Shown below is an incompletepedigree for color blindness within a family. Individuals
with “?” provide no information for their phenotype. For all other individuals their genotype is evident from the diagram. From the information provided determine the genotypes and phenotypes for all individuals. Using the numbers in the diagram below, fill in the table that follows. You must determine which individuals are carriers and or afflicted from the
information provided.
Individual
Gender
Genotype
Phenotype
(Normal/Carrier/Colorblind)
1
2
3
4
5
6
7
8
6. Explain your determination of which individuals are carriers.
Use your textbook, notes and this website to answer the pre lab questions. http://biology.clc.uc.edu/Courses/bio105/sex-link.htmhttp://anthro.palomar.edu/biobasis/bio_4.htmhttp://www.biology.arizona.edu/Mendelian_genetics/problem_sets/sex_linked_inheritance/sex_linked_inheritance.html
PreLab Questions:
1. Define the term sex linked in your own words.
2. List 3 common human sex linked traits.
3. What is the most common sex linked trait in fruit flies.
4. What is the genotype of a colorblind female?
5. Define the genetic usage of the term “carrier”.
6. Can a male be a carrier of an X linked sex linked trait? Why or why not?
Click on the link below to access the online lab.
http://www.mhhe.com/biosci/genbio/virtual_labs_2K8/pages/LinkedTrait.html
Download and print the instructions for reference as you work through the lab. As you work through the lab fill in the table below. Use this information to answer the questions that follow contained in this document.
Begin by clicking on the notebook on the right hand side of the lab table. Explore the genetics of eye color and wing types by crossing various flies. Once you’ve ...
Biology 103 Laboratory Exercise – Genetic Problems
Introduction
Although the science of genetics has become a highly sophisticated discipline dealing
with the interactions of hereditary factors at the molecular level, it has its roots in the
basic laws of heredity initially discovered and presented by Gregor Mendel more than
one hundred years ago. Mendel's success in discovering these laws was due largely to his
application of the simple rules of mathematical probability - the laws of chance - to his
observations concerning the inheritance of certain characteristics in the garden pea plant.
Reginald Punnett and the Punnett Square
The Punnett square is a diagram used by biologists to determine genotypic probability
within the offspring from a particular genetic cross. The Punnett square shows every
possible genotypic combination of maternal alleles with the paternal alleles for a genetic
cross. Punnett squares only give probabilities for genotypes, not phenotypes. The square
diagram was designed by the British geneticist, Reginald Punnett (1865-1967) and first
presented to the science community in 1905. Punnett’s Mendelism (1905) is considered
the first popular science book to introduce genetics to the public.
Solving Genetic Problems
R
R'
R
RR RR'
R'
RR' R'R'
Maternal alleles
A
A
a
Aa
Aa
Paternal
Alleles
a
Aa
Aa
The first step in solving a genetic problem is to establish the genetic symbols you will use
in your problem solution. Stay consistent by using these same symbols throughout the
problem solving process.
Represent dominant and recessive alleles (different forms of a gene) using traditional
genetic symbols. Dominant alleles should be represented with the capital version of an
alphabetic letter while using the lower case version to show recessiveness. For example:
B = black color, b = white color.
Each individual gene or trait is diploid (2n) in nature and therefore, must be represented
with two alleles. Continuing with the alleles mentioned previously, an individual may
have the genetic makeup BB, Bb, or bb when using those alleles.
Remember that gametes (sperm and egg) are haploid (n) and can only provide one allele
per trait. For example: B or b
An individual’s genotype contains the possible gametes that can be expected to be
produced by that individual. Much of genetics revolves around the probability of the
makeup of gametes. If the individual is homozygous, all of the gametes produced will
possess the same kind of allele. For example, an individual with the genotype BB would
be expected to produce only B gametes and individuals with genotype bb would produce
only b gametes.
If the individual is heterozygous, that is the individual’s genotype contains one dominant
allele and one recessive allele (Bb), the gametes produced will possess one or the other of
the two forms of the gene – B or b. ...
Blaine Kitchenware – Case AssignmentMBA – Corporate Finance.docxAASTHA76
Blaine Kitchenware – Case Assignment
MBA – Corporate Finance
You have been hired as a consultant to Victor Dubinski, the CEO of Blaine Kitchenware. You are charged with putting together a written report with supporting numerical analysis that addresses the following items:
1. Is the current capital structure and payout policy for Blaine optimal? Explain and justify your conclusion. Use numbers whenever possible.
2. Should Blaine recommend a large share repurchase to the Board of Directors? What are the advantages and disadvantages of this action? Again, explain and justify your conclusions. Use numbers whenever possible.
3. Consider two specific share repurchase proposals:
a. First Proposal
i. Blaine will issue $50 million in new debt at an interest rate of 6.75%
ii. Blaine will use $209 million of cash from its balance sheet
iii. Blaine will use these two sources of cash to repurchase 14 million shares at $18.50/share.
b. Second Proposal
i. Blaine will issue $95 million in new debt at an interest of 6.875%
ii. Blaine will use $209 million of cash from its balance sheet
iii. Blaine will use the two sources of cash to repurchase 16 million shares at $19.00/share.
c. Third Proposal
i. Blaine will issue $156 million in new debt at 7.125%
ii. Blaine will use $209 million of cash from its balance sheet
iii. Blaine will use the two sources of cash to repurchase 18.5 million shares.
d. How does a share repurchase affect Blaine? Consider the impact on items including (but not limited to) Blaine’s EPS, ROE, interest coverage, debt ratio, debt rating, the family’s ownership interest (proportion of shares outstanding) and the company’s cost of capital (WACC).
e. What is your recommendation for the good of the company based on your analysis of the share repurchase options? Should they stay with the status quo, or go with one of the recapitalization options?
BSC1005 Biology General
Chapter 10
Patterns of Inheritance
1
Chapter 10: Patterns of Inheritance, Mendel Laws
Multiple-Choice Questions
2) Which of the following statements best represents the theory of pangenesis developed by Hippocrates?
A) Pregnancy is a spontaneous event, and the characteristics of the offspring are determined by the gods.
B) Particles called pangenes, which originate in each part of an organism's body, collect in the sperm or eggs and are
passed on to the next generation.
C) Offspring inherit the traits of either the mother or the father, but not both.
D) Fertilization of plants is dependent on an animal.
E) Heritable traits are influenced by the environment and the behaviors of the parents.
3) Which of the following statements regarding hypotheses about inheritance is false?
A) The theory of pangenesis incorrectly suggests that reproductive cells receive particles from somatic cells.
B) Contrary to the theory of pangenesis, somatic cells do not influence eggs or sperm.
C) The blending hypothesis does not explain how trait ...
Study Guide Chapter 15 -Chromosomal Basis of Inheritance-Answers.docxhanneloremccaffery
Study Guide Chapter 15 -Chromosomal Basis of Inheritance-Answers
15.1 Mendelian Inheritance and chromosome theory______________________________
1. Thomas Hunt Morgan identified the first solid evidence associating a specific___gene_____ on a specific chromosome.
2. Why did Morgan choose Drosophila as his experimental organism? (List 3 reasons)
They reproduce quickly, a new generation of adults forms every two weeks!
Prolific (single matting = hundreds of offspring)
Only 4 chromosome pairs = simple genetics
3A. The normal phenotype for a character (phenotype most common in nature) is called___wild type_____.
3B. Provide at least two examples this phenotype in Drosophila.
Red eyes
gray body
4A. An alternative phenotype for a character (phenotype not common in nature) is called____mutant____.
4B. Provide at least two examples of this phenotype in Drosophila.
White eyes
Black body
5. Morgan and his students invented a notation to symbolize alleles in Drosophila that differed from the notation Mendel used to represent alleles. Describe how Morgan’s Drosophila characters are named and the symbol used for the allele type.
You can think of Morgan’s wild-type allele as equivalent to the dominant allele in Mendel’s naming system. And you can think of Morgan’s mutant allele as equivalent to Mendel’s recessive allele. So whenever you have one wild-type allele and one mutant allele, that organism will have the wild-type phenotype.
Mendel
Used the first letter of the dominant character name to represent the dominant allele. This letter was capitalized for the dominant allele, lower case for the recessive allele
Ex: purple flower allele dominant to white flower allele: P= dominant allele, p=recessive allele
Morgan
-Used the first letter of the mutant character name to represent the wild-type allele. The letter for the wild-type and mutant alleles are both lower case. The wild-type allele gets a + sign and the mutant does not.
Ex: red eye allele wild-type to white eye mutant: w+ = wild-type allele, w= recessive allele.
6. Morgan and his students invented a notation to symbolize alleles in Drosophila. Which of the following genotypes would produce a fly that is wild-type for eye color (red vs. white eyes)?
w+ w+
w+ w
w w
w+w+ and w+w
7. Morgan performed an experiment that yielded a 3:1 ratio of offspring in the F2 generation; however, only the males of this F2 generation had white eyes.
White- eye females DO EXIST, so why were only white-eye males observed in this cross?
Complete the following chart and describe why Morgan observed these results and how it allowed him to conclusively determine that the gene for eye color was located on the X chromosome.
Morgan’s experiment (see Figure 15.4) Use notation developed by Morgan and his students
If the gene for eye color is located on the X chromosome….
P generation
Red eye Female genotype for eye color = w+w+
White eye Male genotype for eye color= w
F1 generatio.
Dragon Genetics Hands on LabIntroductionThere are many patterns o.docxmadlynplamondon
Dragon Genetics Hands on LabIntroduction:
There are many patterns of inheritance. This lab will explore Mendelian traits where alleles are dominant or recessive. You will also have to understand genes that express incomplete dominance, codominance and are sex-linked. We will utilize this lab to explore those terms and to use our knowledge of gene combinations to create an organism.
Utilize this link to learn more about Mendelian traits, incomplete dominance and codominance.(Opens in new window) This link helps you to learn more about sex-linked genes. (Opens in new window)Part 1 Procedure: Checking for Understanding (Worth up to 15pts)
A. For each genotype, tell me if it’s homozygous or heterozygous. If it is homozygous also tell me if it is dominant or recessive.
1. BB
2. jj
3. Pp
4. RR
5. ff
B. Dimples (D) is dominant to (d) no dimples. Brown Eyes (B) are dominant to (b) blue eyes. For each genotype indicate the phenotype.
1. Dd
2. DD
3. dd
4. BB
5. bb
6. Bb
C. For each phenotype list the genotype(s). Trait: Red flower (R); White flower (r)
1. Red flower (homozygous)
2. Red flower (heterozygous)
3. White flowerPart 2: Creating the Parents (Worth up to 48 pts)
1. Flip a coin to determine the genotype for each trait and record it in the data table.
a. Heads= allele 1, Tails = allele 2 ( Example: if you flipped heads twice, your dragon will have two copies of allele 1 for her genotype)
2. Determine the phenotype resulting from the allele pair for each trait.
3. Repeat steps 1-2 for each trait and complete the Mama Dragon’s Table 1.
a. Pay attention to the notes under the trait; it means it does not fully follow Mendelian rules.
Example:
For the trait Eye Size. I flipped my coin 2 times. Flip 1= heads (E); Flip 2= heads (E). So I would put EE in the Genotype box and Two Small Eyes in the Phenotype box for that trait.
Table 1: Genotypes and Phenotypes for Mama Dragon
Trait
Allele 1
Allele 2
Genotype
Phenotype
Eye Size & Number
Two Small (E)
One large (e)
Eye Color
(incomplete Dom)
Red (R)
White (r)
Tail Shape
Curly (T)
Straight (t)
Tail Color
Purple (P)
Orange (p)
Teeth
Sharp (S)
Round (s)
Feet
Four toes (F)
Two toes (f)
Horn Color
(Codominant)
Purple (P)
White (W)
Ear Shape
Pointy (Y)
Round (y)
Claws
Long (L)
Short (l)
Skin Color
(Codominant)
Blue (B)
Yellow (Y)
Mane Texture
(incomplete Dom)
Curly (M)
Straight (m)
Dragon Hemophilia
(Sex- Linked)
Hemophilia (XH)
Normal Clotting (Xh)
Table 2: Genotype and Phenotypes for Daddy Dragon
Trait
Allele 1
Allele 2
Genotype
Phenotype
Eye Size & Number
Two Small (E)
One large (e)
Eye Color
(incomplete Dom)
Red (R)
White (r)
Tail Shape
Curly (T)
Straight (t)
Tail Color
Purple (P)
Orange (p)
Teeth
Sharp (S)
Round (s)
Feet
Four toes (F)
Two toes (f)
Horn Color
(Codominant)
Purple (P)
White (W)
Ear Shape
Pointy (Y)
Round (y)
Claws
Long (L)
Short (l)
Skin Color
(Codominant)
Blue (B)
Yellow (Y)
Mane Texture
(incomplete Dom)
Curly (M)
Straigh ...
Human body cells contain 46 chromosomes- The first 22 pairs are called.pdfkrishnac481
Human body cells contain 46 chromosomes. The first 22 pairs are called autosomes, and they
contain numerous genes that affect the traits of the individual. The last pair, number 23, are the
sex chromosomes. The sex chromosomes determine gender (i.e., either male or female), but there
are other genes on this pair of chromosomes as well. Males sex chromosomes are XY, while
female sex chromosomes are XX . The gametes in a human (either egg cells or sperm cells)
contain only 23 chromosomes. Fertilization, the fusion of an egg and a sperm, restores the total
of 46 chromosomes in a human zygote. Non-gamete cells are called somatic cells, and they have
all 46 chromosomes in them. 1. Your sex chromosomes: XX Just as in Mendel's pea plant
experiments, genes in humans can be dominant or recessive, and the results of "crosses" can be
predicted using Punnett squares. A phenotype is the physical expression of a gene (made up of a
pair of alleles). The genotype is the actual genetic makeup of the allele pair. An individual
having two identical alleles for a gene is said to be homozygous. There can be homozygous
dominant or homozygous recessive combinations. Dominant traits are represented by capital
letters; recessive by lower case letters. An individual having non-identical alleles for a gene is
said to be heterozygous. Note that the phenotype of a heterozygous individual is determined by
the dominant gene. Dominant alleles tend to cover up the presence of any recessive alleles. In a
case of alleles that show simple dominance / recessiveness, it is not possible to know if an
individual who possesses a dominant trait has the homozygous dominant or the hatarnzunniic
nanntune haced an nhanntuns tha onlv nne wo knnwe for rertain is the Accessibility: Investigate
Example: What phenotypes and genotypes could one expect from a cross between two pea
plants, one true-breeding for yellow seeds and the other true-breeding for green seeds? Yellow
seeds are dominant to green. Complete the Punnett square below. Y The true-breeding The true-
breeding green seed plant yellow seed plant can only contribute can only contribute a recessive
allele. a dominant allele. Many human traits are controlled by a single pair of alleles and through
simple dominant and recessive rules. Example: Tongue rolling - If you can roll your tongue
lengthwise, you have the trait controlled by the dominant allele. Let " R " represent the dominant
allele in your genotype and r represent the recessive allele. If you have the dominant phenotype,
how do you know if you are homozygous dominant or heterozygous. That depends upon
knowing if one of your parents couldn't roll their tongue. For example, my Dad cannot roll his
tongue but I can. So, my genotype is Rr for this trait. If you do know know about your parents,
then you have to put both possible genotypes for yourself, i.e. RR or Rr . 2. What is your
phenotype (roller or non-roller)? 3. What is your genotype? A. If you and your parents can both
ro.
Gregor Mendel used pea plants to study heredity in a series of exper.docxisaachwrensch
Gregor Mendel used pea plants to study heredity in a series of experiments. Mendel worked by carefully observing and recording traits in successive generations of plants. Knowledge about DNA and chromosomes came later.
This lab will apply genetic laws to human inheritance using Punnett squares.
Recall that DNA is wound tightly into chromosomes. Cells with only one set of chromosomes, such as sex cells, are
haploid
. When two haploid cells fuse during fertilization, a diploid zygote with two full sets of chromosomes is formed. Most cells of a mature individual are diploid.
Homologous chromosomes
have the same genes, but they might have different versions (
alleles
) of those genes.
Diploid
cells have two alleles for each gene. These alleles might be identical (gene A) or different (gene B). Each gene’s
locus
is its location on a chromosome.
Human traits come through dominant or recessive inheritance. For example, the cystic fibrosis traits carried by a dominant allele are always expressed, even if the recessive gene is present (FF or Ff). The recessive is only expressed when two copies of the recessive allele are present (ff).
Mother: Healthy carrier
F
f
Father: Healthy carrier
F
FF
Healthy non-carrier
Ff
Healthy carrier
f
Ff
Healthy carrier
FF
Affected
Human gender is carried on the X and Y chromosomes. Females are XX and males are XY. Heredity traits such as color blindness, which is the inability to distinguish among some colors, are carried on the X chromosome (X
c
). The presence of one normal X
C
will allow normal vision.
In this next set of exercises, you will determine the genotypes of the parents by considering the inheritance patterns of traits in their children. The following is a table of the phenotypes of the family members:
Phenotype
Alleles
Parents
Mother
Not color blind
Freckles
Type B blood
X
c
X
C
Ff
I
B
i
Father
Color blind
No freckles
Type A blood
X
c
Y
Ff
I
A
i
Children
Abby
Color blind
Freckles
X
c
X
c
Ff or ff
Brady
Not color blind
No freckles
X
C
Y
ff
Carly
Not color blind
No freckles
X
c
X
C
ff
Dennis
Color blind
Freckles
X
c
Y
Ff or ff
Exercise 1: Color Blindness
Using the alleles XC (not color blind) and Xc (color blind), distribute the gametes from each parent to the outside of the Punnett square. Drag and drop the child with the correct phenotype to the box within the Punnett square that has the corresponding genotype that would occur from the fusion of egg and sperm as indicated by your placement of the gametes.
Exercise 2: Freckles
Freckles are groups of cells on the skin that produce the pigment melanin, often in response to exposure to ultraviolet (UV) light. The gene for freckles is inherited in a dominant/recessive pattern. A person carrying even a single copy of the dominant allele (F) will have freckles. A person who is homozygous recessive (ff) will have no freckles.
Using the alleles F (freckles) and f (no freckles), distribute the gametes from each parent to the outside of the Punne.
Biology 106 EpistasisSex linked TraitsAnswer each question in.docxhartrobert670
Biology 106 Epistasis/Sex linked Traits
Answer each question in the space provided (in your own words of course). There is more than ample space for the answers. Don’t worry if your answer doesn’t take up the entire space!! Remember to use your name and the assignment name for the file name.
1. Marfan syndrome is a dominant disorder. In the cross Mm x mm, what percentage of the children are expected to inherit the disorder? Explain your answer.
2. How does non-disjunction cause Down syndrome? What is specifically happening in which process to cause this disorder?
3a. Labrador retrievers coat color is controlled by two genes. For the cross bbEe x BbEe, list the gametes produced by each parent.
3b. What color puppies will this cross produce?
3c. List the percentage for each color from this cross.
4a. Why are there more males with red/green colorblindness than females?
4b. If Xc denotes the colorblindness allele and X denotes the normal vision allele; From the cross XcX x XY What percentage of boys will be colorblind? What percentage of girls will be colorblind?
5. Shown below is an incompletepedigree for color blindness within a family. Individuals
with “?” provide no information for their phenotype. For all other individuals their genotype is evident from the diagram. From the information provided determine the genotypes and phenotypes for all individuals. Using the numbers in the diagram below, fill in the table that follows. You must determine which individuals are carriers and or afflicted from the
information provided.
Individual
Gender
Genotype
Phenotype
(Normal/Carrier/Colorblind)
1
2
3
4
5
6
7
8
6. Explain your determination of which individuals are carriers.
Use your textbook, notes and this website to answer the pre lab questions. http://biology.clc.uc.edu/Courses/bio105/sex-link.htmhttp://anthro.palomar.edu/biobasis/bio_4.htmhttp://www.biology.arizona.edu/Mendelian_genetics/problem_sets/sex_linked_inheritance/sex_linked_inheritance.html
PreLab Questions:
1. Define the term sex linked in your own words.
2. List 3 common human sex linked traits.
3. What is the most common sex linked trait in fruit flies.
4. What is the genotype of a colorblind female?
5. Define the genetic usage of the term “carrier”.
6. Can a male be a carrier of an X linked sex linked trait? Why or why not?
Click on the link below to access the online lab.
http://www.mhhe.com/biosci/genbio/virtual_labs_2K8/pages/LinkedTrait.html
Download and print the instructions for reference as you work through the lab. As you work through the lab fill in the table below. Use this information to answer the questions that follow contained in this document.
Begin by clicking on the notebook on the right hand side of the lab table. Explore the genetics of eye color and wing types by crossing various flies. Once you’ve ...
Biology 103 Laboratory Exercise – Genetic Problems
Introduction
Although the science of genetics has become a highly sophisticated discipline dealing
with the interactions of hereditary factors at the molecular level, it has its roots in the
basic laws of heredity initially discovered and presented by Gregor Mendel more than
one hundred years ago. Mendel's success in discovering these laws was due largely to his
application of the simple rules of mathematical probability - the laws of chance - to his
observations concerning the inheritance of certain characteristics in the garden pea plant.
Reginald Punnett and the Punnett Square
The Punnett square is a diagram used by biologists to determine genotypic probability
within the offspring from a particular genetic cross. The Punnett square shows every
possible genotypic combination of maternal alleles with the paternal alleles for a genetic
cross. Punnett squares only give probabilities for genotypes, not phenotypes. The square
diagram was designed by the British geneticist, Reginald Punnett (1865-1967) and first
presented to the science community in 1905. Punnett’s Mendelism (1905) is considered
the first popular science book to introduce genetics to the public.
Solving Genetic Problems
R
R'
R
RR RR'
R'
RR' R'R'
Maternal alleles
A
A
a
Aa
Aa
Paternal
Alleles
a
Aa
Aa
The first step in solving a genetic problem is to establish the genetic symbols you will use
in your problem solution. Stay consistent by using these same symbols throughout the
problem solving process.
Represent dominant and recessive alleles (different forms of a gene) using traditional
genetic symbols. Dominant alleles should be represented with the capital version of an
alphabetic letter while using the lower case version to show recessiveness. For example:
B = black color, b = white color.
Each individual gene or trait is diploid (2n) in nature and therefore, must be represented
with two alleles. Continuing with the alleles mentioned previously, an individual may
have the genetic makeup BB, Bb, or bb when using those alleles.
Remember that gametes (sperm and egg) are haploid (n) and can only provide one allele
per trait. For example: B or b
An individual’s genotype contains the possible gametes that can be expected to be
produced by that individual. Much of genetics revolves around the probability of the
makeup of gametes. If the individual is homozygous, all of the gametes produced will
possess the same kind of allele. For example, an individual with the genotype BB would
be expected to produce only B gametes and individuals with genotype bb would produce
only b gametes.
If the individual is heterozygous, that is the individual’s genotype contains one dominant
allele and one recessive allele (Bb), the gametes produced will possess one or the other of
the two forms of the gene – B or b. ...
Blaine Kitchenware – Case AssignmentMBA – Corporate Finance.docxAASTHA76
Blaine Kitchenware – Case Assignment
MBA – Corporate Finance
You have been hired as a consultant to Victor Dubinski, the CEO of Blaine Kitchenware. You are charged with putting together a written report with supporting numerical analysis that addresses the following items:
1. Is the current capital structure and payout policy for Blaine optimal? Explain and justify your conclusion. Use numbers whenever possible.
2. Should Blaine recommend a large share repurchase to the Board of Directors? What are the advantages and disadvantages of this action? Again, explain and justify your conclusions. Use numbers whenever possible.
3. Consider two specific share repurchase proposals:
a. First Proposal
i. Blaine will issue $50 million in new debt at an interest rate of 6.75%
ii. Blaine will use $209 million of cash from its balance sheet
iii. Blaine will use these two sources of cash to repurchase 14 million shares at $18.50/share.
b. Second Proposal
i. Blaine will issue $95 million in new debt at an interest of 6.875%
ii. Blaine will use $209 million of cash from its balance sheet
iii. Blaine will use the two sources of cash to repurchase 16 million shares at $19.00/share.
c. Third Proposal
i. Blaine will issue $156 million in new debt at 7.125%
ii. Blaine will use $209 million of cash from its balance sheet
iii. Blaine will use the two sources of cash to repurchase 18.5 million shares.
d. How does a share repurchase affect Blaine? Consider the impact on items including (but not limited to) Blaine’s EPS, ROE, interest coverage, debt ratio, debt rating, the family’s ownership interest (proportion of shares outstanding) and the company’s cost of capital (WACC).
e. What is your recommendation for the good of the company based on your analysis of the share repurchase options? Should they stay with the status quo, or go with one of the recapitalization options?
BSC1005 Biology General
Chapter 10
Patterns of Inheritance
1
Chapter 10: Patterns of Inheritance, Mendel Laws
Multiple-Choice Questions
2) Which of the following statements best represents the theory of pangenesis developed by Hippocrates?
A) Pregnancy is a spontaneous event, and the characteristics of the offspring are determined by the gods.
B) Particles called pangenes, which originate in each part of an organism's body, collect in the sperm or eggs and are
passed on to the next generation.
C) Offspring inherit the traits of either the mother or the father, but not both.
D) Fertilization of plants is dependent on an animal.
E) Heritable traits are influenced by the environment and the behaviors of the parents.
3) Which of the following statements regarding hypotheses about inheritance is false?
A) The theory of pangenesis incorrectly suggests that reproductive cells receive particles from somatic cells.
B) Contrary to the theory of pangenesis, somatic cells do not influence eggs or sperm.
C) The blending hypothesis does not explain how trait ...
What is greenhouse gasses and how many gasses are there to affect the Earth.moosaasad1975
What are greenhouse gasses how they affect the earth and its environment what is the future of the environment and earth how the weather and the climate effects.
Nutraceutical market, scope and growth: Herbal drug technologyLokesh Patil
As consumer awareness of health and wellness rises, the nutraceutical market—which includes goods like functional meals, drinks, and dietary supplements that provide health advantages beyond basic nutrition—is growing significantly. As healthcare expenses rise, the population ages, and people want natural and preventative health solutions more and more, this industry is increasing quickly. Further driving market expansion are product formulation innovations and the use of cutting-edge technology for customized nutrition. With its worldwide reach, the nutraceutical industry is expected to keep growing and provide significant chances for research and investment in a number of categories, including vitamins, minerals, probiotics, and herbal supplements.
A brief information about the SCOP protein database used in bioinformatics.
The Structural Classification of Proteins (SCOP) database is a comprehensive and authoritative resource for the structural and evolutionary relationships of proteins. It provides a detailed and curated classification of protein structures, grouping them into families, superfamilies, and folds based on their structural and sequence similarities.
(May 29th, 2024) Advancements in Intravital Microscopy- Insights for Preclini...Scintica Instrumentation
Intravital microscopy (IVM) is a powerful tool utilized to study cellular behavior over time and space in vivo. Much of our understanding of cell biology has been accomplished using various in vitro and ex vivo methods; however, these studies do not necessarily reflect the natural dynamics of biological processes. Unlike traditional cell culture or fixed tissue imaging, IVM allows for the ultra-fast high-resolution imaging of cellular processes over time and space and were studied in its natural environment. Real-time visualization of biological processes in the context of an intact organism helps maintain physiological relevance and provide insights into the progression of disease, response to treatments or developmental processes.
In this webinar we give an overview of advanced applications of the IVM system in preclinical research. IVIM technology is a provider of all-in-one intravital microscopy systems and solutions optimized for in vivo imaging of live animal models at sub-micron resolution. The system’s unique features and user-friendly software enables researchers to probe fast dynamic biological processes such as immune cell tracking, cell-cell interaction as well as vascularization and tumor metastasis with exceptional detail. This webinar will also give an overview of IVM being utilized in drug development, offering a view into the intricate interaction between drugs/nanoparticles and tissues in vivo and allows for the evaluation of therapeutic intervention in a variety of tissues and organs. This interdisciplinary collaboration continues to drive the advancements of novel therapeutic strategies.
Introduction:
RNA interference (RNAi) or Post-Transcriptional Gene Silencing (PTGS) is an important biological process for modulating eukaryotic gene expression.
It is highly conserved process of posttranscriptional gene silencing by which double stranded RNA (dsRNA) causes sequence-specific degradation of mRNA sequences.
dsRNA-induced gene silencing (RNAi) is reported in a wide range of eukaryotes ranging from worms, insects, mammals and plants.
This process mediates resistance to both endogenous parasitic and exogenous pathogenic nucleic acids, and regulates the expression of protein-coding genes.
What are small ncRNAs?
micro RNA (miRNA)
short interfering RNA (siRNA)
Properties of small non-coding RNA:
Involved in silencing mRNA transcripts.
Called “small” because they are usually only about 21-24 nucleotides long.
Synthesized by first cutting up longer precursor sequences (like the 61nt one that Lee discovered).
Silence an mRNA by base pairing with some sequence on the mRNA.
Discovery of siRNA?
The first small RNA:
In 1993 Rosalind Lee (Victor Ambros lab) was studying a non- coding gene in C. elegans, lin-4, that was involved in silencing of another gene, lin-14, at the appropriate time in the
development of the worm C. elegans.
Two small transcripts of lin-4 (22nt and 61nt) were found to be complementary to a sequence in the 3' UTR of lin-14.
Because lin-4 encoded no protein, she deduced that it must be these transcripts that are causing the silencing by RNA-RNA interactions.
Types of RNAi ( non coding RNA)
MiRNA
Length (23-25 nt)
Trans acting
Binds with target MRNA in mismatch
Translation inhibition
Si RNA
Length 21 nt.
Cis acting
Bind with target Mrna in perfect complementary sequence
Piwi-RNA
Length ; 25 to 36 nt.
Expressed in Germ Cells
Regulates trnasposomes activity
MECHANISM OF RNAI:
First the double-stranded RNA teams up with a protein complex named Dicer, which cuts the long RNA into short pieces.
Then another protein complex called RISC (RNA-induced silencing complex) discards one of the two RNA strands.
The RISC-docked, single-stranded RNA then pairs with the homologous mRNA and destroys it.
THE RISC COMPLEX:
RISC is large(>500kD) RNA multi- protein Binding complex which triggers MRNA degradation in response to MRNA
Unwinding of double stranded Si RNA by ATP independent Helicase
Active component of RISC is Ago proteins( ENDONUCLEASE) which cleave target MRNA.
DICER: endonuclease (RNase Family III)
Argonaute: Central Component of the RNA-Induced Silencing Complex (RISC)
One strand of the dsRNA produced by Dicer is retained in the RISC complex in association with Argonaute
ARGONAUTE PROTEIN :
1.PAZ(PIWI/Argonaute/ Zwille)- Recognition of target MRNA
2.PIWI (p-element induced wimpy Testis)- breaks Phosphodiester bond of mRNA.)RNAse H activity.
MiRNA:
The Double-stranded RNAs are naturally produced in eukaryotic cells during development, and they have a key role in regulating gene expression .
This pdf is about the Schizophrenia.
For more details visit on YouTube; @SELF-EXPLANATORY;
https://www.youtube.com/channel/UCAiarMZDNhe1A3Rnpr_WkzA/videos
Thanks...!
1. Pre-AP Biology Genetics Practice Problems
You WILL have problems like these on the test. Practice! Due Monday November 22nd
Part 1 Introduction:
1. Describe the genotypes given (use your notes). The first two are already done.
A. DD homozygous, dominant D. ss ______________________
B. Dd _heterozygous E. Yy ______________________
C. dd __________________ F. WW ____________________
2. In humans, brown eye color (B), is dominant over blue eye color (b). What are the phenotypes
of the following genotypes? In other words, what color eyes will they have?
A. BB ________________________
B. bb ________________________
C. Bb ________________________
The Five (5) Steps Associated With Solving a Genetics Problem:
If you take the time to follow the directions below, you will be able to solve most genetics problems.
1. Determine the genotypes of the parents or whatever is given in problem.
2. Set up your Punnett square as follows: *# sq. based on possible gametes that can be formed.
3. Fill in the squares. This represents the possible combinations that could occur during fertilization.
4. Write out the possible genotypic ratio of the offspring.
5. Using the genotypic ratio determine the phenotypic ratio for the offspring.
.
Part 2: Sample Problem (Just read this over, it is a practice problem)
A heterozygous male, black eyed mouse is crossed with a red eyed, female mouse.
Predict the possible offspring!
Step 1: Determine the genotype of the parents. The male parent is heterozygous which means he
has one allele for black eyes and one allele for red eyes. Since his eyes are black, this means that
black allele must be dominant over the red allele. So the male parents genotype is “Bb” (B =
allele for black eye, b = allele for red eye).
The female parent has red eyes, there is only one way to have this recessive phenotype, so she
must to be homozygous recessive. Homozygous recessive means that her genotype must be
“bb”. Therefore, genotype of the parents is Bb x bb.
2. Step 2:
During meiosis (the formation of sex cells) one member (allele) of each gene pair separate. The male
mouse (Bb) produces some sperm containing “B” (the allele for black eye) and some sperm
containing “b” (the allele for red eyes).
On one axis of the Punnett square you put the two possible gametes for the male.
Repeat this for the other axis for the possible female gametes. Since she is “bb” you must put
“b”and “b”.
Step 3:
During fertilization sperm meets the egg. The Punnett square show us the various possibilities
during fertilization. The offspring must be one of these genotypes listed in the squares.
Repeating the process we can see all of the possible genotypes.
Step 4:
The genotypic ratio is determined by counting each possible genotype. You’ll note there are two
“Bb” for every two “bb”. Therefore, we write the ratio as 2 : 2
Bb : bb
Normally we reduce to the lowest terms: 1 : 1
Bb : bb
3. Step 5:
The Bb will produce a black eyed mouse (phenotype) and the bb will produce a red eyed mouse
(phenotype). The phenotypic ratio is written as 1 : 1
black eye : red eye
Ratios tell you there is an even chance of having offspring with black eyes as there is for having
offspring with red eyes. That would be the same as a 50% probability of having red eyes, or a
50% probability of having black eyes.
Part 3 Monohybrid Cross
When we study the inheritance of a single gene it is called a monohybrid cross.
**On the following pages are several problems.
1. A heterozygous, smooth pea pod, plant is crossed with a wrinkled pea pod plant. There are
two alleles for pea pod, smooth and wrinkled. Predict the offspring from this cross.
a. What is the the genotype of the parents? ________
b. Set up a Punnett square with possible gametes below.
c. Fill in the Punnett square for the resultant offspring.
d. What is the predicted genotypic ratio for the offspring ?___________________
e. What is the predicted phenotypic ratio for the offspring ?__________________
f. If this cross produced 50 seeds how many would you predict to have a wrinkled pod?
2. In humans, acondroplasia “dwarfism” (D) is dominant over normal (d). A homozygous
dominant (DD) person dies before the age of one. A heterozygous (Dd) person is dwarfed. A
homozygous recessive individual is normal. A heterozygous dwarf man marries a dwarf
heterozygous woman……..
a. What is the probability of having a normal child? _________
b. What is the probability that the next child will also be normal? __________
b. What is the probability of having a child that is a dwarf? __________
b. What is the probability of having a child that dies at one from this disorder? __________
4. 3. In humans, free earlobes (F) is dominant over attached earlobes (f). If one parent is
homozygous dominant for free earlobes, while the other has attached earlobes can they produce
any children with attached earlobes? Show your work in the punnett square below.
Answer:
4. In humans widow’s peak (W) is dominant over straight hairline (w). A heterozygous man for
this trait marries a woman who is also heterozygous.
a. List possible genotypes of their offspring.
b. List the phenotypic ratio for their children.
Part 4: Working Backwards
Sometimes we only know about the offspring and we want to learn about the parents. If you have been
paying attention, you should have started to notice a pattern. For example, when both parents are
heterozygous the phenotypic ratio always comes out 3 to 1. If one parent is homozygous recessive and
the other is heterozygous, the phenotypic ratio always comes out 1 to1. Keeping this in mind see if
you can solve the next two problems:
1. In pea plants, yellow seeds (Y) are dominant and green seeds (y) are recessive. A pea plant
with yellow seeds is crossed with a pea plant with green seeds. The resulting offspring have
about equal numbers of yellow and green seeded plants. What are the genotypes of the parents?
Answer:
2. In another cross, a yellow seeded plant was crossed with another yellow seeded plant and it
produced offspring of which about 25% were green seeded plants. What are the genotypes of
both parents?
Answer:
5. Part 5: Back Cross/Test Cross
When an organism has the dominant phenotype, then its genotype can be either heterozygous or
homozygous dominant (you can’t tell by looking at it). In order to find out we must do a test cross
using an homozygous, recessive organism. For example:
In Dalmatian dogs, the gene for black spots is dominant to the gene for liver colored spots. If a
breeder has a black spotted dog, how can she find out whether it is homozygous(BB) or
heterozygous(Bb) spotted dog? *B = black spots and b = liver spots
If the breeder finds a black spotted dog, whose ancestry is not known, she cannot tell by looking at
the dog if it is BB or Bb. She should find a liver spotted dog, whose genotype must be “bb” and mate
it with the black spotted dog in question.
This is the cross of a homozygous (BB) individual:
Notice that all of the offspring will be Bb and
therefore, there is no possibility of having a
liver spotted offspring.
*This would be the resultant Punnett sq. for the
heterozygous(Bb) individual.
If any of the breed offspring has liver spots, then she can say that she had a heterozygous black
spotted dog. If all the offspring had black spots then she can say that the suspect dog was
homozygous.
Practice Problems
1. You found a wild, black mouse. Explain how you would determine the genotype of this mouse.
*Hint in mice, white fur is recessive.
a. Draw Punnett squares for your possible crosses.
b. You have 24 offspring, 23 with black fur and 1 with white fur. What was the genotype of
the mouse? __________
c. If you only had 3 black offspring, can you tell what the genotype was of the suspect
mouse? Explain why or why not.
6. Part 6: Dihybrid Cross
When we study two traits on different chromosomes, at one time, we call this a dihybrid cross.
You still follow the same five step process for Monohybrid crosses but now there will be four
times as many possibilities because we are studying two traits.
Example: A female guinea pig is heterozygous for both fur color and coat texture is crossed with a
male that has light fur color and is heterozygous for coat texture. What possible offspring can they
produce? Dark fur color is dominant (D) and light fur (d) is recessive. Rough coat texture (R) is
dominant, while smooth coat (r) is recessive.
Step 1: The guinea pig that is heterozygous for both color and texture this means it has one
allele for each trait. Therefore its genotype would be “DdRr”. The other guinea pig has light
fur, since that is a recessive trait the genotype for that trait must be “dd”. It is also heterozygous
for fur texture, which means a genotype of “Rr”. All together its overall genotype must be“ddRr”.
Step 2 and 3: The Punnett square will be larger now because there are more possible sperm and
egg combinations. During the formation of sperm a “D” could go with a “R” producing a
sperm “DR”, or a “D” could go with a “r” forming a sperm with “Dr”.
Filling-in the Punnett square it should look like the one we started below . Finish off filling in
the blank squares in the Punnett square.
Parent 1: DdRr
Step 4: After filling-in the Punnett square you should obtain the following genotypic ratio:
*remember the numbers should add up to the number of squares filled in:
4 DdRr : 2 DdRR : 4 ddRr : 2 ddRR : 2 Ddrr : 2 ddrr
Parent 1: DdRr
DR Dr dR dr
DdRr ddRR ddRr
Ddrr ddRr ddrr
DdRr ddRR ddRr
Ddrr ddRr ddrr
Step 5: There will be only four different phenotypes because the 4 DdRr and the 2 DdRR will
have dark fur with rough coat, and the 4 with ddRr and the 2 ddRR will have light fur
with rough coat, while the 2 Ddrr will have dark fur with smooth coat and the 2 ddrr will
have light fur with smooth coat.
Therefore the phenotypic ratio would be:
6 dark, rough (4 DdRr and the 2 DdRR) 6 light rough (4 with ddRr and the 2 ddRR)
2 dark smooth (2 Ddrr) 2 light smooth (2 ddrr).
d r
dR
dr
dR
DdRR
DdRr
d R
dr
DdRR
DdRr
Parent 2: ddRr
Parent 2: ddRr
7. 1. In pea plants, the round seed allele is dominant over the wrinkled seed allele, and the yellow
seed allele is dominant over the green seed allele. The genes for seed texture and those for seed
color are on different chromosomes. A plant heterozygous for seed texture and seed color is
crossed with a plant that is wrinkled and heterozygous for seed color.
*R = round, r = wrinkled, Y=yellow, y = green
a. Construct a Punnett square for this cross.
Parent 1: ______________
b. What is the expected phenotypic ratio for the offspring?
2. In humans there is a disease called Phenylketonuria (PKU)which is caused by a recessive
allele. People with this allele have a defective enzyme and cannot break down the amino acid
phenylalanine. This disease can result in mental retardation or death. Let “E” represent the
normal enzyme. Also in humans in a condition called galactose intolerance or galactosemia,
which is also caused by a recessive allele. Let “G” represent the normal allele for galactose
digestion. In both diseases, normal dominates over recessive.
a. If two adults were heterozygous for both traits (EeGg), what are the chances of having a
child that is completely normal? _____________ b. Has just PKU? _____________
c. Has just galactosemia? _____________ d. Has both diseases? _____________
Parent 2: __________
8. Part 7: Incomplete Dominance or Codominance
In Four o’clock flowers the alleles for flower color are both equal therefore neither dominates
over the other. We call this condition incomplete dominance or codominance and it violates
Mendel’s principle of dominance. A red four o’clock flower (rr) is crossed with a white flower
(ww). Since there is no dominant trait we use two different little letters for the genotype.
Step 1: The genotype of the red flower will be “rr” and the genotype of the white flower is “ww”.
Step 2 and 3: Complete a Punnett square for this cross.
r r
Step 4: All of the offspring will be “rw”. So the genotypic ratio is: 4 : 0 : 0
Step 5: All of the offspring will have one of each allele (rw), so all will be pink.
1. Predict the offspring when two pink Four o’clock flowers are crossed.
a. Complete a Punnett square for this cross.
b. What is the predicted genotypic ratio for the offspring?
b. What is the predicted phenotypic ratio for the offspring?
2. In humans straight hair (ss) and curly hair (cc) are codominant traits, that result in hybrids who
have wavy hair (sc). Cross a curly hair female with a wavy haired male.
a. Complete a Punnett square for this cross.
b. What are the chances of having a curly haired child?
rw rw
rw rw
w
w
9. Part 8: Multiple Allele
So far we have studied traits or genes that are coded for by just two alleles. Like in rabbits, there was
one allele for brown hair color and one allele for white hair. However, some traits are coded for by
more than two alleles. One of these is blood type in humans. This is a violation of Mendel’s Principle
of unit characteristics.
In humans, there are four types of blood; type A, type B, type AB, and type O. The alleles A and B are
codominant to each other and the O allele is recessive to both A and B alleles. So a person with the
genotype AA or AO will have A type of blood.
a. What possible genotypes will produce B type of blood? ________________________
b. What is the only genotype that will produce O type of blood? ____________________
c. What is the only genotype that will produce AB type of blood? ___________________
1. You are blood type O and you marry a person with blood type AB.
a. Complete a Punnett square for this cross.
b. List the possible blood types (phenotypes) of your offspring. ____________________
2. In the 1950’s, a young woman sued film star/director Charlie Chaplin for parental support of her
illegitimate child. Charlie Chaplin’s blood type was already on record as type AB. The mother of
the child had type A and her son had type O blood.
a. Complete a Punnett square for the possible cross of Charlie and the mother.
b. The judge ruled in favor of the mother and ordered Charlie Chaplin to pay child support
costs of the child. Was the judge correct in his decision based on blood typing evidence?
Explain why or why not. *refer to any Punnett squares to support your answer.
3. Suppose a newborn baby was accidentally mixed up in the hospital. In an effort to determine the
parents of the baby, the blood types of the baby and two sets of parents were determined.
Baby 1 had type O Mrs. Brown had type B Mr. Brown had type AB
Mrs. Smith had type B Mr. Smith had type B
a. Draw Punnett squares for each couple (you may need to do more than 1 square/ couple)
b. To which parents does baby #1 belong? Why? Refer to your Punnett squares.
10. Part 9 SEX LINKED TRAITS
As many of you know boys are different than girls. In humans sex is determine by the twenty third
pair of chromosomes known as “sex chromosomes”. If you have two x-shaped (XX) chromosomes
you are destined to be a female. If you have an x and a Y-shaped (XY) chromosomes you are destined
to be a male. Since the X and Y chromosomes carry different information, any genes found on the X
chromosomes are referred to as sex-linked genes. Therefore, women will have two alleles for these
genes because they have two (XX) chromosomes. On the other hand, men have only one allele for
each of these genes because they have only one X chromosome (XY). This is clearly a violation of
Mendel’s Principle of Unit Characteristics, which implies that you receive one set of alleles from
each parent.
Example: In fruit flies, the gene for eye color is carried on the X chromosome which is a sex
chromosome (sex-linked). The allele for red eyes is dominant over the allele for white eyes. If a
white-eyed female fruit fly is mated with a red-eyed male, predict the possible offspring.
Step 1: Since the female has white eyes, she must be “X r
Xr
”. The male is red-eyed and because he
has only one X chromosome, he has only one allele for eye color. His eyes are red so he must be RY.
means he only has one allele for eye color, so he must be “XR
Y”. Since the allele “R” is present on
the X chromosome only, and there is no other allele for eye color because the male other sex
chromosome is a Y chromosome.
Step 2: For sex-linked traits we need to list the genotype in a different fashion. We must identify the
individual as being male or female according to their sex chromosomes. Females are XX, and males
are XY. Sex-linked traits are only found on the X chromosome, therefore the letters are placed as
superscripts (above) the X chromosome. Therefore the genotype for the female fly is X r
X r
and the
male is XR
Y. You can use shorthand notation rr x RY, but sometimes this can be confusing.
Step 3: The Punnett square for the parent flies are shown below.
Step 4: The genotypic ratio is 1 : 1
XR
Xr
Xr
Y
Step 5: The individual XR
X r
will be a female because she has two X chromosomes. She will have red
eyes because she has Rr. The individual with Xr Y will be a male because he has the X and Y
chromosomes. He will have white eyes because he has only one allele and it is “r”. So from this cross
you would expect all of the females to have red eyes and all of the males to have white eyes.
11. 1. Hemophilia is a sex-linked trait. A person with hemophilia is lacking certain proteins that are
necessary for normal blood clotting. Hemophilia is caused by a recessive allele so use “N” for
normal and “n” for hemophilia. Since hemophilia is sex-linked, remember a woman will have
two alleles (NN or Nn or nn) but a man will have only one allele (N or n). A woman who is
heterozygous (a carrier) for hemophilia marries a normal man:
a. What are the genotypes of the parents?
b. Make a Punnett square for the above cross.
c. What is the probability that a male offspring will have hemophilia? __________
d. What is the probability of having a hemophiliac female offspring? __________
2. Can a color blind female have a son that has normal vision? Color blindness is caused by a
sexlinked recessive allele. N = normal vision and n = color blind
3. Baldness is a sex-linked trait. What parental genotypes could produce a bald woman?
H =normal hair, and h = bald
12. Part 10: Pedigree Charts:
In genetics, traits can be traced over several generations similar to a family tree. This family
tree is called a Pedigree chart. Pedigree charts are useful in gathering background genetic
information that can be used for medical reasons. Horse race enthusiasts also rely heavily on
pedigree charts to predict a horse’s success. sWhen interpreting pedigree charts remember
square are male and circles are females
1. Use the below pedigree chart to answer the following three questions. Muscle type is not a sex
linked characteristic.
a. Place the genotypes of each individual below its symbol.
b. What is the genotype of individual #3 and 4?.
c. Can either individual #8 or 9 be homozygous?
d. Explain the family relationship that #12 has with #2.
2. Label the genotype for each of the individuals below its symbol on the pedigree chart (note: eye
color is not a sex-linked trait).
13. 3. List the possible genotypes of the following hemophilia pedigree chart below. Remember
hemophilia is a sex linked trait that is caused by a recessive allele, therefore you must denote the
individuals sex chromosomes (XN
Xn
and Xn
Y or Nn and nY) as well as the hemophilia allele (n).
4. Examine the following pedigree chart of color-blindness. In humans, color blindness is caused
by a recessive sex-linked allele. On the diagram, label the genotypes of the individuals 1-16.
14. 5. A blue-eyed man (1) whose parents were brown eyed (2 & 3), marries a brown eyed woman (4),
whose father was brown eyed (5) and whose mother (6) was blue eyed. They have one female
child who is blue eyed (7). Blue eyes are recessive.
a. Make a pedigree chart based on the above information.
b. Label the genotypes of the individuals in the chart.