Posted by: mrborden | May 20, 2012

Medelian Genetics


Wed May 30
film clips – free fall from space
Cloning humans
Tues May 29Handout on dyhybrid crosses
Thursday May 24 and Friday May 25
Handout on monohybrid and dyhybrid crosses
80 points mono 70 points dyhybrid crossses
Two-Factor Crosses (Di-hybrid) EC problems
1. In man, assume that spotted skin (S) is dominant over non-spotted skin (s) and that wooly hair (W) is
dominant over non-wooly hair (w). Cross a marriage between a heterozygous spotted, non-wooly man
with a wooly-haired, non-spotted woman. Give genotypic and phenotypic ratios of offspring.
2. In horses, black is dependent upon a dominant gene, B, and chestnut upon its recessive allele, b. The
trotting gait is due to a dominant gene, T, the pacing gait to its recessive allele, t. If a homozygous black
pacer is mated to a homozygous chestnut trotter, what will be the appearance of the F1 generation?
3. In snapdragon flowers, red color is not completely dominant over white color and tall plants are
dominant over short plants. What would expect to get from a genetic cross of a homozygous tall
red snapdragon with a short white plant? Give genotypic and phenotypic ratios of the offspring.

Name:________________________________ Date:________________

Genetic Dihybrid Worksheet – show your answers on the back of this paper

Problems: Monohybrid, Dihybrid, and Test Crosses1. In hamsters, black coat is dominant to white coat and rough coat is dominant
to smooth coat. If two heterozygous black-rough coated hamsters where mated
a. What would be the phenotype ratio of the F 1 generation?
b. What would be the genotype ratios of the F 1 generation?

2. In garden peas, axial flower position is dominant and terminal flower position is recessive; tall vine is dominant and short vine recessive. A plant known tobe purebred for tall vine and axial flowers is crossed with a plant having short vines and bearing terminal flowers
a. What is the phenotype of the offspring?
b. What is the genotype of the offspring?
c. Predict the types of offspring (their phenotypes) that would appear in the F
2 generation and their ratio.

3. In dogs, wirehair is a dominant trait, smooth hair is recessive. Two wire-haired dogs produce a male puppy that is wire-haired. To find out most quickly whether he carries the allele for smooth hair, he should be mated to what type of female?

4. In a certain animal, a breed is known that always has a hairy tail; another breed is known that always has a naked tail. How would you determine which trait is dominant?

5. In cocker spaniels, black is a dominant trait (B), red is recessive (b). Solid color is dominant (S) and white spotting is recessive (s). A red male was mated to a black-and-white female. They had five puppies as follows: 1 black,1 red, 1 black-and-white, and 2 red-and-white. What are the genotypes of the parents?

6. Two plants having red flowers are mated. When their F 1 offspring are studied, it is found that 95 have red flowers and 31 have white flowers. What is the genotype of the parents and diagram a cross to support your answer.

Genotype calculator
Wed May 23
Students will learn that offspring inherit entire chromosomes from their
parents.
• Students will recognize that if two genes are parts of the same chromosome,
they will inherit them together.
• Students will recognize that if two genes are parts of different chromosomes,
they will inherit them independently.
• Students will distinguish the difference in inheritance patterns between two
characteristics that are parts of one chromosome and two characteristics that
are determined by genes that are parts of different chromosomes.
• Students will utilize the principles of probability to predict the outcome of
genetic crosses for two traits.
• Students will utilize the principles of probability to explain genetic crosses
involving more than one trait.
o Students will use Punnett Square and Pedigree Charts to examine
patterns of heredity.
3.0 Standards Alignment
Alignment to National Math and Science Standards (NCTM or NSES)
Objective Standards
Students will be able to determine
whether the genes for 2 traits are parts
of the same or different chromosomes
through breeding experiments and
pedigree analysis.
Students will use representations to
model and interpret physical, social,
and mathematical phenomena.
Students will extend their
understanding of probability to the
inheritance of two traits.
Mathematical tools and models guide
and improve the posing of questions,
gathering of data, constructing
explanations and communicating
results.
Dihybrid Punnett squares practice

For dihybrid cross we study the inheritance of two genes. For dihybrid cross the Punnett squares only works if the genes are independent of each other, which meomans when form a maternal and paternal gametes – each of them can get any allele of one pair, along with any allele of the another pair. This principle of independent assortment was discovered by Mendel in experiments on dihybrid and polyhybrid crosses.

The following example illustrates Punnett square for a dihybrid cross between two heterozygous pea plants. We have two genes Shape and Color. For shape: “R” is dominant allele with round phenotype and “w” is recessive allele with wrinkled phenotype. For color: “Y” is dominant allele with yelloy phenotype and “g” is recessive allele whith green phenotype. Maternal and paternal organisms have some genotype- “RwYg”.

First you need to determine all possible combinations of gametes, for this you can also use Punnett squares:

Then they can produce four types of gametes with all possible combinations: RY, Rg, wY, wg. And now form the Punnett square for genotypes:

From punnett square in the offspring we have genotype ratio and probability: 1(6,25%)RRYY : 2(12,5%)RwYY : 1(6,25%)wwYY : 2(12,5%)RRYg : 4(25%)RwYg : 2(12,5%)wwYg : 1(6,25%)RRgg : 2(12,5%)Rwgg : 1(6,25%)wwgg.

Since dominant traits mask recessive traits, from punnett square we have phenotypes combinations whith ratio and probability: 9(56,25%)R-Y-(round, yellow) : 3(18,75%)R-gg(round,green) : 3(18,75%)wwY-(wrinkled, yellow) : 1(6,25%)wwgg(wrinkled, green). The ratio 9:3:3:1 is typical for a dihybrid cross.

Trihybrid Punnett squares practice

Make punnett square for trihybrid cross between two heterozygous plants is more complicated. To solve this problem, we can use our knowledge of mathematics. To determine all possible combinations of gametes for trihybrid cross we have to remember the solution of polynomials:
■Let make polynomial for this cross: (A + a) X (B + b) X (C + c).
■We multiply the expression in the first bracket on the expression of a second and we get : (AB + Ab + aB + ab) X (C + c).
■ Now multiply this expression by the expression in the third bracket and we get: ABC + ABc + AbC + Abc + aBC + aBc + abC + abc.

Then they can produce eight types of gametes with all possible combinations. This solution can be illustrated by the Punnett square:
Now form the Punnett square for genotypes:
Genotypes ratio and probability for Trihybrid Punnett Squares
But how do we calculate the ratio of genotypes from this punnett square. Again use the polynomials.
■ We know the genotype ratio for monohybrid cross: 1AA: 2Aa: 1aa.
■ Now we form a polynomial for our case: (1AA + 2Aa + 1aa) X (1BB + 2Bb + 1bb) X (1CC + 2Cc + 1cc).
■Multiply the first two expressions we get : (1AABB + 2AABb + 1AAbb + 2AaBB + 4AaBb + 2Aabb + 1aaBB + 2aaBb + 1aabb) X (1CC + 2Cc + 1cc).
■Multiplying this expression on the third we get the results for genotypes (second column – ratio, third column – probability):

Parents genotypes – AaBbCc X AaBbCc crossing genotypes:

genotype calculator
Tuesday May 22
1)One cat carries heterozygous, long-haired traits (Ss), and its mate carries homozygous short-haired traits (ss). Use a Punnett square to determine the probability of one of their offspring having long hair.
A) 100%
B) 25%
C) 75%
D) 50%

2 One cat carries heterozygous, long-haired traits (Ss), and its mate carries homozygous short-haired traits (ss). Use a Punnett square to determine the probability of one of their offspring having short hair.
A) 25%
B) 50%
C) 75%
D) 100%

3 One flower is heterozygous red (Rr) and it is crossed with a homozygous white (rr) plant. Use a Punnett square to determine the probability of one of their offspring having a red color.
A) 75%
B) 25%
C) 50%
D) 100%

4 One flower is heterozygous red (Rr) and it is crossed with a homozygous white (rr) plant. Use a Punnett square to determine the probability of one of their offspring having a white color.
A) 100%
B) 25%
C) 75%
D) 50%

5 In a certain species of plant, the color purple (P) is dominant to the color white (p). According to the Punnett Square, what is the probability of an offspring being white?
A) 25%
B) 0%
C) 75%
D) 100%

6 In a certain species of plant, the color purple (P) is dominant to the color white (p). According to the Punnett Square, what is the probability of an offspring being purple?
A) 25%
B) 0%
C) 100%
D) 75%
7 In a certain species of plant, the color purple (P) is dominant to the color white (p). According to the Punnett Square, what is the probability of an offspring being white?
A) 50%
B) 25%
C) 75%
D) 100%
8 In a certain species of plant, the color purple (P) is dominant to the color white (p). According to the Punnett Square, what is the probability of an offspring being purple?
A) 75%
B) 25%
C) 50%
D) 100%
9 In a certain species of pine trees, short needles (S) are dominant to long needles (s). According to the Punnett square, what is the probability of an offspring having short needles?
A) 50%
B) 25%
C) 75%
D) 100%
10 In a certain species of pine trees, short needles (S) are dominant to long needles (s). According to the Punnett square, what is the probability of an offspring having long needles?
A) 25%
B) 50%
C) 75%
D) 100%

Monday May 21
heredity, the sum of all biological processes by which particular characteristics are transmitted from parents to their offspring. The concept of heredity encompasses two seemingly paradoxical observations about organisms: the constancy of a species from generation to generation and the variation among individuals within a species. Constancy and variation are actually two sides of the same coin, as becomes clear in the study of genetics. Both aspects of heredity can be explained by genes, the functional units of heritable material that are found within all living cells. Every member of a species has a set of genes specific to that species. It is this set of genes that provides the constancy of the species. Among individuals within a species, however, variations can occur in the form each gene takes, providing the genetic basis for the fact that no two individuals (except identical twins) have exactly the same traits.

The set of genes that an offspring inherits from both parents, a combination of the genetic material of each, is called the organism’s genotype. The genotype is contrasted to the phenotype, which is the organism’s outward appearance and the developmental outcome of its genes. The phenotype includes an organism’s bodily structures, physiological processes, and behaviours. Although the genotype determines the broad limits of the features an organism can develop, the features that actually develop, i.e., the phenotype, depend on complex interactions between genes and their environment. The genotype remains constant throughout an organism’s lifetime; however, because the organism’s internal and external environments change continuously, so does its phenotype. In conducting genetic studies, it is crucial to discover the degree to which the observable trait is attributable to the pattern of genes in the cells and to what extent it arises from environmental influence.

Because genes are integral to the explanation of hereditary observations, genetics also can be defined as the study of genes. Discoveries into the nature of genes have shown that genes are important determinants of all aspects of an organism’s makeup. For this reason, most areas of biological research now have a genetic component, and the study of genetics has a position of central importance in biology. Genetic research also has demonstrated that virtually all organisms on this planet have similar genetic systems, with genes that are built on the same chemical principle and that function according to similar mechanisms. Although species differ in the sets of genes they contain, many similar genes are found across a wide range of species. For example, a large proportion of genes in baker’s yeast are also present in humans. This similarity in genetic makeup between organisms that have such disparate phenotypes can be explained by the evolutionary relatedness of virtually all life-forms on Earth. This genetic unity has radically reshaped the understanding of the relationship between humans and all other organisms. Genetics also has had a profound impact on human affairs. Throughout history humans have created or improved many different medicines, foods, and textiles by subjecting plants, animals, and microbes to the ancient techniques of selective breeding and to the modern methods of recombinant DNA technology. In recent years medical researchers have begun to discover the role that genes play in disease. The significance of genetics only promises to become greater as the structure and function of more and more human genes are characterized.

This article begins by describing the classic Mendelian patterns of inheritance and also the physical basis of those patterns—i.e., the organization of genes into chromosomes. The functioning of genes at the molecular level is described, particularly the transcription of the basic genetic material, DNA, into RNA and the translation of RNA into amino acids, the primary components of proteins. Finally, the role of heredity in the evolution of species is discussed.
genetics%20wkst
Punnett%20Square%20worksheet
Punnett Squares Worksheet
Use Punnett Square to answer the following questions. Show your work.
1. Widow’s peak is dominant to no widow’s peak. Determine the genotype and
phenotype ratios for a homozygous dominant female and a homozygous recessive male.
2. Dimples is dominant to no dimples. Determine the genotype and phenotype ratios for
a heterozygous female and a homozygous dominant male.
3. Short hair is dominant to long hair in mice. Determine the genotype and phenotype
ratios for a homozygous recessive female and a heterozygous male.
4. Cleft chin is dominant to no cleft chin. Determine the genotype and phenotype ratios
for a heterozygous female and a heterozygous male.
5. Brown eyes are dominant to blue eyes. Determine the genotype and phenotype ratios
for a homozygous dominant female and a homozygous dominant male.
6. Purple flowers are dominant to white flowers in pea plants. Determine the genotype
and phenotype ratios for a homozygous dominant female and a heterozygous male.
7. Brown hair is dominant to gray hair in mice. Determine the genotype and phenotype
ratios for a homozygous recessive female and a homozygous recessive male.
8. Free ear lobes are dominant to attached ear lobes. Determine the genotype and
phenotype ratios for a heterozygous female and a homozygous recessive male.
9. Green peas are dominant to yellow peas. Determine the genotype and phenotype
ratios for a heterozygous female and a heterozygous male.
10. Tall plants are dominant to short plants. Determine the genotype and phenotype
ratios for a homozygous recessive female and a homozygous dominant male

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.
E.g. 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.
Page #7
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
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 : 6 light rough : 2 dark smooth : 2 light smooth.
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.
website
Name: ___________________________
Practice Dihybrid Crosses
1. In certain plants, yellow (Y) is dominant over green (y) and disk shaped (D) is dominant over sphere shaped (d).
Complete a punnett square to find the phenotypic ratio of the following cross:
ddYY × DDyy
2. In rabbits, black color is due to the dominant gene B and brown color is due to its recessive allele, b. Short hair is due to the dominant gene H and long hair is due to its recessive allele, h. In a cross between a rabbit that is heterozygous for both black and short with a homozygous brown and long haired female, what would the expected phenotypic ratio be of the offspring?
3. In cattle, black coat R is dominant to red coat, r. Hornless H is dominant to horned, h. A certain bull was mated to four cows. Cow 1, black and hornless, gave birth to a hornless red calf. Cow 2, red and hornless, gave birth to a horned black calf. Cow 3, black and horned, gave birth to a red hornless calf. Cow 4, red and horned, gave birth to a hornless calf.
What are the genotypes of each of the cows and their calves?
What is the genotype of the bull?
Cow 1 __________ Calf __________
Cow 2 __________ Calf __________
Cow 3 __________ Calf __________
Cow 4 __________ Calf __________ Bull __________
Name: ___________________________
4. In mice, black is dominant to tan and short tails are dominant over long.
Use B for black and b for tan.
Use T for short tail and t for long tail.
Write the genotype for a heterozygous black and short tailed mouse. ______________
Cross two of these heterozygous mice.
From the Punnett Square, describe the phenotype of the offspring by writing the phenotypic ratio.
5. In certain bacteria, an oval shape, R, is dominant to round. Thick cell walls, T, are dominant to thin. Cross a heterozygous oval, thick cell walled bacteria with a round thin cell walled bacteria. Describe the phenotypes that can be produced from this cross.
6. In guinea pigs, black coat color is dominant over white. Short hair is dominant over long, and rough coat is dominant over smooth. What would the offspring look like if you crossed a guinea pig homozygous for black coat, short hair, and rough coat, with a guinea pig that is white, long haired and smooth coat?
Name: ___________________________
Practice Dihybrid Crosses
1. In certain plants, yellow (Y) is dominant over green (y) and disk shaped (D) is dominant over sphere shaped (d).
Complete a Punnett square to find the phenotypic ratio of the following cross:
ddYY × DDyy
Dy
Dy
Dy
Dy
dY
DdYy
DdYy
DdYy
DdYy
dY
DdYy
DdYy
DdYy
DdYy
dY
DdYy
DdYy
DdYy
DdYy
dY
DdYy
DdYy
DdYy
DdYy
P1

16 disk, yellow : 0 disk, green : 0 sphere, yellow : 0 sphere, green
2. In rabbits, black color is due to the dominant gene B and brown color is due to its recessive allele, b. Short hair is due to the dominant gene H and long hair is due to its recessive allele, h. In a cross between a rabbit that is heterozygous for both black and short with a homozygous brown and long haired female, what would the expected phenotypic ratio be of the offspring?
BbHh x bbhh

p5
bh
bh
bh
bh
BH
BbHh
BbHh
BbHh
BbHh
Bh
Bbhh
Bbhh
Bbhh
Bbhh
bH
bbHh
bbHh
bbHh
bbHh
bh
bbhh
bbhh
bbhh
bbhh
4 black, short : 4 black, long : 4 brown, short : 4 brown, long
3. In cattle, black coat R is dominant to red coat, r. Hornless H is dominant to horned, h. A certain bull was mated to four cows. Cow 1, black and hornless, gave birth to a hornless red calf. Cow 2, red and hornless, gave birth to a horned black calf. Cow 3, black and horned, gave birth to a red hornless calf. Cow 4, red and horned, gave birth to a hornless calf.
What are the genotypes of each of the cows and their calves?
What is the genotype of the bull?
Cow 1 RrH? Calf rrH?
Cow 2 rrHh Calf Rrhh
Cow 3 Rrhh Calf rrHh
Cow 4 rrhh Calf ?rHh Bull RrHh
Name: ___________________________
4. In mice, black is dominant to tan and short tails are dominant over long.
Use B for black and b for tan.
Use T for short tail and t for long tail.
Write the genotype for a heterozygous black and short tailed mouse. BbTt
Cross two of these heterozygous mice.
From the Punnett Square, describe the phenotype of the offspring by writing the phenotypic ratio.
BT
Bt
bT
bt
BT
BBTT
BBTt
BbTT
BbTt
Bt
BBTt
BBtt
BbTt
Bbtt
bT
BbTT
BbTt
bbTT
bbTt
bt
BbTt
Bbtt
bbTt
bbtt
9 black, short : 3 black, long : 3 tan, short : 1 tan, long
5. In certain bacteria, an oval shape, R, is dominant to round. Thick cell walls, T, are dominant to thin. Cross a heterozygous oval, thick cell walled bacteria with a round thin cell walled bacteria. Describe the phenotypes that can be produced from this cross.
RrTt x rrtt
rt
rt
rt
rt
RT
RrTt
RrTt
RrTt
RrTt
Rt
Rrtt
Rrtt
Rrtt
Rrtt
rT
rrTt
rrTt
rrTt
rrTt
rt
rrtt
rrtt
rrtt
rrtt


P4


4 oval, thick : 4 oval, thin : 4 round, thick : 4 round, thin
6. In guinea pigs, black coat color is dominant over white. Short hair is dominant over long, and rough coat is dominant over smooth. What would the offspring look like if you crossed a guinea pig homozygous for black coat, short hair, and rough coat, with a guinea pig that is white, long haired and smooth coat?
BBHHRR x bbhhrr  BbHhRr

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Responses

  1. i forgot the homework for tonight what was it again.


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