Primate Anatomy & Taxonomy

Primate Anatomy & Taxonomy. Name: ______________________________________  Section: ___________

ANT 3514C – Introduction to Biological Anthropology

Lab 6: Primate Anatomy & Taxonomy

Lab Objectives:

• Evaluate the dental formula of an unknown primate and place it within a major clade

• Interpret the difference between gradistic and cladistic methods of grouping primates

• Identify the anatomical synapomorphies that distinguish the major primate clades

• Draw a cladogram to illustrate the modern, broadly-accepted primate phylogeny

Purpose: To examine the skeletal traits that distinguish the major primate clades.

The study of non-human primates has been recognized since ancient times as relevant to understanding human anatomy. This was perhaps best recognized by a wide audience of scholars for the first time in 1735 when Carolus Linnaeus, despite his strong creationist views, included humans with other apes and monkeys in the group Anthropomorpha. By the tenth edition of his Systema Naturae in 1758 he had abandoned this term and began calling the group by the familiar name we now use: Order Primates. Linnaeus was motivated to group humans with other primates because of the many anatomical similarities that he perceived uniting them. In modern biological terms, we now refer to these structures as synapomorphies, or ‘shared derived traits.’ For instance, all primates have a broad, flat nail on their big toe, which is a structure unlike any of the narrow claws found in other mammals. We use synapomorphies like these to reconstruct patterns of shared ancestry and build cladograms to better understand the pattern of primate evolution and where humans belong in it. This method of reconstructing relationships between taxa based on shared derived characteristics is known as cladistics.

While morphology and phylogeny have shared a close relationship for centuries, the reliance on synapomorphies to group organisms is relatively recent. In the early 20th century, primatologists such as Wilfrid Le Gros Clark (famous for helping to debunk the Piltdown Man fraud) grouped primates based on their overall similarity in appearance. This method was reminiscent of Aristotle’s “Great Chain of Being,” with primitive primates at the base and humans at the apex. This way of thinking has been called gradistic, because it suggests primate evolution proceeds in a simple, uniform direction for all traits from primitive to derived. While this system has intuitive appeal, it does not correspond with the way we think evolution proceeds. Modern primatologists and anthropologists use cladistic methods based on a nested hierarchy of synapomorphies, because we believe these more accurately reflect how evolution works.

Although phylogenetic trees are built today using cladistic methods, it is clear that gradistic thinking still subconsciously underlies much of our approach to reconstructing evolution. You may have already noticed that phylogenetic trees that include humans tend to place them at one extreme end of the tree, implying some directionality or end goal to evolution, even though there is no reason they need to be placed there! Within the primate order the shift from gradistic to cladistic thinking has impacted how we perceive the relationships of many taxa, most notably the tarsier, which we will investigate more in Station 3. You will be exposed to more examples of gradistic thinking when looking at the human fossil record, where many features (such as brain size) are continuous, and identifying synapomorphies can be particularly difficult. For this lab we will investigate many of the important skeletal synapomorphies that define the largest primate clades.

Station 1: What defines a primate? (0.6 pt.) 

A college happened upon a mystery skull while looking through a mammalian skeletal collection. She thinks it may be a primate and comes to you for your expert opinion. She cannot mail you the skull so she emails youaeveral photos. Examine the photos below. Use the list of primate features (found in the lab reading for this week) to help you make the distinctions.

!1

Station 2: Dentition (2 pts.) Different primate clades can be identified by their dental formulae. Primates have 2 incisors, and 2-3 premolars (except the aye-aye, which has a very unusual dentition). Most other mammals have either more or fewer teeth. For the following questions, first determine the dental formula, then consider the cusp pattern.

! !

!   !

1. Write the dental formula for each of the craniums or mandibles provided:

A)

B)

C)

D)

E)

2. Answer the following questions using the above dental formulae and the written/illustrated materials provided:

a) Which specimen is not a primate? How do you know?

b) Which specimen is a New World monkey? How do you know?

c) Does specimen “E” have the bilophodont or Y-5 molar cusp pattern? Based on this cusp morphology and its dental formula, what primate group does this specimen belong to?

d) Which mandible is human? What traits did you use to make your identification?

e) Which specimens are apes? How do you know? (Hint: you are an ape)

Station 3: Strepsirrhini and Haplorhini (2 pts.)

Using the handouts, images, and websites, complete the following table illustrating some of the important morphological differences between each primate group.

Strepsirrhine: http://eskeletons.org/boneviewer/nid/12540/region/skull/bone/cranium  http://humanorigins.si.edu/evidence/3d-collection/primate/loris-malaysia-usnm-84389  http://humanorigins.si.edu/evidence/3d-collection/primate/loris-malaysia-usnm-84389-0  Haplorrhine: http://eskeletons.org/boneviewer/nid/12538/region/skull/bone/cranium

http://humanorigins.si.edu/evidence/3d-collection/primate/gorilla-rwanda-usnm-396937-beringeicranium 

with primitive primates in a grade called “Prosimia.” They are now grouped with monkeys and apes in a

Examine the tarsier skull (http://www.eskeletons.org/boneviewer/nid/12544/region/skull/bone/cranium) and study the table. Tarsiers can be placed in the prosimian grade or in the haplorhine clade. Name one primitive, prosimian feature that tarsiers retain and one derived, haplorhine feature that they possess. Add which of these features is a synapomorphy, and which is a symplesiomorphy.

Prosimian feature:         Haplorhine feature:

 

2) Does the mystery skull at this station belong to a strepsirrhine or a haplorhine primate? List at least one trait which helped you determine this.

!

 

Station 4: Platyrrhini and Catarrhini (1.6 pts.)

The Haplorhine suborder is divided into two infraorders: Anthropoidea (Monkeys and Apes) and Tarsiiformes (tarsiers). Anthropoidea is further divided into two parvorders: Platyrrhini and Catarrhini. Platyrrhines are native to Central and South America (the ‘New World’) and Catarrhines are native to Africa, Europe, and Asia (the ‘Old World’).

Platyrrhine: http://eskeletons.org/boneviewer/nid/12546/region/skull/bone/cranium  Catarrhine: http://eskeletons.org/boneviewer/nid/12547/region/skull/bone/cranium   http://humanorigins.si.edu/evidence/3d-collection/primate/baboon-usnm-258502 http://humanorigins.si.edu/evidence/3d-collection/primate/baboon-usnm-258502-0

1) Based on what you’ve learned so far, identify what group the following “mystery primate” skulls belong to. To receive credit, list the character(s) you used to make your identification. 

!

!

A) Is “A” a platyrrhine or catarrhine? How do you know?

B) Is “B” a platyrrhine or catarrhine? How do you know?

2) What advantages might there be to having a prehensile tail for an arboreal primate?

Station 5: Cercopithecoidea and Hominoidea (1.8 pts.)

Within Catarrhini are the two superfamilies Cercopithecoidea (Old World monkeys) and Hominoidea (apes). Use the table below to describe the features of each in relation to the other.

Cercopithecoid: http://eskeletons.org/boneviewer/nid/12547/region/skull/bone/cranium  http://humanorigins.si.edu/evidence/3d-collection/primate/baboon-usnm-258502 http://humanorigins.si.edu/evidence/3d-collection/primate/baboon-usnm-258502-0 Hominoid: http://eskeletons.org/boneviewer/nid/12549/region/skull/bone/cranium

http://humanorigins.si.edu/evidence/3d-collection/primate/siamang-indonesia-usnm-114497

1) Which two traits in the above table would be the most useful for determining if an animal was a cercopithecoid or a hominoid in the fossil record? Hint: think about discrete (traits which are either present or absent) vs. continuous traits.

2) List one human autapomorphy – a trait that humans have to the exclusion of all the other primates. Hint:

think about what makes humans unique within the order Primates.

Exercise 2: Systematics and Primate Phylogeny (2 pts.) 

Below is a hypothetical phylogeny for six different taxa (A–F). In the phylogeny, the appearance of a new character is represented as a number in a circle. For instance, Character 3 evolved sometime after the common ancestor of Taxa D, E, and F diverged from the common ancestor these taxa share with Taxon C. Character 3 would therefore be a shared, derived trait, or synapomorphy of taxa D, E, and F.

A) Which character is a synapomorphy of E and F?

B) Is Character 1 a synapomorphy or a symplesiomorphy for taxa C and D?

C) Is Character 1 useful for reconstructing the relationship between C and D? Why or why not?

D) Of the 5 characters listed, which represents an autapomorphy?

Study the primate phylogeny in your textbook and fill in the blanks below. Be mindful of spelling: some names are very similar, but have different meanings!  If you are having difficulty filling out the phylogeny, you may print out the last page, neatly handwrite the answers in the blanks, and paste a picture of the phylogeny back into the document.

Primate Anatomy & Taxonomy

 
"Looking for a Similar Assignment? Get Expert Help at an Amazing Discount!"

BIO – INET Lab Genetics Worksheet Report Tempate

BIO – INET Lab Genetics Worksheet Report Tempate. INET Lab
Genetics Worksheet Report Template
Student:
Email:
Date:

For each correctly conducted genetics problem, you receive 5 points for a maximum 20 points.  All Punnett squares must
be shown.
1. There are 3 alleles controlling the ABO blood types. IA and IB are codominant genes so that the combination IAIB
produces the AB blood type. The third allele, (i) is recessive to the other two alleles.  Show your work for each cross.
Indicate which of these parents could produce the given child.
The first one is provided with answers as an example.
(HINT: Do not forget that a heterozygous versus a homozygous blood type may yield different results. )
a)

Parent
A x AB

Child
IB
B

IA

IA

IA IA

IA IB

IA

IA IA

IA IB

IA

IB

IA

IA IA

IA IB

i

IA i

IB i

Produce Child of Type?

No

Produce Child of Type?

© Access Learning Systems

2

b)

Parent

Child

A x O

A

Produce Child of Type?

Produce Child of Type?

© Access Learning Systems

3
c)

Parent

Child

A x B

O

Produce Child of Type?

Produce Child of Type?

Produce Child of Type?

Produce Child of Type?

d)

Parent

Child

A x AB

O

© Access Learning Systems

4

Produce Child of Type?

Produce Child of Type?

© Access Learning Systems

5
e)

Parent

Child

B x B

O

Produce Child of Type?

Produce Child of Type?

Produce Child of Type?

f)

Parent

Child

AB x AB

A

Produce Child of Type?

© Access Learning Systems

6
2. The allele for albinism (a) is recessive to the allele for normal pigmentation (A). A normally pigmented  woman whose
father is an albino marries an albino man whose parents are normal. They have three children, two normal and one
albino. Give the genotypes for each person listed. Prove your answer.

8

Maternal Grandmother
Father

6

9

Maternal Grandfather

Mother
Child 1

8

7

Paternal Grandmother

5

6

Paternal Grandfather

7

5

4

1

4

3

Relationship

3

2

#
2

1

Child 2

9

Genotype(s)

Child 3

a) Show the cross of the paternal grandparents.

(HINT: What must their genotypes be to produce and albino child? )

b) What are the possible genotypes of the maternal grandmother?

© Access Learning Systems

7

© Access Learning Systems

8
Show the cross of the mother and father.

3. In horses, black coat color is influenced by the dominant allele (B), and chestnut coat color by the recessive allele (b).
a) What color horse would you use to find out the genotype of a black trotter?

b) Give the genotype and phenotype. Show your work and discuss the reason you would know the genotype of the
black horse.

Possible Allele Pairings with Homozygous Dominant Trotter

F1 offspring probabilities:

Possible Allele Pairings with Heterozygous Trotter

F1 offspring probabilities:

Reasoning:

© Access Learning Systems

9
4. In Drosophila, the fruit fly, white eyes are determined by a recessive X­linked gene, and the wild­type or normal brick­
red eyes are due to its dominant allele. Use symbols of the following types: X rY = a white­eyed male; XRXR = a
homozygous normal red female.
a) What offspring can be expected from a cross of a white­eyed male and a homozygous normal female?

b) Show the genotypes and list the phenotypes of the F1 offspring.
F1 Cross

F1 offspring probabilities:
Male

Female

Red­Eyed
White­Eyed

c) Now, cross the F1 offspring.  Show the genotypes and list the phenotypes of the F 2 offspring.
F2 Cross

F2 offspring probabilities:
Male
Red­Eyed
White­Eyed

Female

INET Lab

Genetics Worksheet Report Template

1

 

© Access Learning Systems

 

Student:

Email:

Date:

 

 

 

For each correctly conducted genetics problem, you receive 5 points for a maximum 20 points. All Punnett squares must be shown.

 

1. There are 3 alleles controlling the ABO blood types. IA and IB are codominant genes so that the combination IAIB produces the AB blood type. The third allele, (i) is recessive to the other two alleles. Show your work for each cross.

 

Indicate which of these parents could produce the given child.

 

The first one is provided with answers as an example.

(HINT: Do not forget that a heterozygous versus a homozygous blood type may yield different results.)

 

a) Parent Child
  A x AB B

 

 

  IA IB
IA IA IA IA IB
IA IA IA IA IB
  IA IB
IA IA IA IA IB
i IA i IB i

 

 

Produce Child of Type? No
Produce Child of Type? Yes

 

b) Parent Child
  A x O A

 

 

 

     
     
     
     
     
     

 

 

Produce Child of Type?  
Produce Child of Type?  

 

 

 

c) Parent Child
  A x B O

 

 

 

     
     
     
     
     
     

 

 

Produce Child of Type?  
Produce Child of Type?  

 

     
     
     
     
     
     

 

Produce Child of Type?  
Produce Child of Type?  

 

 

d) Parent Child
  A x AB O

 

 

 

 

     
     
     
     
     
     

 

 

Produce Child of Type?  
Produce Child of Type?  

 

 

 

e) Parent Child
  B x B O

 

 

 

     
     
     
     
     
     

 

 

Produce Child of Type?  
Produce Child of Type?  

 

     
     
     

 

Produce Child of Type?  

 

 

 

f) Parent Child
  AB x AB A

 

 

 

 

     
     
     

 

 

Produce Child of Type?  

 

 

2. The allele for albinism (a) is recessive to the allele for normal pigmentation (A). A normally pigmented woman whose father is an albino marries an albino man whose parents are normal. They have three children, two normal and one albino. Give the genotypes for each person listed. Prove your answer.

 

 

 

# Relationship Genotype(s)
1 Paternal Grandfather  
2 Paternal Grandmother  
3 Maternal Grandfather  
4 Maternal Grandmother  
5 Father  
6 Mother  
7 Child 1  
8 Child 2  
9 Child 3  

 

3 4 1 2 5 6 7 9 8

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

a) Show the cross of the paternal grandparents.

(HINT: What must their genotypes be to produce and albino child?)

 

 

 

     
     
     

 

 

 

b) What are the possible genotypes of the maternal grandmother?

 

 

 

     
     
     
     
     
     

 

 

 

c) Show the cross of the mother and father.

 

 

 

     
     
     

 

 

 

3. In horses, black coat color is influenced by the dominant allele (B), and chestnut coat color by the recessive allele (b).

 

a) What color horse would you use to find out the genotype of a black trotter?

6

 

 

 

 

 

5

 

 

 

b) Give the genotype and phenotype. Show your work and discuss the reason you would know the genotype of the black horse.

 

 

Possible Allele Pairings with Homozygous Dominant Trotter

 

     
     
     

 

 

F1 offspring probabilities:
 

 

 

Possible Allele Pairings with Heterozygous Trotter

 

     
     
     

 

 

F1 offspring probabilities:
 

 

 

 

Reasoning:

4

 

 

 

 

 

5

 

 

 

4. In Drosophila, the fruit fly, white eyes are determined by a recessive X-linked gene, and the wild-type or normal brick-red eyes are due to its dominant allele. Use symbols of the following types: XrY = a white-eyed male; XRXR = a homozygous normal red female.

 

a) What offspring can be expected from a cross of a white-eyed male and a homozygous normal female?

6

 

 

 

 

 

7

 

 

 

b) Show the genotypes and list the phenotypes of the F1 offspring.

 

 

F1 Cross

 

     
     
     

 

 

F1 offspring probabilities:
  Male Female
Red-Eyed    
White-Eyed    

 

 

 

c) Now, cross the F1 offspring. Show the genotypes and list the phenotypes of the F2 offspring.

 

 

F2 Cross

 

     
     
     

 

 

F2 offspring probabilities:
  Male Female
Red-Eyed    
White-Eyed    

BIO – INET Lab Genetics Worksheet Report Tempate

 
"Looking for a Similar Assignment? Get Expert Help at an Amazing Discount!"

Foundations of Biological Sciences I Evolutionary Agents

Foundations of Biological Sciences I Evolutionary Agents.

Foundations of Biological Sciences I Evolutionary Agents – 1

A quick recap…. There are several terms that need to be clarified so that you can more easily follow the exercise. A gene is a

piece of DNA that directs the expression of a particular characteristic (trait). Genes are located on

chromosomes, and the location where a particular gene is found is referred to as the locus (plural: loci) of that

gene. An allele is a gene for which there is an alternative expression, which can lead to the alterative form of a

trait. For example, a diploid organism carries the allele “A” on one homologous chromosome, and the allele “A”

on the other. The genotype of this organism is then AA and it is said to be homozygous. An organism may also

carry two different alleles. For example on one chromosome it could carry the allele “A” and on the other it

could carry the allele “a”. The genotype of such an organisms is then Aa, and it is described as heterozygous for

this chromosomal locus.

The genotype of an organism is the listing of the two alleles for each trait that it possesses. The phenotype of an

organism is a description of the way a trait is displayed in the structure, behavior, or physiology of the organism.

Some alleles are dominant to others and mask the presence of other alleles. The dominant condition is indicated

by uppercase letters (e.g., “A”). The alleles that are masked are called recessive alleles. The recessive condition

is indicated by lowercase letters (e.g., “a”). When both dominants are present in the genotype (AA), the organism

is said to be homozygous dominant for the trait, and the organisms will show the dominant phenotype (trait

expression A). When both recessives are present in the genotype (aa), the organism is said to be homozygous

recessive for the trait, and the organisms will show the recessive phenotype (trait expression a). In the case of

complete dominance, the dominant allele completely masks the recessive allele, and an organism with a

heterozygous genotype (Aa) will show the dominant phenotype (trait expression A).

 

Evolutionary Agents

Evolution is a process resulting in changes in gene frequencies (= the genetic make-up) of a population over

time. The mechanisms of evolution include selection (which can cause change over time & adaptation), and

forces that provide variation and cause change over time (but not adaptation). Factors that change gene

frequencies over time are referred to as evolutionary agents.

A powerful way to detect the presence of evolutionary agents is the use of the Hardy–Weinberg model. This

model can be applied to traits that are influenced by several loci; the simplest case is for a trait that is regulated

by one locus with two alleles.

With the Hardy–Weinberg model, the frequency of genotypes in the population can be predicted from the

probability of encounters between gametes bearing the different alleles. With alleles R and B occurring at

frequencies p and q, respectively, the frequency of genotypes in the population is described by the formula:

 

p 2

+ 2pq + q 2

= 1

Hardy-Weinberg

equilibrium

If p is the frequency of one allele, and q

is the frequency of the other allele, then:

 

p + q = 1

 

 

Foundations of Biological Sciences I Evolutionary Agents – 2

If certain conditions are met, the proportions of genotypes that make

up a population remain constant from generation to generation, and

can be predicted from the Hardy-Weinberg equilibrium.

For example, if flower color is controlled by two alleles (R & B),

and the allele for red color is present in the population 80% of the

time, than the other allele for blue color must be present 20% of the

time. Consequently, the allele frequencies in the population are p =

0.8 and q = 0.2. (0.8 + 0.2 = 1).

From this we can calculate the expected genotype frequencies in the

population. Since p = 0.8, we would expect 64% of the flowers in

the populations to be homozygous for red flower color (expected

genotype frequency for p 2

= 0.8 x 0.8 =0.64). 32% of the

populations would be heterozygous for flower color. They would

have one R allele (p = 0.8) and one B allele (q = 0.2), and if neither

allele were dominant they would appear purple. The expected

genotype frequency of these purple individuals is 2pq = 2 x 0.8 x 0.2 = 0.32. Finally, 4% of the population would be homozygous for the

blue flower color (q 2

= 0.2 x 0.2 = 0.04).

The Hardy–Weinberg model applies

when the following conditions are met:

1) No genetic drift

2) No selection

3) No mutation

4) No migration

 

By contrast, there will be change in gene

frequencies in a population when at least

one of these conditions occur:

1) Genetic drift

2) Selection

3) Mutation

4) Migration

 

In today’s lab, you will do a series of exercises that illustrate the effect of the different evolutionary agents on the

genetic structure of a model population. You will work with populations composed of individuals that are

represented by colored beads. White beads represent individuals that are homozygous for the white allele (WW);

red beads are individuals that are homozygous for the red allele (RR), and pink beads are heterozygous (WR).

These beads live in a habitat – a plastic dishpan filled with smaller beads. The larger beads of our population will

be retained by the mesh, while the smaller beads pass through the mesh.

When the individuals are recovered with the help of the mesh, the frequencies of the color alleles are determined

using the Hardy-Weinberg model. The alleles in our populations are codominant – each white individual

possesses two white alleles, each red individual two red alleles, and the pink individuals have one red and one

white allele. Consequently, the total number of color alleles in a population of 40 individuals is 80. If such a

population contains 10 white individuals, 20 pink individuals, and 10 red individuals, the frequency of white

alleles (p) is

(2 x number white beads) + number pink beads

p = —————————————————————-

(2 x number of beads total)

(2 x 10) + 20

p = ———————– = 0.5

80

Because p + q = 1, the frequency of the red allele (q) must also be 0.5.

1. NAT URAL S E L E CT I O N

Natural selection disturbs the Hardy-Weinberg equilibrium by discriminating between individuals with respect to

their ability to survive and reproduce. Individuals that are better at surviving to produce young will contribute

more genes to the next generation; they are said to have greater fitness than those individuals that leave no or

fewer offspring.

In this experiment you will test the hypothesis that individuals are more likely to survive and reproduce when

their coloration makes it easier to hide from predators in the environment.

 

 

Foundations of Biological Sciences I Evolutionary Agents – 3

1. Work in groups of four. Each group member assumes one of the following roles

Predator: Search for prey (large beads)

Data Recorder / Timer: record numerical results and time the predation sessions

Calculator: use a calculator to calculate the allele frequencies

Caretaker: look after and manipulate the experimental setup

2. Create a white habitat by filling the dishpan with small white beads. Establish an initial population by

adding 10 large white beads, 20 large pink beads and 10 large red beads into the habitat. “Hide” the

individuals in the habitat by mixing the large and small beads. The predator will prey on the large beads,

removing as many individuals as possible in a set amount of time. The survivors will reproduce a new

generation, upon which the predator will prey again. This cycle will be repeated several times. Make a

prediction as to how the frequency of red alleles in the populations will change over time.

Prediction:

 

3. The predator hunts for prey (large beads) in the habitat, and uses the pair of forceps to catch as many prey

items as possible in 30 seconds.

4. After the predation (selection) episode, strain the habitat with the sieve and count the remaining red, pink

and red individuals. Record the numbers in the second row in Table 1. Calculate the frequencies of the white

(p) and red (q) alleles remaining in the population, and record them in Table 2 (under First generation). For

example, if 6 white, 8 pink and 8 red individuals remain, the frequency of the white alleles is

To calculate p, use the observed numbers of each color within the formula:

(2 x number white beads) + number pink beads

p = ——————————————————–

(2 x number of beads total)

(2 x 6) + 8

p = ———————– = 0.45

44

 

Table 1: Large- Bead counts before and after four rounds of Natural Selection (Predation)

 

Population

White Beads

Pink Beads

Red Beads

Total Beads

 

Initial

Before 10 20 10 40

After

 

Second Generation

Before

After

 

Third Generation

Before

After

 

Fourth Generation

Before

After

 

 

Foundations of Biological Sciences I Evolutionary Agents – 4

Table 2: Allele and genotype frequencies due to Natural Selection (Predation)

Population p q p 2

2pq q 2

Initial 0.5 0.5 0.25 0.5 0.25

 

First Generation

 

Second Generation

 

Third Generation

 

Fourth Generation

 

5. Based on the new values (after selection) for allele frequencies, calculate the genotype frequencies for the

homozygous white (p 2 ) and red (q

2 ) individuals, and for the heterozygous pink individuals (2pq). Record the

new allele frequencies in Table 2. For example, if p now equals 0.43, the frequency of the homozygous

white individuals is

p 2

= (0.48) 2

= 0.23

6. Assuming that 40 individuals comprise the next and all succeeding generations, calculate the number of

white, red and pink individuals to create the next generation, and record the numbers in the Before row

under Second generation in Table 1. For this, and all future calculations, round up or down to the nearest

whole number. For example, if p 2 =0.23, the number of white individuals for the next generation is

p 2

x 40 = 0.23 x 40 = 9.2 or 9 white individuals

7. Calculate the numbers of white, red and pink individuals you need to construct the new generation, and

introduce them into the habitat for a new round of selection.

8. Repeat the selection and reproduction steps for three more rounds, filling in the remaining rows in Tables 1

and 2. When you are done, use the frequencies of the red allele from Table 2 to construct a histogram in the

appropriate space in Figure 1 below. Remember to label your axes and complete the figure caption.

Figure 1: Changes in frequency of the red allele (q) due to selection. ……………………………………………………………………..

……………………………………………………………………………………………………………………………………………………………………………….

……………………………………………………………………………………………………………………………………………………………………………….

 

 

 

 

 

 

 

 

 

 

0 1 2 3 4

1.0

0.5

0.0

 

 

Foundations of Biological Sciences I Evolutionary Agents – 5

What is your conclusion as to the prediction you made in point 2?

There are 3 different patterns of selection. Directional selection favors one extreme phenotype over the other and

causes allele frequencies to change in a predictable direction. Stabilizing selection favors an intermediate

phenotype, rather than one at the extremes. Disruptive selection disfavors intermediate phenotypes, and favors

the extreme ones. Which kind of selection is illustrated by predation of white, pink and red individuals in a white

habitat?

 

If two identical populations inhabited different environments (e.g. white and red habitats), how would the

frequency of the color genes in each habitat compare after a large number of generations?

 

When two populations become genetically different through time (divergence), individuals can lose the ability

to interbreed, and two new species are formed. This process is called speciation.

 

2. EFFECT OF GENE FLOW ON NATURAL SELECTION

New members may join populations (immigrations) or leave the population (emigration). As they do, the

frequencies of alleles in the population change. This gene flow due to migration can be a powerful force in

evolution.

1. Establish a new population as described in the previous section.

2. Begin the selection process as before, but this time 5 red individuals will immigrate into the population

before the new allele frequencies are determined. Write down a prediction of the hypothesis that gene flow

resulting from migration of individuals into a population undergoing predation affects the change in allele

frequencies expected from selection alone. Focus your prediction on the change in the frequency of red

alleles in the population.

3. Conduct 4 cycles of predation with migration. For each generation, write down the number of surviving

individuals in Table 3, and the allele frequencies in Table 4. When you are done, use the frequencies of the

red allele from Table 4 to construct a histogram as your homework.

Homework: Write a hypothesis and prediction for this evolutionary model based on your understanding of

gene flow. Create a histogram that displays the change through four generations of natural selection with

migration. Remember to include a figure caption and axis labels (10 pts). *Hint: see Figure 1

 

 

Foundations of Biological Sciences I Evolutionary Agents – 6

 

Table 3:

Large- Bead counts before and after four rounds of simulated Natural Selection and Gene Flow

 

Population

White Beads

Pink Beads

Red Beads

Total Beads

 

Initial

Before 10 20 10 40

After (First Gen) (survivors + 5)

 

Second Generation

Before

After (survivors + 5)

 

Third Generation

Before _________ _________ __________ _________

After _________ _________ (survivors + 5)

 

Fourth Generation

Before _________ _________ __________ _________

After _________ _________ (survivors + 5)

Table 4: Allele and Genotype frequencies due to Natural Selection and Gene Flow

Population p q p 2

2pq q 2

Initial

First Generation

Second Generation

Third Generation

Fourth Generation

 

How did migration influence the effectiveness of selection in this example?

If white individuals would have immigrated into the population instead of the red ones – how would this have

influenced the change in gene frequencies?

 

Through immigration, new genetic information is introduced into the population. Gene flow thus maintains

genetic variation in a population. Barriers to gene flow can decrease genetic variation within populations, and

also accelerate divergence between populations.

 

 

Foundations of Biological Sciences I Evolutionary Agents – 7

3. MUTATION

New genetic information can also be introduced into a population through mutation.

1. Establish a new population by placing 10 large white beads, 10 large red beads and 20 large pink beads in

the bowl. Do not add the small beads this time.

2. Designate one group member to pick 20 large beads from the bowl (without looking!). Use these 20 beads to establish a new generation. Then replace one white bead with a silver bead. This represents a mutation in the

gamete that one parent contributes to this generation.

3. Calculate the allele frequencies of the new generation, including the frequency of the new color allele (r),

and write them down in Table 5.

 

Table 5: Change in Allele Frequencies due to Mutation

Population p

q

r

 

Initial

 

New Generation

 

After the mutation, three alleles are present in the population (p + q + r = 1). Consequently, the Hardy-Weinberg

equation must be expanded to

p 2 + 2pq + q

2 + 2pr + 2qr + r

2 = 1.0.

In addition to white, pink and red phenotypes, there are now silver, silver-red, and in subsequent generation

potentially black phenotypes. If the next generation contains 50 individuals, how many offspring of each

phenotype would be found in the population? Use Table 6 to calculate the genotype and phenotype frequencies.

 

Table 6: Phenotypes two generations after a mutation

 

Phenotype Genotype Frequency

x 50

 

# Individuals

White p 2

 

 

Pink 2pq

Red q 2

 

 

Silver 2pr

Silver-Red 2qr

Black r 2

 

 

Imagine a population made up of individuals of the color phenotypes in these proportions. What effect will

natural selection have on the phenotypes in a white habitat?

 

Under which conditions would a rare black allele be favored?

 

 

Foundations of Biological Sciences I Evolutionary Agents – 8

4. GENETIC DRIFT

Gene frequencies can change over generations as a result of chance (Genetic Drift). Genetic drift is often a

problem for small populations, because it can result in a loss of genetic variability. In very small populations,

chance can even eliminate an allele from a population.

1. Establish a new population by placing 10 large white beads, 10 large red beads and 20 large pink beads in

the bowl. Do not add the small beads this time.

2. If all 40 members of this population have an equal chance of getting to reproduce, the allele frequencies for

the colors in the population are p = 0.5 and q = 0.5 (see first column of Table 7).

3. Now let’s see what happens when only a subset of the population gets a chance to reproduce. Choose,

without looking, 10 beads from the bowl. They will make up a small group of individuals that reproduce.

Record the allele frequencies in this cluster in the second column of Table 7.

4. Replace the 10 beads to the population and mix well. Then pick 30 new beads. They represent a larger group

of individuals that get to reproduce. Calculate the allele frequencies in this group, and add them to the third

column in Table 7.

 

 

Table 7: Allele frequencies resulting from Genetic Drift

Observed Frequency in

Expected Frequency Small Group Large Group

n

40

 

p

0.5

 

q

0.5

 

Compare the allele frequencies of the three groups of reproducing individuals. How does group size affect the

makeup of the next generation?

 

When members of an old population emigrate to establish a new population, the allele frequencies in the new

population can be heavily influenced by chance.

5. Reestablish the old 10/20/10 beads population you have worked with before. Then pick (without looking) 6

individuals which will represent the 6 founding members of a new population.

6. Move these individuals to a new habitat. Calculate the allele frequencies and record them in Table 8.

 

Table 8: Allele Frequencies in a Founder Population

 

Allele Frequency

Initial population Founder Population

 

p

0.5

 

q

0.5

 

How do the allele frequencies of the small founder population compare to the larger original population?

Foundations of Biological Sciences I Evolutionary Agents

 
"Looking for a Similar Assignment? Get Expert Help at an Amazing Discount!"

Monohybrid crosses

Monohybrid crosses. How to Proceed

  • Read through the introductory materials below.
  • Open the Unit 6 Experiment Answer Sheet and complete the following Experiment exercises this unit:
    • Experiment 6 Exercise 1 – Monohybrid crosses (~45 min)
    • Experiment 6 Exercise 2 – Dihybrid crosses (~30 min)
    • Experiment 6 Exercise 3 – Inheritance of Human Traits (~30 min)
  • Save your completed Unit 6 Experiment Answer Sheet and submit it no later than Sunday midnight (CT).

Monhybrid and Dihybrid Crosses – Introduction

Mendel crossed true-breeding pea plants in order to develop and understanding of how traits are inherited. True-breeding means that if a plant was crossed with itself, it always generated offspring that looked like the parent. Although Mendel didn’t know this at the time, it meant that the parent plant was homozygous or had two copies of the same allele that controlled the appearance of the trait.

Mendel noticed that when he crossed two true-breeding plants exhibiting different versions of a trait (e.g., green and yellow); the offspring (F1) always looked like only one of the parent plants. We know now that the F1 individuals looked like the parent that carried the dominant trait. But what surprised Mendel, was that when he crossed the F1 individuals with each other, the F2 offspring exhibited BOTH traits! Based on this observation, he concluded that the F1 individuals were hybrids, meaning they carried both alleles for a given trait. Only the dominant trait was expressed in the F1 individuals and the recessive trait, although present, was masked.

monohybrid cross is when you are interested in crossing individuals that vary in only a single trait (e.g., flower color, seed color, stem length). In a dihybrid cross, we are crossing individuals that differ at two traits (e.g., flower color and seed color, flower color and stem length). Obviously, the more traits that vary, the more complex the crosses become!

By examining the distribution of the various traits obtained following different types of crosses, Mendel was able to describe the general pattern of genetic inheritance. Be sure to review the online lecture this unit on Genetics and pp 146-153 in your book before starting these first two exercises.

We will be using the following website for the first exercise. Be sure that you can access it and use it before beginning:

Glencoe-McGraw Hill. No date. Punnett Squares
http://glencoe.mcgraw-hill.com/sites/dl/free/0078759864/383934/BL_05.html (Links to an external site.)

You will need to complete the Tables and answer the questions in the Unit 6 Experiment Answer Sheet for Exercises 1 and 2.

Inheritance of Human Traits – Introduction

Some human traits are controlled by a single gene that has only two alternative alleles. If a characteristic is determined by the dominant allele, one or both parents express that trait and many of the children will as well. Dominant characteristics will most likely be present in every generation, since the expression of these traits requires only one of the dominant alleles in order to be expressed. If the characteristic is determined by the recessive allele, then neither parent may express the trait nor few of the children. This is because two copies of the recessive allele must be present in order for the recessive trait to be expressed. If a trait is X-linked recessive; meaning the gene for the trait is found on the X chromosome, it will be expressed primarily in males.

The application of human genotypes in medicine and genetic counseling is becoming more and more necessary as we discover more about the human genome. Despite our increasing ability to decipher the chromosomes and their genes, an accurate family history remains one of the best sources of information concerning the individual. In this exercise you will determine your genotype for certain characteristics that are controlled by a single gene with two alleles based on your phenotype. We will not be looking at any X-linked traits in this exercise.

Use the information about the traits of interest in the Unit 6 Experiment Answer Sheet to answer the questions found there.

WEEK 6 EXPERIMENT ANSWER SHEET Please submit to the Week 6 Experiment dropbox no later than Sunday midnight.

SUMMARY OF ACTIVITIES FOR WEEK 6 EXPERIMENT ASSIGNMENT

· Experiment 6 Exercise 1 – Monhybrid Crosses

· Experiment 6 Exercise 2 – Dihybrid Cross

· Experiment 6 Exercise 3 – Inheritance of Human Traits

Experiment 6 Exercise 1: Monohybrid Crosses

You will be conducting monohybrid crosses using fruit flies. Open in the following website:

Glencoe-McGraw Hill. No date. Punnett Squares http://glencoe.mcgraw-hill.com/sites/dl/free/0078759864/383934/BL_05.html

Procedure

A. Open the above website and click on the VCR to listen to the introduction. Close the window when done.

B. Click on the Lab Notebook on the lab bench. A breeding scenario will be presented to you that you will need to carry out. Here are the possible phenotypes and genotypes you will using:

a. Normal wings (LL or Ll) or vestigial wings (ll)

b. Gray body (GG or Gg) or black body (gg)

C. Enter the Scenario number in Table 1 below.

D. Based on the Scenario, use the down arrows beneath the Parent 1 and Parent 2 boxes to select the appropriate parents. Look carefully at the flies so that you know which ones to select.

E. Before proceeding, click on the Check Parents button. If necessary, make corrections. If you are correct, the maternal and paternal alleles will be added to the Punnett Square.

F. Next, drag the correct allele combinations and the corresponding fly types to the boxes in the Punnett Square.

G. When you are done, click on Check Offspring. If necessary, make corrections.

H. Record your data in Table 1 below. An example has been given, but note that the example is not using the genotypes and phenotypes used in this exercise.

I. Click on Reset. Repeat steps B – H four more times for a total of five crosses. If a scenario is presented that has already been completed, click Reset again. DO NOT REPEAT a given scenario!

Note that the scenario number you need to record in the Table below is the number associated with the specific scenario you completed.

Table 1. Results of crosses.

 

Parent Genotypes Offspring Genotypes Offspring Phenotype
Scenario # Parent 1 Parent 2 # % # %
Example Rr rr 2 Rr

2 rr

50% Rr

50% rr

2 red

2 white

50% red

50% white

             
             
             
             
             

Questions

1. Which type of cross gave you the greatest number of genotypes? Was the number of phenotypes the same as the number of genotypes or different? If different, why (2 pts)?

2. Can the genotype for a gray-bodied fly be determined? If so, how? (3 pts)?

Experiment 6 Exercise 2: Dihybrid Cross

We will continue to use flies for our crosses, but this time we will examine the inheritance of TWO different traits: body color (gray or black) and wing type (long or vestigial). As with our first crosses, the gray body color is dominant (GG or Gg) over the black body color (gg). And the long wing type is dominant (LL or Ll) over vestigial (ll). Be sure you have reviewed our online Genetics lecture and this week’s reading before proceeding. An example of a dihybrid cross is shown on p 150 in your book.

Recall our flies from the previous exercise. We have the following traits:

· Gray body (GG or Gg) is dominant over black body (gg)

· Long wings (LL or Li) is dominant over vestigial wings (ll)

We will cross a gray bodied fly with long wings which has the genotype GGLl with a gray bodied fly with long wings with a genotype of GgLl . Note that even though the phenotypes are the same, the genotypes of the two parents are different.

Identify the four possible gametes produced by these two individuals. Note that each gamete must consist of two alleles (G or g and L or l):

Parent 1 Parent 2

image1.png image2.png

GGLl GgLl

Parent 1 (GGLl) Gametes: _______ ______ _______ _______ (1 pts)

Parent 2 (GgLl) Gametes: _______ ______ _______ _______ (1 pts)

Create a Punnett square to show the outcome of a cross between these two individuals (GGLl and GgLl) using the gametes you identified above (3 pts).

    Parent 1
           
Parent 2          
           
           
           

Questions

1. What are the possible F1 genotypes (these must now have four alleles) and their percentages (4 pts)?

2. Recall that GG and Gg individuals are gray bodied, while gg individuals are black bodied and that LL and Li individuals have long wings, while ll individuals have vestigial wings. What are the phenotypes of the resulting offspring and what are the percentages of these phenotypes (2 pts)?

Experiment 6 Exercise 3: Inheritance of Human Traits

Read over the Inheritance of Human Traits Introduction under the Week 6 Experiment link in our course before beginning.

Procedure

A. For each of the heritable traits describe below, determine which form you have (dominant form or recessive form). This is your phenotype.

B. Record your phenotype information in Table 2 below. Then, enter the possible genotype(s) you have based on your phenotype.

C. Answer the questions found following Table 2 below.

Description of Heritable Traits

Trait Possible Alleles Dominant Form Recessive

Form

Examples
Ear lobes E or e Detached (Free) Attached  

image3.png 

Hairline W or w Widows peak Straight image4.jpg Widow’s peak Straight
Tongue rolling T or t Able to roll Unable to roll image5.jpg
Hand folding R or r Right thumb on top Left thumb on top image6.jpg
Chin C or c Cleft chin No cleft chin image7.jpg
Tongue folding F or f Can fold tongue backwards Cannot fold tongue backwards image8.jpg
Thumb H or h Straight thumb (cannot bend backwards) Hitchhiker’s thumb (can bend it backwards) image9.jpg
Little Finger B or b Bent inwards Straight image10.jpg
Mid-digital hair M or m Hair on fingers No hair on fingers  

image11.png 

An example is shown as to what should be entered in RED. Please correct the entry for “Ear lobes” based on your personal data. For the Genotypes, please use the letters provided above (8 pts).

Table 2. Your phenotypes and genotypes.

Trait Phenotype Genotype
Ear lobes Unattached OR Detached EE/Ee OR

ee

Hairline    
Tongue Rolling    
Hand Folding    
Chin    
Tongue Folding    
Thumb    
Little Finger    
Mid-digital Hair    

Questions

1.  Which traits did you have that were dominant (1 pts)?

2.  Which traits did you have that were recessive (1 pts)?

3. What does it mean to be homozygous for a trait? Cite source(s) used (1 pts).

4.  What does it mean to be heterozygous for a trait? Cite source(s) used (1 pts).

5.  Define genotype and phenotypeCite source(s) used (1 pts).

6.  Which traits do you know for sure that you were homozygous (1 pts)?

Week 6 Experiment Grading Rubric

Component Expectation Points
Experiment 6 Exercise 1 Correctly perform and record the outcome of five monohybrid crosses (Table 1).

5
 

Demonstrate an understanding of the possible outcomes of monohybrid crosses with respect to genotypes and phenotypes (Questions 1-2).

5
Experiment 6 Exercise 2 Determine the correct parental gametes and conduct a dihybrid cross.

5
 

Correctly evaluate the outcome of a dihybrid cross (Questions 1-2).

6
Experiment 6 Exercise 3

Correctly recognize one’s phenotype and assigns the correct genotype (Table 2).

8
 

Demonstrate an understanding of dominant and recessive traits, genotype vs phenotype and homozygous vs heterozygous (Questions 1-6).

6
TOTAL  

35 pts

Updated October 2013

Monohybrid crosses

 
"Looking for a Similar Assignment? Get Expert Help at an Amazing Discount!"