This document defines key genetics concepts and outlines learning objectives related to inheritance and evolution. It describes how DNA encodes genes, which determine physical traits. Genes are passed from parents to offspring, with dominant alleles usually determining visible traits. In stable populations, allele frequencies remain constant due to random mating and lack of evolutionary pressures. However, mutations, genetic drift, gene flow and natural selection can drive changes in allele frequencies over generations. The document introduces a simulation model where students can observe inheritance and evolution by tracking the rim color of "model bugs" over multiple generations.

1. Learning Objectives Define inheritance Define gene, allele and trait
Describe
Learning ObjectivesDefine inheritanceDefine gene, allele and traitDescribe the role of DNA
in protein synthesis (Central Dogma)Define genotype and phenotypeExplain the
relationship between genes, alleles and physical traits (phenotype)Define dominant and
recessive traitsConstruct a Punnett square for a dominant and recessive trait for one
generationExplain how traits are inheritedEvaluate how population genetics is used to
study the evolution of populationsExplain expectations about allele frequencies in a stable
population (i.e., a population that is not experiencing evolutionary forces and that meets
Hardy-Weinberg assumptions)Compare and contrast the mechanisms of evolutionary
forces (mutations, natural selection, gene flow, and genetic drift)Discuss the influence an
evolutionary force can have on allele frequencies within a populationDNA, Genes and
InheritanceHow is it that offspring resemble their parents? You are probably aware that the
answer somehow relates to DNA and genes. In this set of laboratory exercises, you will
explore the transfer of traits from parents to offspring and see expressions of variations in
genes.The theory of chromosomal inheritance1 states that DNA nucleotides are the code for
genes that determine all physical traits such as proteins like insulin or tissues like skin. DNA
make up chromosomes. Genes are sequences of DNA that serve as the code for proteins. The
code is first read and a strand of nucleotides complementary to DNA (called messenger
RNA, or mRNA) is synthesized in the cell nucleus then the complementary code is read by
ribosomes in the cell cytoplasm. The second reading sets the stage to produce proteins
according to the nucleotide sequence in the gene. This sequence from genes to proteins is
referred to as the Central Dogma of genetic code of life.Central Dogma of Life SciencesGene
(DNA)−→−−−−−−−−transcribed intoMessenger RNA code−→−−−−−−−translated intoAmino
acids, proteinsGene (DNA)→transcribed intoMessenger RNA code→translated intoAmino
acids, proteinsGenes are made up of DNA and vary in length from a few hundred DNA
nucleotide bases to more than two million nucleotide bases. Humans have about 20,000
genes in the human genome.2A gene is responsible for all parts of an organism. Products of
a gene can be thought of as ‘traits.’ (Note: Traits can be a physical appearance at a
microscopic or macroscopic level, or they can be behaviors.) Physical traits or behavioral
traits are also known as the phenotype of the organism, whereas the term ‘genotype’ refers
to genes that code for those traits.Gene (Genotype)−→−−−−codes forTrait (Phenotype)Gene
(Genotype)→codes forTrait (Phenotype)For a living organism to exist, a gene was necessary

2. to code for every macromolecule, protein, or enzyme that makes up that individual
organism. A gene can be described as a segment of DNA that is “the fundamental physical
and functional unit of heredity, which carries information from one generation to the next.”
3Deoxyribonucleic acid, or DNA, is the genetic blueprint for living organisms. The
information in the blueprint is a code made of four chemicals called bases. Bases are
attached to a sugar molecule and a phosphate molecule and, together, form a nucleotide.4
Genetic material consists of strands of the nucleotides adenine (A), thymine (T), cytosine
(C), and guanine (G). The complementary base pairs A=T and C≡G are bound by weak
hydrogen bonds. Together, the strands form a structure known as a double helix. The two
strands are bound together by hydrogen bonds that pair the nucleotide guanine (G) with
cytosine (C) and the nucleotide adenine (A) with thymidine (T).The helices are coiled tightly
and form chromosomes. In humans, there are 23 paired chromosomes in somatic cells—
cells that are not egg and sperm. The paired nature of these chromosomes are referred to as
diploid. Sex chromosomes in somatic cells are paired in the form of XX or XY pairs,
depending on whether the organism is female or male, respectively. In egg and sperm cells,
however, the chromosomes—more correctly, DNA—undergo meiosis which “unpairs” the
chromosomes, making them haploid (non-paired) strands of chromosomes.Dominant and
Recessive TraitsAll living organisms inherit traits in the form of DNA in alleles in sex
chromosomes of their parents’ DNA. Alleles are variations in the nucleotide sequence of a
gene; genes are in the same location on the chromosome for the parents’ alleles. Offspring
inherit genes and specific variations in each gene (alleles) from their parents. For example,
Gregor Mendel studied the gene for the color of peas. One allele of the gene resulted in
yellow peas while another allele resulted in green peas.Some alleles dominate and a
particular trait might be more common in a population due to a dominant allele. Examples
of dominant genes in the human population are right-handedness, dark hair, and brown
eyes. Other alleles are recessive and their trait is not visible unless both parents carry the
recessive allele and the offspring inherits the recessive alleles from mother and father.Let’s
consider the Punnett square shown on the left. In the male, the genotype for the gene in
question is heterozygous; that is, one of his alleles is dominant and the other is recessive.
For the female, the genotype for the gene in question is homozygous; that is, both of her
alleles are dominant. If a large number of offspring are born to these parents, none (0%) of
the offspring in the first generation would have the physical features associated with the
recessive trait (phenotype). Confirm that conclusion by studying the Punnett square.In the
case of two heterozygous parents (as in the Punnett square on the left), such as might occur
in later generations and in descendants of these two parents, statistically speaking, 25% of
the offspring would show the physical features associated with the recessive allele (see the
Punnett square on the left). As with all statistical probabilities, the genotypes of the actual
offspring will more closely resemble the expected percentages in the Punnett square as the
number of births grow. In a realistic simulation as in this set of laboratory exercises, you
will be able to run many trials and collect the data for several “births” of offspring, which
will permit you to collect data to test hypotheses based on your expectations about the
proportion of the alleles for the genotypes of the offspring.Genes (and alleles) in the egg and
sperm cells of parents essentially re-combine and sort randomly during meiosis and

3. fertilization so that no two offspring are genetically identical (except for identical twins).
Random assorting drives the statistical nature of the Punnett square. Consider a Punnett
square of allele probabilities for offspring for as many as 20,000 different genes—which is
the estimated number of genes in the human genome. The possible phenotypic outcomes
would vary accordingly. Random sorting of genes between two parents as well as random
sorting of genes in a whole population of a species lead to genetic and phenotypic
diversity.The reproductive success of parents or a given population of a species drives the
opportunity to contribute to generations far into the future. Parents and populations that
survive to reproduce in an environment where they live and who have offspring who repeat
that cycle will contribute to future generations.Population Genetics and the Hardy-
Weinberg EquilibriumAll females and males of the same species together make a
population. The genes of populations typically do not vary much. However, changes in the
genes in a population can occur through random mutation. A random mutation in the
nucleotide sequence of gene in an egg or sperm cell could be the origins of a new allele in
the population if the offspring survive and reproduce and pass on that new allele to future
generations. A mutation in the nucleotide sequence of a gene is the only way a new allele
can occur. Otherwise, parents simply pass on their unmutated allele—an allele common to
all members of the entire, stable population—to their offspring.Most populations of a given
species have a stable set of genes and alleles. The primary origin of a new, unique allele,
different from ones already in the population, is a random mutation. In fact, most random
mutations in the nucleotide sequence of genes lead to death of the organism.Scientists study
how and why alleles vary within populations of organisms. The following characteristics
apply to alleles and genotypes in stable populations:Individuals of all genotypes within a
given population have equal rates of survival and equal reproductive success. No selection
process is at work in the stable population.Changes in a nucleotide sequence of a gene (a
mutation) do not create a new alleleIndividuals do not migrate into or out of the
populationThe population is large enough that random changes in alleles are on a small
scale and do not come to dominate within the populationIndividuals within the population
are not selective about their mates; they mate randomly5These characteristics are central
tenets of the Hardy-Weinberg formulation which is a mathematical representation of alleles
(and traits) within a stable population of the same species. The allele frequency for a given
gene within a stable population can be readily quantified using the Hardy-Weinberg
equation (shown below).p2+2pq+q2=1,p2+2pq+q2=1,where pp represents the frequency of
the dominant allele of a gene within a population, qq represents the frequency of the
recessive allele within a population, and 2pq2pq represents the frequency of heterozygotic
individuals within the population. The Hardy-Weinberg model relies on probability as the
computational strategy to determine how prevalent an allele (or trait) is within a
population. The proportional total of all frequencies of homozygotic genotypes (dominant
and recessive, such as BB and bb) and all frequencies of heterozygotic genotypes (a mixture
of dominant and recessive such as Bb) totals to 1.To gain further insight into the Hardy-
Weinberg equation, consider the Punnett square below where the allele frequencies of a
gene are presented along with the genotype of the offspring.Example: Consider two alleles
of a gene, B and b, in a population where the frequency of B is p=0.7p=0.7 and the frequency

4. of b is q=0.3q=0.3. Notice that the total frequency is p+q=1p+q=1, which means that there
are only two different alleles for this gene in the population; they are all accounted for.An
assumption for a stable population is that mating is random. Mating probabilities can be
easily computed using the probability formula:
p2+2pq+q2=1p2+2pq+q2=1.Probabilities:Both parents have B alleles, which means a
frequency of occurrence of BB genotype offspring in the population
isp×p=0.7×0.7=0.49p×p=0.7×0.7=0.49, or 49%49% frequency of BB genotype.Both parents
have b alleles, which means a frequency of occurrence of bb genotype offspring in the
population isq×q=0.3×0.3=0.09q×q=0.3×0.3=0.09, or 9%9% frequency of bb genotype.One
parent contributes a B allele and the other parent contributes a b allele, which means a
frequency of occurrence of Bb genotype offspring is
2×p×q=2×0.7×0.3=0.422×p×q=2×0.7×0.3=0.42 or 42%42% frequency of Bb
genotype.Notice that the Hardy-Weinberg equation is satisfied, that is,
p2+2pq+q2=0.49+0.42+0.09=1p2+2pq+q2=0.49+0.42+0.09=1.Evolutionary Forces that
Disrupt the Hardy-Weinberg EquilibriumTo use the Hardy-Weinberg equation, it is
necessary to assume that populations are stable and that allele frequency does not change
over time. These assumptions are generally accurate for large populations of a given
species; however, there are natural disruptions that occur which break these assumptions.
Consider the scope of life on Earth if populations never changed. Would many of the species
on Earth today exist if there were no changes to allele frequency within large populations?
Biodiversity is completely due to natural selection, and natural selection is due to several
specific processes that involve dynamic changes in the allele frequencies for a
species.Processes that contribute to natural selection—that is, changes in allele frequency
within a population—include (1) random mutations in the nucleotide sequence of gene, (2)
the tendency of some members of a population who happen to have a specific set of alleles
to become physically separated from larger populations (gene flow or migration), and (3)
the survival and reproductive success of a small sub-population whose gene pool does not
reflect the same allelic composition of the larger population (genetic drift).Consider the
Punnett Square allele frequency values above for the Hardy-Weinberg allele frequencies in a
special context in which, for example, a natural disaster causes the physical separation—
such as by crevasse—of a smaller population having the allele bb from the larger population
with dominant allele BB. The offspring of the smaller population will have a new set of
alleles—all alleles would be bb and the formerly recessive allele would become dominant—
more frequent—in this splinter population.The model system discussed in the next section
(below) consists of (simulated) bugs where the alleles (genotype) are BB and Bb and the
phenotype is the expression of the color blue for the exoskeleton. A Punnett square can be
constructed for a mating pair or for an entire population of mating pairs. The allele
frequencies of the offspring across all mating pairs in the population vary according to the
potential influence of processes that govern natural selection. If the population is stable,
then the allele frequencies stay the same (Hardy-Weinberg assumptions hold). If natural
selection (mutation, gene flow, or genetic drift) exert an influence, allele frequencies will
change.Orientation to the Model SystemThe simulation shown in the image below is
representative of the genetics experiments you will conduct in this set of laboratory

5. exercises. For one set of exercises, you select the alleles of the parents and will examine the
alleles of offspring from mating pairs (“Baby Bugs”). In the other set of exercises, you will
explore the impact of natural selection on many mating pairs (the population) over several
generations (“Bug Pop”).Two forms of data will be provided at the conclusion of each data
run:(1) A tally of the number of offspring for the three different allele possibilities and(2) a
corresponding display of offspring phenotypes that represent a visual counterpart of the
allele distribution for the offspring.The allele (gene) that will be of interest in this set of
experiments is the rim (or outer ring) color of each bug. There are two alleles, blue and
yellow. Our model bugs are shown below. Bugs can have a blue outer rim (i.e. blue-rimmed)
or a yellow outer rim (i.e. yellow-rimmed). Rim color is determined by the genotype (allele)
of the bug. The blue-rimmed trait is dominant (B allele).Model Bug Key:Phenotype (trait):
blue-rimmed, Genotype: BB, alleles: B and BPhenotype (trait): yellow-rimmed, Genotype:
bb, alleles: b and bPhenotype (trait): blue-rimmed, Genotype: Bb, alleles: B and bIn your
second exercise several generations of bugs will be born and die.Dead bugs of parent
generations: Gray ghost-like rims and bodiesProcedure I OverviewWhen you conduct the
“Baby Bugs” activities, you will be working with a blue allele that is dominant and noted by
B and a yellow allele that is recessive and noted by b. This activity will simulate breeding
between bugs. You will investigate the offspring that result from specific breeding pairs.
Your work will include comparing probabilities calculated from your data and from Punnett
Squares.Procedure II OverviewWhen you conduct the “Bugs Pop” activities, you will explore
what happens when evolutionary forces act on a population, and the dominance and
recessive qualities of the alleles are not driving the survival and reproductive success of the
parents and offspring. For this activity, you will observe potential composition changes to a
small isolated breeding population of bugs over time. Your work will include comparing
probabilities calculated from your data and from the Hardy-Weinberg equation.Summary of
Formulas Needed for CalculationsCalculating Average Values from DataExample: Determine
the average BB baby count from ten data run values: 3, 6, 4, 4, 3, 3, 5, 6, 4, 5.BB
average=3+6+4+4+3+3+5+6+4+510=4310=4.3BB
average=3+6+4+4+3+3+5+6+4+510=4310=4.3Calculating Probabilities (frequencies) and
Percentages from DataProbabilities and percentages can be determined from a single data
point or from an average of multiple data runs. Average data values tend to be more
representative of a population when there is variation between individual data
runs.Example: Determine the genotype probabilities (frequencies) and percentages from
the following information:data averages: BB average = 4.3, Bb average = 2.3, bb average =
3.4total number of babies = 10.For the BB genotype we haveBB probability=BB averagetotal
number of babies=4.310=0.430BB probability=BB averagetotal number of
babies=4.310=0.430BB percentage=100%×BB probability=100%×0.430=43.0%.BB
percentage=100%×BB probability=100%×0.430=43.0%.For the Bb genotype we haveBb
probability=Bb averagetotal number of babies=2.310=0.230Bb probability=Bb averagetotal
number of babies=2.310=0.230Bb percentage=100%×Bb
probability=100%×0.230=23.0%.Bb percentage=100%×Bb
probability=100%×0.230=23.0%.For the bb genotype we havebb probability=bb
averagetotal number of babies=3.410=0.340bb probability=bb averagetotal number of

6. babies=3.410=0.340bb percentage=100%×bb probability=100%×0.340=34.0%.bb
percentage=100%×bb probability=100%×0.340=34.0%.Example: Determine the phenotype
probabilities and percentages from the results above.For the blue phenotype we haveblue
phenotype probability=BB probability+Bb probability=0.430+0.230=0.660blue phenotype
probability=BB probability+Bb probability=0.430+0.230=0.660blue phenotype
percentage=100%×blue phenotype probability=100%×0.660=66.0%.blue phenotype
percentage=100%×blue phenotype probability=100%×0.660=66.0%.For the yellow
phenotype we haveyellow phenotype probability=bb probability=0.340yellow phenotype
probability=bb probability=0.340yellow phenotype percentage=100%×yellow phenotype
probability=100%×0.340=34.0%.yellow phenotype percentage=100%×yellow phenotype
probability=100%×0.340=34.0%.Determine Percentages from a Punnett SquareExample:
Punnett Square for two heterozygous parents (both genotype Bb)Punnett Square Example
for Two Heterozygous Parents (both genotype Bb)Both parents are genotype Bb
(heterozygous).Insert the parents alleles along the top and left sideFill in the genotypes in
the first row (use one allele from each parent)Fill in the genotypes in the second row (use
one allele from each parent)What does the completed Punnett Square tell us? There are two
equally valid interpretations.Expected probabilities of a single birth between Bb
parents.25% chance (1/4) the child is genotype BB25% chance (1/4) the child is genotype
bb50% chance (2/4) the child is genotype BbWhat does this mean in terms of the expected
phenotype from a single birth between these parents?BB and Bb genotypes share the same
phenotype and so there is a 75% chance (3/4) of a child being born into that
phenotype.Likewise, there is a 25% chance (1/4) of a child being born into the phenotype
associated with the bb genotype.Expected probabilities of births in a population composed
only of Bb parents.on average 25% of births (1/4) are genotype BBon average 25% of births
(1/4) are genotype bbon average 50% of births (2/4) are genotype BbWhat does this mean
in terms of the expected births in a population composed only of Bb parents?BB and Bb
genotypes share the same phenotype and so, on average, 75% of births (3/4) are that
phenotype.Likewise, on average, 25% of births (1/4) belong to the phenotype associated
with the bb genotype.Using the Hardy-Weinberg EquationThe Hardy-Weinberg equation
can be used to understand the proportions of alleles, genotypes, and phenotypes in a stable
population.Example: A small breeding population of bugs initially consists of 12 blue-
rimmed bugs of genotype BB and 8 yellow-rimmed bugs (8 genotype bb). After several
generations of births and deaths, the population is observed to consist of 17 blue-rimmed
bugs (7 BB and 10 Bb) and 3 yellow-rimmed bugs (3 bb). Is this population consistent with
the expectations of the Hardy-Weinberg model, that is, is this population stable?Population
DataInitial Population Composition — 12 blue bugs of genotype BB and 8 yellow bugs (8
genotype bb)Final Population Composition — 17 blue bugs (7 BB and 10 Bb) and 3 yellow
bugs (3 bb)Step 1: Determine pp and qq from the initial population composition
datap=number of B allelestotal number of alleles=2440=0.600p=number of B allelestotal
number of alleles=2440=0.600q=number of b allelestotal number of
alleles=1640=0.400q=number of b allelestotal number of alleles=1640=0.400Step 2: Use
the Hardy-Weinberg equation to predict the future population compositionThe Hardy-
Weinberg equation is p2+2pq+q2=1p2+2pq+q2=1, where q2q2 is the proportion of yellow

7. bugs and p2+2pqp2+2pq is the proportion of blue bugs. However, once we know the
proportion of yellow bugs it is easier to the determine the blue bug proportion using
1−q21−q2.proportion of yellow bugs=q2=(0.400)2=0.160⇒16.0%proportion of yellow
bugs=q2=(0.400)2=0.160⇒16.0%proportion of blue
bugs=1−q2=1−0.160=0.840⇒84.0%proportion of blue
bugs=1−q2=1−0.160=0.840⇒84.0%Step 3: Determine the observed final population
pheotype percentages from dataobserved proportion of blue bugs=number of blue
bugstotal number of bugs=1720=0.850⇒85.0%observed proportion of blue bugs=number
of blue bugstotal number of bugs=1720=0.850⇒85.0%observed proportion of yellow
bugs=number of yellow bugstotal number of bugs=320=0.150⇒15.0%observed proportion
of yellow bugs=number of yellow bugstotal number of bugs=320=0.150⇒15.0%Step 4:
Compare the Hardy-Weinberg predicted and observed final population phenotype
percentagesBlue — predicted: 84.0%84.0%; observed: 85.0%85.0%Yellow — predicted:
16.0%16.0%; observed: 15.0%15.0%Conclusion: This population is consistent with the
expectations of the Hardy-Weinberg model.1. Klug, WS, Cummings, MR, Spencer, CA, and
Palladino, MA, Concepts of Genetics, 10th edition (2012), page 5.2.
https://ghr.nlm.nih.gov/primer/basics/gene.3.
https://www.ncbi.nlm.nih.gov/books/NBK21962/.4. What is DNA? Genetics Home
Reference: Your Guide to Understanding Genetic Conditions from U.S. National Library of
Medicine, National Institute of Health, https://ghr.nlm.nih.gov/primer/basics/dna.
Retrieved on 7/23/2019.5. Klug, WS, Cummings, MR, Spencer, CA, and Palladino, MA,
Concepts of Genetics, 10th edition (2012), page 702.