L6

L6 Notes

Learning Objectives

· Define inheritance

· Define gene, allele and trait

· Describe the role of DNA in protein synthesis (Central Dogma)

· Define genotype and phenotype

· Explain the relationship between genes, alleles and physical traits (phenotype)

· Define dominant and recessive traits

· Construct a Punnett square for a dominant and recessive trait for one generation

· Explain how traits are inherited

· Evaluate how population genetics is used to study the evolution of populations

· Explain 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 population

DNA, Genes and Inheritance

How 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 Sciences

Gene (DNA)−→−−−−−−−−transcribed intoMessenger RNA code−→−−−−−−−translated intoAmino acids, proteinsGene (DNA)→transcribed intoMessenger RNA code→translated intoAmino acids, proteins

Genes 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.2

A 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 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.” 3

Deoxyribonucleic 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 Traits

All 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 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 Equilibrium

All 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:

1. 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.

2. Changes in a nucleotide sequence of a gene (a mutation) do not create a new allele

3. Individuals do not migrate into or out of the population

4. The population is large enough that random changes in alleles are on a small scale and do not come to dominate within the population

5. Individuals within the population are not selective about their mates; they mate randomly5

These 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 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 is
p×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 is
q×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 Equilibrium

To 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 System

The simulation shown in the image below is representative of the genetics experiments you will conduct in this set of laboratory 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. (1) A tally of the number of offspring for the three different allele possibilities and

2. (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 B

 Phenotype (trait): yellow-rimmed, Genotype: bb, alleles: b and b

 Phenotype (trait): blue-rimmed, Genotype: Bb, alleles: B and b

In your second exercise several generations of bugs will be born and die.
   Dead bugs of parent generations: Gray ghost-like rims and bodies

Procedure I Overview

When 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 Overview

When 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 Calculations

Calculating Average Values from Data

Example: 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=4.3

Calculating Probabilities (frequencies) and Percentages from Data

Probabilities 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 = 3.4
total number of babies = 10.

For the BB genotype we have

BB probability=BB averagetotal number of babies=4.310=0.430BB probability=BB averagetotal number of babies=4.310=0.430

BB percentage=100%×BB probability=100%×0.430=43.0%.BB percentage=100%×BB probability=100%×0.430=43.0%.

For the Bb genotype we have

Bb probability=Bb averagetotal number of babies=2.310=0.230Bb probability=Bb averagetotal number of babies=2.310=0.230

Bb percentage=100%×Bb probability=100%×0.230=23.0%.Bb percentage=100%×Bb probability=100%×0.230=23.0%.

For the bb genotype we have

bb probability=bb averagetotal number of babies=3.410=0.340bb probability=bb averagetotal number of babies=3.410=0.340

bb 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 have

blue phenotype probability=BB probability+Bb probability=0.430+0.230=0.660blue phenotype probability=BB probability+Bb probability=0.430+0.230=0.660

blue 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 have

yellow phenotype probability=bb probability=0.340yellow phenotype probability=bb probability=0.340

yellow 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 Square

Example: 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).

1. Insert the parents alleles along the top and left side

2. Fill in the genotypes in the first row (use one allele from each parent)

3. 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 BB

· 25% chance (1/4) the child is genotype bb

· 50% chance (2/4) the child is genotype Bb

What 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 BB

· on average 25% of births (1/4) are genotype bb

· on average 50% of births (2/4) are genotype Bb

What 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 Equation

The 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 Data

Initial 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 data

p=number of B allelestotal number of alleles=2440=0.600p=number of B allelestotal number of alleles=2440=0.600

q=number of b allelestotal number of alleles=1640=0.400q=number of b allelestotal number of alleles=1640=0.400

Step 2: Use the Hardy-Weinberg equation to predict the future population composition

The Hardy-Weinberg equation is p2+2pq+q2=1p2+2pq+q2=1, where q2q2 is the proportion of yellow 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 data

observed 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 percentages

Blue  —  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.
3.
4. What is DNA? Genetics Home Reference: Your Guide to Understanding Genetic Conditions from U.S. National Library of Medicine, National Institute of Health, Retrieved on 7/23/2019.
5. Klug, WS, Cummings, MR, Spencer, CA, and Palladino, MA, Concepts of Genetics, 10th edition (2012), page 702.