Weeks 4-6 Practice Quiz Flashcards
Why can’t we measure genetic variation in a population using observable traits (phenotypes)? (Select all that apply.)
Many traits are encoded by multiple genes.
Phenotypes are not determined by genes.
The environment can also affect the phenotype.
All traits are encoded by a single gene.
Many traits are encoded by multiple genes.
The environment can also affect the phenotype.
Why:
Phenotypes are physical characteristics and genotypes are genetic makeup but phenotypes can be changed/ altered by genes. Traits can be coded by more than one gene and the genotype as well as the environment can affect the phenotype
For reciprocal altruism to work, individuals must be able to:
recognize one another and remember previous interactions
.
calculate the fitness value of individual interactions.
recognize one another.
remember previous interactions.
recognize one another, remember previous interactions, and calculate the fitness value of individual interactions.
recognize one another and remember previous interactions
Why:
Reciprocal altruism is similar to doing a favor for a neighbor and them now owing you a favor. So in order to collect on that favor you need to remember which neighbor it was and what you did something for them.
In a population of Mendel’s garden peas, the frequency of green-flowered plants (genotype aa) is 49%. The population is in Hardy-Weinberg equilibrium. What are the frequencies of the AA and Aa genotypes?
33% AA, 18% Aa
42% AA, 9% Aa
9% AA, 42% Aa
The frequencies cannot be determined from the information provided.
49% AA, 2% Aa
9% AA, 42% Aa
Why:
If aa is 49%, then so is q^2. In order to find p^2 you can square root the q^2 value and substract that from one to deteremine what p is because p+q=1. Once you find p you square it to get the p^2 value. Now to find what 2pq is for the heterozygous allele, you can plug in the values you found for p and q in and mulitply 2pq.
Imagine the following genotype frequencies in a population: p2 = 0.49, 2pq = 0.42, q2 = 0.09. Now assume that there is NON-RANDOM MATING where individuals with one genotype will only mate with individuals that also have their genotype. Assume this pattern of mating goes on until the frequency of heterozygotes is effectively zero. In addition, there is also inbreeding depression such that individuals with the genotype represented by p2 die before they can reproduce. What will be the frequency of allele q?
1.0
- 3
- 91
- 7
- 49
1.0
Why:
If there is non random mating then hardy-weinberg equilibrium is no longer applicable. Due to non random mating the heterozygous allele becomes zero due to no mating with different alleles. This cannot continue the heterozygous frequency. So now the homozygous dominant AA allele represented as p^2 has been affected by inbreeding depression which is when that allele has been getting mutaions that decrease the fitness of the allele type. That leaves the q frequency as the only allele left to reproduce so they become one since they only mate among themselves.
The accumulation of different mutations in genetically separated populations is known as _____ and is the key to speciation.
reinforcement
genetic divergence
reproductive fitness
intersexual selection
genetic divergence
Why:
Mutations add to genetic divergence and help diversify individuals in the same species/population. After some time if the diversification becomes high enough then new species can form.
Male birds of different species sing species-specific songs to attract mates. Females only mate with males that sing their species-specific song. This is an example of:
temporal isolation.
reinforcement.
pre-zygotic isolation.
post-zygotic isolation.
pre-zygotic isolation.
Why:
Pre-zygotic isolation is behavior that prevents fertilization or mating. So the birds only mating with the gender of the same song as them is isolating them from mating with other birds that have different songs..
New species can form through allopatric or sympatric speciation. Which of the following mechanisms will act more strongly on populations that are initially separated in allopatry than on those initially separated in sympatry?
gene flow reinforcement genetic drift disruptive selection
genetic drift
Why:
Genetic drift is when there are changes in the allele frequency because of a small population size and if they are separate species geographically then it is easy for a mutation or unfavorable trait to spread to the whole population if there is a small amount.
The graph in Figure 22.9, below, shows that the extent of the adaptive radiation of the Galápagos finches is correlated with the number of islands present in the archipelago. What can be concluded from the graph?
Picture: As the number of islands increases so does the number of finch species over the time frame of 5 million years. Once the islands stop increasing the finch soon followed and stopped increasing. The finch pop was always slightly below the number of islands.
This shows that each island hosts sympatrically speciating populations. This shows that allopatric speciation tends not to occur on islands. This shows that opportunities for geographic isolation are a key component of the speciation process. This shows that rates of speciation are correlated with the net total land area available. This shows that the balance between rates of sympatric and allopatric speciation is governed by mutation rates.
This shows that opportunities for geographic isolation are a key component of the speciation process.
Why:
Due to the number of new islands causing more diversification which in turn creates new species of finches. This proves speciation is thriving because of the barriers of being on islands that are separated allow the species to change.
We expect the fossil record to be incomplete for all of the following reasons EXCEPT:
Sediments accumulate irregularly. Organisms with hard parts tend to fossilize more readily than soft-bodied organisms. Fossilization requires burial in sediment. Under normal conditions, organisms are broken down by biological and physical processes. All biomolecules, including lipids and pigment molecules, decay immediately after death.
All biomolecules, including lipids and pigment molecules, decay immediately after death.
Why:
All biomolecules begin to decay after death. But they do not fully decay right after, it is a process.
A population of rodents, called population A, lived together in harmony on a large landmass until one group of the population dispersed to a nearby island. Two million years later, the island population is split into two smaller, equal-sized populations when a river formed across the middle of the island. Now two new species have evolved on the island, A1 and A2. They have replaced the population from which they were derived. Which represents the phylogeny of the populations discussed in this scenario?
diagram K diagram M diagram H
diagram k
Why:
This phylogeny reflects that species A is now an outgroup because when there was a split into another species. That species split on its own again into two separate species and took that one species place.So those two species A1 and A2 are sister groups.
Excluding the outgroup, taxon A, how many equivalent sister-group relationships are depicted between the two phylogenies given below?
four three one There are no equivalent sister-taxa shown between the two phylogenies. two
There are no equivalent sister-taxa shown between the two phylogenies.
Explain: Two taxa being in the same branching pattern are not necessarily related (sister groups). Also the two phylogenies show sister taxa but are not equivalent to one another because they are different species.
The concordance of the two great patterns in the history of life—the branching order of the tree of life and the sequence of forms in the fossil record—is powerful evidence in support of the theory of evolution. Which of the following imaginary examples of evidence would DISPROVE the theory of evolution?
Lack of evidence of a transitional form between fish and tetrapods. A fossil of a dinosaur footprint in 70-million-year-old rock. A fossil of a mammal that is older than fossils of the first reptiles. A fossil of a DNA molecule that is older than a fossil of the first fish.
A fossil of a mammal that is older than fossils of the first reptiles.
Why:
This disproves the theory of evolution because ?????
What does the term 2pq represent in the Hardy-Weinberg relation?
the frequency of deleterious mutations the frequency of homozygous dominant individuals the frequency of homozygous recessive individuals the frequency of heterozygotes
the frequency of heterozygotes
Imagine the following genotype frequencies in a population: p2 = 0.49, 2pq = 0.42, q2 = 0.09. Now assume that there is nonrandom mating where individuals with one genotype will only mate with individuals that also have their genotype. Assume this pattern of mating goes on until the frequency of heterozygotes is effectively zero. What will the frequency of allele p be in the population?
0. 7 0. 3 0. 91 0. 49 1. 0
0.7
Why:
Allele frequencies do not change with inbreeding. This requires that they calculate the frequency of p from the genotype frequencies. Many students harbor the misconception that q2 individuals always decrease in a population—something common where students often think the allele in lower frequency is recessive and therefore “bad.” 0.91 may also be a common answer because students will assume that all individuals will become p2 (usually associated with dominant, and the extreme form of the misconception would be that the p allele will go to fixation.
If a population is not in Hardy-Weinberg equilibrium, we can conclude that:
one of the assumptions of the Hardy-Weinberg equilibrium has been violated. evolution has occurred because one or more of the assumptions of the Hardy-Weinberg equilibrium has been violated. All of these choices are correct. nonrandom mating has occurred. natural selection has occurred.
evolution has occurred because one or more of the assumptions of the Hardy-Weinberg equilibrium has been violated.
