Download Key for Exam 2 Part 2 - Evolutionary Biology

EXAM II 2012
EVOLUTIONARY BIOLOGY – PART II
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Lab Day & Time
Negative Points
Positive Points
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B. Definition Questions. Define the following terms or phrase giving an example and answer
the follow up question. Do not be superficial in your responses; give details (4 pts. each)
1) Pleiotropy
=The determination of more than one character of an organism by a single gene.
An example would be Marfan’s syndrome where a person with this condition has several disabilities
such as wandering eyes, weak heart valves, elastic arteries, loose skeletal joints
How can the principle of pleiotropy be used to explain a trait that appears to be
disadvantageous to its owner?
Since a single gene controls many traits then it is possible that one trait is very advantageous but
another is disadvantageous and the advantages may outweigh the disadvantages so the allele that is
controlling the trait will be retained.
2) Codon
= three nucleotides in messenger RNA that determines which amino acid will be positioned
next in a polypeptide chain.
For example, three uracils in a row will code for the amino acid phenylalanine
Why are some mutations more deleterious than others?
If we consider just point mutations, then a simple substitution of one nucleotide for another would
normally just mean that a single amino acid might be changed; that might be a minor change or
perhaps no change in enzyme structure and function. But if the change were to code for a stop or
start codon, the entire gene would be disrupted. The same would be true if either an addition or
subtraction (deletion) mutation occurred; this would cause a frameshift in the entire mRNA code
and the entire “enzyme” structure would be altered and it would be non-functional. If this enzyme
were vital to the survival, then the effects would be drastic.
3) Sexual dimorphism
= This is the condition where the physical appearance of the two sexes is different from one another.
For example, male deer in many species have horns and the females don’t. The males of bird species
are frequently highly colored and the females are drab.
What is the selective advantage, if any, for this trait?
When sexual dimorphism is present it indicates that there are different selective pressures on the sexes.
Typically, it means that the males are contesting for females and therefore have displays or
structures (e.g. antlers, fangs, large size, etc.) that are used in battle to gain breeding rights. Or the
male attributes (e.g. color patterns) are used to attract females.
4) Lysogenic life cycle
= This is the life cycle of a virus where the viral nucleotide sequence becomes integrated into the host
DNA and remains there indefinitely until it suddenly becomes active and begins to reproduce.
Examples include HIV, chicken pox, and herpes.
Why is it so difficult to treat people that have diseases that go through this life cycle?
It is difficult to treat people with these viruses because the viral DNA is hiding in the host genome and
treatment can potentially harm the host.
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C. Short Answer Questions (3 pts each) State whether the statement is true or false and then
give the evidence supporting your claim
1) Reproductive isolation cannot be achieved without natural selection
False. Several other possibilities occur other than natural selection to bring about
reproductive isolation. These would be mutation, or genetic drift, or hybrid polyploidy. All of
these could alter the alleles in such a way that a population might no longer be able to
interbreed with another population of the original species.
2) Sexual selection involves male competition.
Maybe. It is true that one of the main ways that sexual selection can occur is if males
fight over the breeding rights for access to females. But the other major way that it can
occur is by female choice; that is, females choosing males that have characteristics
they favor.
3) Disruptive selection can lead to sympatric speciation.
True. Disruptive selection is the type of natural selection where there is a survival advantage
for individuals in the population that are at the extremes of a particular trait. I.e. the average
individuals are at a disadvantage. For example, birds that have either large or small beaks
might be able to survive in times of food shortage. If this continued for very long, and if their
were a tendency for birds with large beaks to breed with others with similar beaks and small
beak birds to select mates with similar sized beaks, then this could potentially lead to distinct
species.
4) Somatic mutations can never have an impact on evolution.
True. Somatic mutations are ones that occur in non reproductive cells such as the skin,
digestive tract or muscles. These mutations might be deleterious to their owners but cannot
influence evolution unless they occur to the cells that produce the eggs in the ovary or the
sperm in the testes. Only then can these mutations be passed on to the next generation and
influence evolution.
5) Panda bears are more likely to become extinct than grizzly bears
True. There are many reasons why species become extinct, but one of the major reasons is a
failure of the food supply. Panda bears are specialist feeders, eating only bamboo for their
food and that food supply is seriously threatened as deforestation continues in China.
But the grizzly bears are generalists in their food consumption and will eat a variety of animal
and plant food. And if one food source is threatened they can switch to an alternate food
source.
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D. Long Questions
1) William Paley made the argument that organisms look like they are designed, i.e. well adapted to
survive in their environment. He ascribed this to a higher power that produced a perfect organism.
What were at least 3 weaknesses of his argument? And how would evolutionists explain these 3
problems today? (6 points)
The arguments are (1) Organisms aren’t perfect. There are many poor designs and imperfect
structures (eyes, throat, prostate). Also, the fact that most organisms that have lived on earth are
extinct; they weren’t perfect in design, obviously.
(2) The presence of vestigial organs cannot be adequately explained by creationists. E.g. Why would a
perfect designer create eyes that cannot see in blind cave animals?
(3) There are embryological structures that are formed and then are modified into adult shapes that do
not make sense for a perfect creator. (e.g. whales have blowholes for breathing as adults but in the
embryo the nostrils are in the normal position. Humans have tails as embryos)
(4) There are philosophical difficulties since the intelligent designer argument relies on miracles.
Creationists are argue that any complex structure must be due to a miracle if scientists can not
currently explain its evolution. This isn’t science.
(5) The political/legal argument: intelligent design has been ruled as unconstitutional to teach in a
public school science classroom as it relies on the designer as a god and government cannot impose
religion.
Evolutionists state that organisms are designed by natural selection operating on the available
variations that are present in the population; the best phenotypes/genotypes generally survive. But they
are not perfect, only adequate. Vestigial organs are easily explained as the ancestors had these organs
and they were then functional; but as they are no longer used, the selective pressure has been reduced
and they are in the process of being lost. Similarly, the embryological anomalies are present because
the ancestral genes are still present. As for the philosophical difficulties, relying on miracles is not
science; and scientists continue to explore problems until they are solved by known physical and
chemical processes.
As for the legal argument, evolutionists agree with the courts: Intelligent Design doesn’t belong in the
science classroom.
2) The Death of Baby Pierre. On March 7, 1964, the baby known as Pierre was born in a remote part of Quebec Province
in Canada. He appeared to be a healthy six-pound twelve-ounce child, except he did not eat well. Over the weeks after
his birth, he became progressively more lethargic, vomiting periodically. Most peculiarly, his urine smelled of rotten
cabbage, and soon the smell permeated his clothes and body. By the time he was admitted to the hospital on September
14, his muscles were weak and his ribs were showing. Baby Pierre had only gained half a pound in the six months since
his birth. The doctors kept him alive by feeding him through a tube threaded through his nose and into his stomach. He
gained weight and strength for a while, and then suddenly took a turn for the worse. On November 30, baby Pierre
vomited blood and died.
It soon became increasingly apparent that other babies in the Chicoutimi area of Quebec Province had similar
symptoms, and people recalled similar deaths in this remote area 120 miles north of Quebec City. Some families lost
several children to Pierre's disease. In those families stricken, it soon became clear that the parents were normal, but
about one quarter of their children were afflicted. Boys and girls were equally afflicted. Specialists soon concluded that
all of the facts indicated that this was a genetic disorder.
Baby Pierre and the other stricken children were victims of hereditary tyrosinemia. This is caused by an autosomal
recessive disease. The children lacked the normal gene which produces a liver enzyme that breaks down the amino acid
tyrosine. Without the enzyme, tyrosine builds up in the liver and kidneys leading to the cabbage-like smell of the urine.
Lethal side-effects follow. (A liver transplant is the only long-term treatment of the disease as of 1997)
Both parents avoid this fate because, although they carry one copy of the defective gene, they carry a normal gene
which produces more than enough enzymes for normal liver function. The parents are unwitting carriers of the disease;
they are heterozygotes, while Baby Pierre was a homozygous recessive.
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Below is a pedigree of three generations of Canadians. (Males are represented by squares and
females by circles) Two parents A and B gave birth to 4 children, D, E, G, and I. The latter
died of the tyrosinemia. The others married and the fates of their offspring are show below
1. Using the symbols of "a" for recessive allele and "A" for the dominant normal allele write
the genotype for the following individuals. (4pts)
A ____Aa_____________
E. _AA (or Aa since F probably Aa
I _______aa___________
R_Aa (67% chance) vs. AA (33%chance)
2. Is the above pedigree consistent with a sex-linked trait? (1 pt.)
No, since males and females are equally affected in this case. In the typical sex-linked pattern
such as color blindness or muscular dystrophy, males are normally the only sex afflicted. These
traits are carried on the X chromosome. Since males have only one X chromosome, if the trait is
present, they will surely show it. Females, on the other hand, will usually not display the trait even
if they are carriers. They have two X chromosomes and the normal X will compensate for the
“defective” X. If the trait shows up in a female (a very rare occurrence), both X chromosomes will
have the trait.
3. What is the likelihood that female K will have a normal child if she marries a normal
person who is a carrier for tyrosinemia? (1 pt.)
If K is AA, then all children will be normal (50% will be AA and 50% will be Aa)
If K is Aa, then Aa x Aa gives a probability of 3 normal children per 1 diseased child or 75%
normal
4. What is the likelihood of female K having a normal child if she marries her cousin M?
(1pt.)
Cousin M could be either AA or Aa just like K could be AA or Aa.
With combinations:
AA x AA} all normal children will result
AA x Aa} all normal children will result
Aa x AA} all normal children will result
Aa x Aa} 75% normal
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Hereditary tyrosinemia is usually quite rare, affecting only 1 in 100,000 newborns. The
situation in the French Canadians in Chicoutimi, Quebec, is dramatically different; 1 birth in
685 can be expected to produce a child with the disorder.
5.
From the answers below, what is the most likely reason for the large number of
tyrosinemia cases in Chicoutimi? Explain the reason for your answer. (1 pt.)
(A) A high mutation rate.
(B) A selective advantage for tyrosinemia in this part of Quebec.
(C) Founder effect.
(D) Nutritional patterns in the people; large quantities of tyrosine in the diet.
(E) Pleiotropy where the allele for tyrosinemia has beneficial effects as well as
harmful effects.
Reason for your choice:
There is no reason to expect a high mutation rate (answer A) or selective advantage (answer
B) for tyrosinemia in Quebec. Since it is a genetic disease, answer D is unlikely. Answer E is
conceivable but why should it show up only in Chicoutimi? (C) is the most likely answer since
it is reasonable to predict Chicoutimi is a small community. (In fact, the Canadian geneticist
Dr. Claude Leberge found the Chicoutimi region was settled by a few dozen families who
migrated north from Quebec's Charleuoix County. Most people living there are descendants of
the original settlers. The tyrosinemia victims can be traced back to one couple, Louis and
Marie Gagne, who emigrated from France. It seems likely that either Louis or Marie had the
gene for tyrosinemia and passed it to some of their nine children, and innumerable
grandchildren, at least two of whom moved to Charleuoix. Their progeny later moved to
Chicoutimi. The significant inbreeding produced a high proportion of people who carried the
tyrosinemia gene. This is a classic example of the founder effect.)
6. The Hardy-Weinberg equation allows us to make some calculations about the local population. (5 pts.)
(A) What is the frequency of "aa" individuals?
Given 1/685 children have the disease and are aa, this means q2 = .00146 \
(B) What is the frequency of the "a" allele in the gene pool? q=.0382
(C) What is the frequency of the "A" allele in the gene pool?Since p + q = 1 and q = .0382, then 1 - .0382
= p = 0.9618
(D) What is the frequency of heterozygous individuals? 2 pq = .073 or 1 person in 14 are Aa individuals.
The frequency of aa (homozygous recessive) q2 = .00146
The frequency of AA (homozygous dominants) p 2 = 0.925
(E) Assuming random breeding, what are the chances that two carriers will marry?
.073 x .073 = .0053 or 1 marriage in 190 will involve two carriers
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