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Exercise 11 - Genetics
Introduction
Many features or characteristics of plants and animals are acquired during its lifetime. For example, a
person may get a tattoo and although the mark is, for all intents and purposes, permanent on their skin
their children will not be born with tattoos. Other characteristics, however, are inheritable and can be
passed from one generation to the next. These inheritable characteristics are encoded in DNA in units
called genes.
Genes may exist as alternative forms known as alleles. An allele of a gene is nothing more than a gene
choice (blue or brown for eye color, unattached or attached for earlobes, widow’s peak or no widow’s
peak, etc.). Some alleles are dominant because, when present, they control the appearance or
phenotype of the organism. Other alleles are recessive and only phenotypically expressed when no
dominant allele for that gene is present.
Diploid organisms receive one set of chromosomes each from the maternal (female) and the paternal
(father) parent. If the alleles for a given gene inherited from the parents are the same, the organism is
considered homozygous for that trait and heterozygous if the alleles are different. The actual
combination of alleles inherited by the offspring for a trait is known as its genotype.
This laboratory exercise will investigate how dominant and recessive alleles affect an organism and
demonstrate how these effects occur in a predictable pattern.
Materials
Equipment
ears of purple : yellow genetic corn
straight pins
PTC taste paper
color vision diagnostic charts
Part A: Monohybrid Cross in Corn (Zea mays)
Corn kernels are actually the fruit of the corn plant, each containing an individual corn embryo. Around
the embryo are a number of structures providing nourishment and protection. Between the endosperm
(where starch is stored) and the pericarp (covering of the kernel) is a layer of cells called the aleurone.
The color of the aleurone is controlled by several genes. One gene produces a purple aleurone, the
other the yellow kernels with which we are most familiar. The purple allele is dominant, the yellow
allele being recessive. The dominant purple allele can be symbolized by “R” while “r” represents the
recessive yellow allele. Therefore, corn kernels with the homozygous genotype “RR” are purple as are
those kernels who are heterozygous (“Rr”). Only kernels homozygous for the yellow allele (“rr”) will
express the recessive phenotype, yellow.
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Exercise 11 – Genetics
Procedure
1. Obtain an ear of genetic corn. Notice it contains a mixture of purple and yellow kernels. This
ear of corn represents the F2 generation from the following cross:
Rr
x
Rr
RR, Rr, rr
The parents (“Rr” and “Rr”) are the F1 generation and were obtained from the P generation
cross of a homozygous dominant (“RR”) and recessive (“rr”) individual. The offspring (“RR, ‘Rr”,
“rr”) are then the F2 generation. This cross results in a mixture of phenotypes in the F2
generation. Most of the kernels are purple, a fewer number are yellow. Punnett square
diagrams are used to diagram genetic crosses. Using this mating, fill in the Punnett square in
Table 11.1
Table 11.1 Monohybrid Cross Punnett Square for Kernel Color in Corn (Zea mays)
F1 gametes
R
r
genotype __________
genotype __________
phenotype __________
phenotype __________
genotype __________
genotype __________
phenotype __________
phenotype __________
R
r
Based upon the Punnett square in Table 11.1, about what proportion of the kernels in the ear of
corns should be purple? _____
What proportion should be yellow? _____
2. Place a straight pin at the end of the row being counted to serve as a place holder. Count the
number of purple kernels and yellow kernels in one row only. Enter data in Table 11.2
Table 11.2 Kernel Counts in F2 Generation Corn (Zea mays) – One Row
# of kernels
Number
Proportion*
purple
yellow
total
*divide the number of kernels of each color by the total number of kernels in that row
Is the proportion of purple to yellow kernels close to the predicted 3:1 (75% purple, 25% yellow)
ratio? __________
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Exercise 11 – Genetics
While the numbers might be close (and even if they are not) to the expected phenotypic ratio of
3:1, statistical theory says the bigger the sample size the closer our actual data should fit
theoretical predictions.
3. Increase the sample size for this dataset by counting all rows in the ear of corn and fill in Table
11.3
Table 11.3 Kernel Counts in F2 Generation Corn (Zea mays) – All Rows
# of kernels
Number
Proportion*
purple
yellow
total
*divide the number of kernels of each color by the total number of kernels in that row
Did counting more kernels make the numbers come more closely to the expected 3:1 ratio of
purple to yellow color? __________
Part B: Phenylthiocarbamide (PTC) Taste Test
Phenylthiocarbamide is an organic compound which interacts with the taste buds on the tongue and
mouth of some people but not in others. Therefore, some people can taste PTC and are considered
“tasters.” Other people cannot taste PTC and are considered “non-tasters.” Being able to taste is the
dominant condition (“T”), non-tasters the recessive (“t”).
Procedure
1. Obtain a piece of paper treated with a small amount of PTC and place it on the tongue. Tasters
will sense a very strong flavor, non-tasters will taste only “paper”
2. Collect results on the number of tasters and non-tasters from the population in class and use
that data to fill in Table 11.4
Table 11.4 PTC Tasting
Phenotypes
Number
Proportion*
taster
non-taster
total
*divide the number of phenotypes by the total class size
What is the genotype of a taster? __________
What is the genotype of a non-taster? __________
What is the predicted phenotypic ratio for a cross between two heterozygous (Tt) individuals?
__________
Were the resultant proportions close to this ratio? __________
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Exercise 11 – Genetics
If the proportions were not close (and even if they were), provide an explanation as to why they
would be (hint: sample size).
Part C: Multiple Alleles – Human Blood Types
Some genes have more than two alleles. Although each organism can only have two copies of a gene
(why? _________________), within an entire population there may be several alleles. One example of
this is seen with the ABO blood types in human beings. Blood types are produced by multiple alleles.
On the membranes of red blood cells are proteins which can stimulate an immune response. These
proteins are called antigens. The dominant allele “IA” produces the A antigen protein, the dominant
allele “IB” produces the B antigen protein. The recessive allele “i” produces no protein. When a genetic
trait has more than one dominant allele, a situation of codominance may exist. The genetic trait for
human ABO blood typing contains three alleles (IA, IB, i).
Procedure
1. Fill in Table 11.5 with the possible genotypes for the listed phenotypes
Table 11.5 Human ABO Blood Typing Genotypes and Phenotypes
Genotype(s)
_____ _____ or _____ _____
Phenotype
A
_____ _____ or _____ _____
B
_____ _____
AB
_____ _____
O
Part D: X-Linked Characteristics
In humans, all somatic cells (typical body cells) contain 23 pairs of chromosomes. Of these, 22 pairs are
autosomes, the last pair are the sex chromosomes. The sex chromosomes are related to the gender of
the individual and are called X and Y. Women have two X chromosomes, men have an X and a Y.
Haploid sex cells (gametes) produced by women (ova) have only the X chromosome, male gametes
(sperm) have either X or Y. Sex of the offspring is determined by the male.
The X chromosome is large and carries many genes such as those for essential muscle proteins and
retinal pigments. The Y chromosome, on the other hand, is quite small and carries only a few genes,
mostly related to male gender development. A defective gene on the X chromosome will be
phenotypically expressed in a male because there is no other X chromosome to compensate. However,
a woman with the same defective gene will not express it phenotypically if her other X chromosome is
normal. She will be a carrier though in that she has the defective gene but does not express it.
Red-green colorblindness is caused by a mutation in a gene for retinal pigments on the X chromosome.
The defective allele is recessive to the normal one so a woman with one normal X chromosome (X) and
one colorblind carrying chromosome (XC) will have normal color vision because the X chromosome is
dominant for this trait over the XC chromosome. Men, however will be colorblind if they possess the XC
chromosome since there is no other X to be dominate over it and will exhibit red-green colorblindness.
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Exercise 11 – Genetics
Procedure
1. Obtain a color vision diagnostic chart
2. Test for colorblindness in yourself and among lab partners
What is your gender? __________ (if you don’t know, excuse yourself to the restroom to find
out)
What is your sex chromosome genotype? __________
Do you exhibit red-green colorblindness? __________
Do any lab partners exhibit colorblindness? __________
If so, what are their sexes and genotypes? ____________________
Part E: Dihybrid Crosses
Dihybrid crosses are used to examine the inheritance patterns in more than one gene (di = two). As an
example, consider fur color and texture in guinea pigs. Among these organisms, black fur color is
dominant (“B”) over white (“b”) and rough fur coat is dominant (“R”) over smooth (“r”).
Procedure
Use a Punnett Square to answer the questions regarding the offspring in the F1 generation from the
P generation crosses indicated
1.
BBRR
x
bbrr
What will the genotype be of all of the offspring from this cross?
What will be the phenotypes of all of the offspring from this cross?
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Exercise 11 – Genetics
2.
Bbrr
x
bbRr
What proportion of the guinea pigs will be
3.
black, rough? __________
black, smooth? __________
white, rough? __________
white, smooth? __________
BbRr
x
bbRr
What proportion of the guinea pigs will be
black, rough? __________
black, smooth? __________
white, rough? __________
white, smooth? __________
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Exercise 11 – Genetics
4.
BbRr
x
BbRr
What proportion of the guinea pigs will be
black, rough? __________
black, smooth? __________
white, rough? __________
white, smooth? __________
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Exercise 11 – Genetics
Practice Problems and Review Questions
1. Define or describe the following:
monohybrid cross
dihybrid cross
testcross
allele
dominant allele
recessive allele
genotype
phenotype
homozygous
heterozygous
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Exercise 11 – Genetics
2. What are the expected phenotypic and genotypic ratios in a monohybrid cross between two
heterozygous individuals?
3. What is the expected phenotypic ratio in a dihybrid cross between two organisms that are
heterozygous for both traits?
4. List all the possible types of gametes that can be produced from an organism with the genotype
AaBb.
5. Humans as well as many other mammals show a genetic condition known as albinism. A
recessive allele interferes with the ability to produce the brown pigment melanin which colors
eyes and hair in addition to protecting the skin from the harmful effects of UV light. People who
are homozygous recessive produce little or no melanin and have very pale eyes, white skin, and
yellow or white hair. Normal pigmentation is produced by a dominant allele (“A”). The albino
allele is recessive (“a”). Use Punnett Squares to calculate genotypic and phenotypic proportions
for the parental generation crosses indicated and denote which F1 genotypes (if any) will be
carriers.
a. AA x aa
b.
Aa x aa
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Exercise 11 – Genetics
c.
Aa x Aa
d.
AA x Aa
e.
aa x aa
6. A woman with normal pigmentation, whose mother was an albino, mates with an albino. What
is the probability their first child will be an albino?
7. Normal parents have an albino child. Give the genotypes for the parents and the child.
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Exercise 11 – Genetics
8. A person is a taster as is their mother. The father is a non-taster. What is this person’s
genotype?
9. A non-taster has two taster parents. What is the genotype of this person and their parents?
10. For ABO blood typing in humans, for which phenotypes is the genotype definitive?
11. A person has blood type O. Their mother is B and father is A. What is the genotype of this
person and their parents?
12. A paternity suit involves a child whose blood type is AB. The mother is blood type B, the alleged
father is O. Make a ruling on this case as to whether it is reasonably possible this is the
biological father.
13. Is it possible for a female human to be colorblind? What would her genotype be?
14. Can two normal vision parents produce a colorblind son? Explain with a diagram.
15. Can two normal vision parents produce a colorblind daughter? Explain with a diagram.
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Exercise 11 – Genetics
16. Can two colorblind parents, produce a normal vision son? Explain with a diagram.
17. A normal vision woman, whose father was colorblind, mates with a colorblind man. What
proportion of their sons would be colorblind? What proportion of their daughters would be
colorblind and what proportion would be carriers? If one of their normal vision sons mates with
a homozygous, normal vision woman, would it be possible for them to have a colorblind child?
18. A guinea pig that is heterozygous for fur color and texture mates with another guinea pig
heterozygous for both traits. They produce a total of 96 offspring. Use a Punnett Square to
diagram this cross then answer the questions below.
How many of the 96 offspring will phenotypically be
black with rough fur? __________
black with smooth fur? __________
white with rough fur? __________
white with smooth fur? __________
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Answers
Answers to Lab Exercise Practice Problems and Review Questions
Exercise 1
#1
Surface area: 124.9 cm2
Volume: 78.4 cm3
Density: 1.5 g/cm3
Sinks
#2
Surface area: 31.2 cm2
Volume: 9.8 cm3
Density: 0.9 g/cm3
Floats
#3
0.26 L = 260 ml
Volume = 252 ml (cm3)
Density = 0.2 g/cm3
Floats
#4
Squirrels are homeothermic (warm-blooded) mammals. In northern latitudes, one of the more obvious
environmental pressures is the cold during the winter months. For southern latitudes, the opposite is
true. Staying cool in the hot summer months is most important there. So, squirrels in the north should
evolve adaptations that would provide for the retention of heat while their southern counterparts
would find ways to stay cool in the summer. Recall the change in surface area to volume ratio observed
in the block of various sizes. The highest ratio was in the smallest block. This means the smallest block
had more surface area exposed to the environment per unit of volume. When applied to animals this
means the smaller ones expose more of their skin (surface area) to the environment (hot, cold, wet,
etc.) per unit of body mass than larger animals. In homeotherms like mammals, this means smaller
animals lose heat from their body through their skin faster than larger ones. This would be beneficial in
the southern squirrels while being able to hold onto body heat longer would allow northern versions to
survive the cold winters.
Exercise 2
#1
a. Stock: 0.5 ml
b. Stock: 5.0 ml
c. Stock: 2.5 ml
DH2O: 9.5 ml
DH2O: 5.0 ml
DH2O: 7.5 ml
a. Stock: 8.0 ml
b. Stock: 2.0 ml
c. Stock: 12.0 ml
DH2O: 12.0 ml
DH2O: 18.0 ml
DH2O: 8.0 ml
#2
#3
Stock: 04.5 ml
DH2O: 4.55 ml
#4
Stock: 2.8 ml
DH2O: 2.2 ml
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#5
Stock: 0.04 ml
DH2O: 4.96 ml
#6
Stock: 18.75 ml
DH2O: 6.25 ml
#7
Stock: 7500 ml
DH2O: 1500 ml
#8
Stock: 0.24 L
DH2O: 0.36 L
#9
50 ml
#10
Stock: 10 ml
DH2O: 40 ml
#11
Buffers are substances that absorb H+ when they are in excess which happens when the environment is
too acidic and release H+ when the environment is too basic. They work to stabilize pH which is crucial
for living organisms in order to maintain homeostasis. For example, excessively acidic or basic
conditions in the body may cause enzyme denaturation and a resultant loss of physiological function.
#12
Glucose – aldehyde and hydroxyl (linear), hydroxyl (ring)
Fructose – ketone and hydroxyl (linear), hydroxyl (ring)
Glycine – amine, carboxylic acid
Glycerol - hydroxyl
#13 (examples, not necessarily just these)
Glucose – starch, maltose, glycogen
Fructose – sucrose
Glycine – proteins
Glycerol – triglycerides, phospholipids
Exercise 3
#1
In this lab, qualitative tests are used to simply detect the absence or presence of a substance. There is
no particular indication of amount just a "yes" or "no" to whether the substance exists or not. For
example, a positive test for starch using IKI only tells you starch is present, but not how much starch. A
quantitative test determines the actual amount of something present. If wanted to know how much
starch was present in a solution we would use a test that would return a value (for example, 3 mg of
starch per liter).
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#2 (a & b)
Since it can reasonably be assumed that distilled water does not contain lipids or proteins it makes a
perfect control. Notice it is distilled water and not tap water. Tap water contains many other things and
would not make a good control. Distilled water is just water, nothing else.
#3
Test Substance
Reagent
Test Procedure
Color of Positive
Result
Color of Negative
Result
Starch
IKI
1 drop
dark blue / black
Amber
Sugar
Benedict’s
1 ml
red, orange,
green, yellow
Blue
Lipid
Sudan IV
Soak 3 mins
pink / red
no color or very
light pink
violet
no color change
1 ml NaOH,
Protein
Biuret
0.5 ml reagent
#4
The IKI test must be performed first. If a Benedict's is done first, the heat will hydrolyze any starch
present and it will go undetected.
#5
They differ in number of monomers. Monosaccharides, like glucose, consist of one monomer. Two
saccharide monomers (for example sucrose which is made up of a glucose and a fructose
monosaccharide), would be a disaccharide. Oligosaccharides are typically three to ten monosaccharides
long. Polysaccharides are typically >10 monosaccharides (for example, starch and glycogen).
#6
Glycerol and palmitic (fatty acids)
#7
Amino acids
#8
Primary - order of the amino acids in the protein. Secondary - a protein's "orientation in space"; evident
as either an α helix (spiral) or ß sheet. Tertiary - protein folding back on itself due to chemical
interactions among the amino acid side (or "R") groups. Quaternary - occurs when two or more
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Answers
polypeptide chains associate (for example, hemoglobin which is made up of four polypeptide chains
bonded together). Proteins also exhibit further complexities of structure and form motifs (for example,
ß-α-ß motif) and domains which are functional units of a protein.
#9
They differ in their particular arrangement of their amino acids. This, of course, affects all other levels
(secondary, tertiary, quaternary, motifs, domains). Organisms ingest proteins, hydrolyze them into
amino acids and then re-assemble those amino acids into organism specific proteins.
#10
DNA, RNA, ATP all contain nitrogen
#11
The word "hydrolysis" literally means "water" ("hydro"), "splitting" ("lysis"). The opposite of this is when
organic molecules are combined to form polymers. This process is called dehydration synthesis because,
in this process, water is lost.
Exercise 4
#1
900x
#2
100x
#3
0.2 mm
#4
750 μ
#5
Length: 125 μ
Width: 37.5 μ
#6
6 mm
#7
Length: 125 μ
Width: 50 μ
#8
250x
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Answers
Exercise 5
#1
33.0%
2
4.7%
#3
-11.1%
#4
-17.0%
#5
Since the bag lost weight, water must have flowed from inside of the bag to the beaker. Water always
flows to the side of the membrane with the greatest solute concentration, which means the solution in
the beaker was hypertonic to the solution inside the bag.
#6
Prokaryotic cells are members of the Domains Archaea and Bacteria do not have a nucleus, are much
less complex than eukaryotic cells, and possess only a single chromosome. Eukaryotic cells (all other
kingdoms) have a nucleus, exhibit much greater complexity of structure and have multiple
chromosomes.
Plant cells have a cell wall, chloroplasts, and typically a large central vacuole typically used to store
water. They also do not have centrioles. Animal cells have no cell wall or chloroplast and, if they have
it, their vacuoles are much smaller. Animal cells have centrioles.
Protists differ from animal cells in that they often possess a cilia or flagella (although some animal cells
have a flagella - sperm), and are the complete organism (that is, protozoans are single celled creatures).
#7
A semi-permeable membrane is one that allows only certain substances to pass through it. Cell
membranes are semi-permeable as they regulate what passes through them letting some things through
while excluding others. The dialysis bag was an example of a semi-permeable membrane because it only
allowed items to pass that could fit through the pores in the tubing.
#8
Hypertonic: environments that contain a higher concentration of solutes when compared to the
environment on the other side of the membrane. Water always flows to the side with the greater
concentration of solutes; therefore cells in hypertonic environments would lose water.
Hypotonic: environments with a lesser concentration of solutes when compared to the environment on
the other side of the membrane. In this instance, cells in these environments would gain water.
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Isotonic: condition of equal solute concentration across the membrane and although water flows back
and forth across the membrane it does so equally. Cells in isotonic conditions exhibit neither a net loss
nor gain of water.
#9
When two aqueous solutions are separated by a semi-permeable membrane, the net water movement
is always from a hypotonic to a hypertonic solution.
Exercise 6
#1
An organic catalyst is an organic molecule (compounds that contain carbon) that “spurs on” or helps to
promote a chemical reaction. Enzymes are examples of organic catalysts. Enzymes enter reactions the
same way they them which is beneficial because enzymes are metabolically "expensive" molecules and
nature would favor them being used over and over again. Of course, like any system they eventually
suffer from entropy and become useless.
#2
Enzymatic reactions are affected by
(1) substrate concentration - increased substrate concentration give the enzyme more to work
on so reaction rate increases
(2) temperature - temperature increases reaction rate until the enzyme has been denatured
(3) pH – an environmental pH different from an enzyme's optimum will cause denaturation and
a subsequent decrease in reaction rate
#3
Low temperatures exhibit low enzymatic activity levels because the molecules (enzyme and substrate)
are moving slowly which decreases the probability they will contact one another. Higher temperatures
and a pH that is substantially different from the enzyme's optimum will cause enzyme denaturation.
Denaturation affects the shape of the enzyme and its active site. A change in the shape of an enzyme's
active site will not allow a substrate to bind and if no substrate can bind to the enzyme then no product
can be produced.
#4
The enzymatic reaction in this lab degrades a more complex molecule (H2O2) into more simple ones (H2O
and O2). This is an example of a catabolic reaction. Catabolic reactions are exergonic and release energy.
#5
E + S Æ ES complex Æ E + P
#6
The rate of enzymatic reaction is directly proportional to substrate concentration.
At optimum, enzymatic reaction rate is greatest.
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Exercise 7
#1
C6H12O6 + O2 Æ CO2 + H2O + ATP
#2
Aerobic - aerobic respiration occurs when the final electron acceptor is oxygen. Aerobic respiration is
respiration with oxygen.
Anaerobic - anaerobic respiration is when the final electron acceptor is something other than oxygen.
When this final electron acceptor is an organic molecule (e.g. acetaldehyde) it is a type of anaerobic
respiration known as fermentation.
NADH, FADH2 - nicotinamide adenine dinucleiotide (NADH) and flavin adenine dinucleotide (FADH2) are
electron carriers. Through a series of oxidation and reduction reactions, these molecule shuttle
electrons throughout the respiration processes.
ATP – adenosine triphosphate is the energy currency created in respiration and used by cells for cellular
work. It is created by using the energy released in respiration to add an inorganic phosphate (Pi) to
adenosine diphosphate (ADP).
#3
Glycolysis (total): 4
Glycolysis (net): 2
Krebs cycle: 2
Aerobic respiration (eukaryotes): 36
Anaerobic respiration (net): 2
#4
Alcoholic beverages are produced by yeast cells undergoing respiration in the absence of oxygen. This
type of anaerobic respiration utilizes acetaldehyde as a final electron acceptor (fermentation). During
this process CO2 is also created. It is the accumulation of this carbon dioxide gas that gives these wines
their "sparkle" or fizziness.
#5
There is a direct relationship between the amount of heat in an environment and the speed of chemical
reactions. As temperature increases, the rate of reaction increases. Homeothermic animals are "warm
blooded" and keep their body at a constant temperature regardless of environmental temperatures.
This constant warmth allows chemical reactions to occur at elevated rates in spite of the temperature in
the environment. On the other hand, the metabolic rate of “cold blooded” (poikilothermic) creatures is
dependent upon the temperature of the environment. Since they are unable to generate body heat in
colder temperature their metabolic rates slow down. Often times, poikilothermic animals use
"behavioral thermoregulation" to increase their body temperature and therefore the rate of their
physiological reactions by laying on warm surfaces (e.g. rocks, concrete, etc.). This also explains why
there are no amphibians in the Arctic or Antarctic.
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#6
Smaller mammals lose heat faster than larger mammals due to their increased surface area to volume
ratio. Just like in cells, as homeothermic organisms get smaller, their amount of surface area (essentially
their skin) increases in relation to their volume. Since organisms lose heat to the surrounding
environment through their skin, smaller "warm blooded" animals lose heat faster than larger ones. To
compensate, they have evolved increased metabolic rates.
Exercise 8
#1
CO2 + H20 Æ C6H12O6 + O2 (this reaction requires an input of energy from sunlight)
#2
Light reactions - those occurring on the thylakoid membrane. These reactions strip the electrons and
protons (H+) from water, energize the electrons using sunlight and then use those energy rich electrons
to activate a proton pump in the cytochrome complex. The pumping of protons across the membrane
sets up an unbalance. To reach equilibrium, protons flow back through the membrane passing through
an ATP synthase protein generating ATP to be used in the dark reactions. The electrons originally
stripped from water are then transferred to NADP+ reducing it to NADPH. Like ATP, NADPH is then
transferred to the dark reactions.
Dark reactions - occur in the stroma of chloroplasts. In this reaction, atmospheric CO2 is reduced
through a series of reactions creating glucose. The dark reactions require the ATP as the source of
energy and electrons to complete this task.
NADPH – the electron carrier responsible for shuttling electrons from the light to the dark reactions to
be used in the reduction of CO2 to glucose.
Rubisco - (ribulose bisphosphate carboxylase/oxygenase), the enzyme that catalyzes the reaction which
uses carbon dioxide to form 3-phosphoglycerate (PGA) thereby initializing the Calvin cycle.
Absorption spectrum - describes the range and efficiency of light absorbed by a particular molecule. The
molecule chlorophyll's absorption spectrum shows two peaks. One occurring at the shorter
wavelengths, the other at longer wavelengths. Minimal absorption occurs among the wavelengths in
between (the greens and yellows).
#3
The dark reaction (in order) consist of
(1) carbon fixation - the enzyme rubisco uses RuBP and atmospheric CO2 to form PGA. It is
called "fixation" because it is in this step where CO2 is "captured" or "fixed."
(2) reduction - energy in the form of ATP and electrons from NADPH created in the light
reactions are used to reduce molecules creating glyceraldehyde 3-phosphate (G3P)
which is eventually converted into glucose.
(3) regeneration - ATP is used to create RuBP necessary for the next cycle round.
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#4
There were 2 peaks in the chlorophyll absorption spectrum occurring at the shorter wavelengths (blue,
purples) and longer wavelengths (red, orange). The least amount of absorption occurred in the
intermediate wavelengths (green, yellow). Since the green wavelengths of light are not absorbed by
plants, it would be this color they would reflect.
#5
Phenol red is yellow in the presence of an acid. The blowing of CO2 into the solution at the beginning of
the lab caused the formation of carbonic acid (H2CO3); hence the yellow color. Through time as the
Elodea photosynthesized it used the CO2 in the water. This decreased the amount of CO2 in the solution,
thereby decreasing the amount of carbonic acid, causing the pH to rise and the phenol red solution to
turn red.
#6
Photosynthesis consists of two sets of reactions. The ATP and NADPH created in the light reactions are
used to drive the dark reactions where CO2 is converted into sugar. During daylight, the abundance of
energy from the sun drives these reactions, but at night the absence of sun causes a slow-down to
occur. The result is that during the day, CO2 is being used by the dark reactions at a much more rapid
rate than at night. Because of this, the amount of CO2 would be greater at night than during the day.
Therefore, the pH in these aquatic systems would be more acidic at night due to the build-up of carbonic
acid formed from the CO2 and water. During the day, the rapid uptake of CO2 by the dark reactions
would remove it from the water and cause the pH to rise (i.e., the water would become less acidic.).
#7
The empty tube contained everything the tube with the Elodea did except for the one variable that was
being tested. In this case, that one variable was the Elodea itself. If the solution in the empty tube
changed color just like the tube with the Elodea then it could not be conclusively stated that is was the
Elodea alone that caused the change. A control is an experimental set up that is the same as the
treatment experimental set up but with only one variable different.
#8
Oxygen
Exercise 9
#1
Cytokinesis - process by which cell cytoplasm is divided between two daughter cells. In animals, it occurs
through cleavage furrowing, in plants through cell plate formation.
Spindle fiber - microtubule structures used to separate sister chromatids and/or homologous
chromosomes.
Synapsis - alignment and pairing of homologous chromosomes during prophase I of meiosis. This forms
a tetrad (4) of sister chromatids.
Homologous chromosomes - in diploid cells, these are the pairs of the same type of chromosomes. One
homologous chromosome is inherited from each parent.
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Chromosome vs. chromatid - a chromatid is a duplicated chromosome joined to its copy by a single
centromere.
#2
During metaphase in mitosis, sister chromatids line up on the middle of cell (cell equator). In metaphase
I of meiosis, homologous chromosomes (each made up of a pair of sister chromatids), line up on the cell
equator.
#3
52
#4
21
#5
Mitosis creates diploid cells from diploid cells and is used to replace old cells or to create new ones. The
process of growing in a multicellular organism involves the creation of new cells. Therefore, we would
see the most mitotic activity during the early developmental (fetal) stages of an organism.
#6
A chiasma is a structure that occurs during prophase I of meiosis and is formed when two chromatids
from homologous chromosomes cross over one another. During anaphase I as homologous
chromosomes separate, chiasma break, swapping genetic material. This swapping of genetic
information will eventually serve to form genetically non-identical gametes (sex cells - sperm, egg).
#7
Genetic variation is crucial for the survival of a species. Environmental conditions on Earth have changed
considerably throughout Earth's history. These different conditions exert a "pressure" on species to
survive. Those that survive best, exhibit this by leaving behind more offspring which perpetuates the
parent's genetics into the future. What constitutes which organism survives best cannot be predicted so
variability in a species means a particular set of genetic information, which has arisen simply through
chance, may impart an advantage to an organism allowing it to leave behind more offspring than its
counterparts.
Exercise 10
#1
DNA - deoxyribonucleic acid, genetic molecule found in all organisms. Genetic traits are coded in this
molecule via the sequence of base pairs.
STR - short tandem repeats, repeating sequences of base pairs found in the introns (noncoding portions)
of DNA of an individual. DNA fingerprinting strives to enumerate the number of times these sequences
repeat themselves yielding a genotype for that particular gene locus.
PCR - polymerase chain reaction, the process by which a DNA sample is copied over and over again in
order to create enough DNA to conduct an analysis.
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Locus (pl. loci) - the location of a particular gene on a chromosome.
Annealing - the complementary base-pairing of a DNA primer to a strand of DNA that is being copied.
Annealing is followed by extending when DNA polymerase starts at the primer and begins adding new
bases to the DNA molecule.
#2
In DNA, adenine binds to thymine, guanine binds to cytosine through hydrogen bonds.
#3
The FBI requires a minimum of 13 autosomal loci to confirm an identification. The odds of two people
sharing the same genotype at all 13 loci is, for all intents and purposes, practically impossible. Just using
three loci as an example will demonstrate. If the odds of two people sharing the same genotype for
locus #1 is 1/20, and 1/40 for locus #2, and 1/100 for locus #3, the total probability would be 1/20 x 1/40
x 1/100 = 1/80,000! That means the odds of two people sharing identical genotypes at 3 loci is 1 in 80
thousand. Increasing that up to 13 loci would decrease this probability considerably. Therefore, while it
is technically possible for two people to match at all 13 loci, the odds are definitely against it.
#4
Suspect #2's banding pattern is an identical match to the banding pattern from the blood found on the
victim's clothing. Although one of fragments in the victim's DNA matches a fragment from the blood
stain, the second band from the victim is nowhere near close. Therefore, suspect #2 is the most likely
candidate for a match.
#5
An offspring will receive one chromosome from each parent. Therefore, the genotype at each loci for
the baby would be a combination of the mom and dad's genotype. To determine if this man is the baby's
father, examine each loci independently. One genotype will come from the mother (obviously) . . . the
other should match the father. In this particular case, each fragment length making up the genotype at
each loci has a match in the mother and the father. At 13 loci, this makes the odds that this man is not
the father very unlikely . . . unless he has an identical twin.
#6
a.
b.
c.
d.
e.
12
13
21
12
24
Exercise 11
#1
Monohybrid cross - cross examining only one trait.
Dihybrid cross - cross examining two traits simultaneously.
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Testcross - cross between an organism that exhibits the phenotypically dominant trait and one that is
phenotypically recessive. The goal is to determine whether the phenotypically dominant organisms is
true-breeding (i.e., homozygous) or not (i.e., heterozygous).
Allele - alternative form of a gene. For example, in Mendel's pea plants the genetic trait of flower color
have two alleles, purple and white.
Dominant allele - allele that is always phenotypically expressed.
Recessive allele - allele that is phenotypically expressed only when the organism is homozygous for
recessive alleles.
Genotype - actual genetic makeup of an organism. Organisms are homozygous dominant, heterozygous,
or homozygous recessive.
Phenotype - physical or physiological expression of a genetic trait.
Homozygous - when an organism contains identical alleles. Homozygous organisms can be homozygous
dominant when both alleles are the dominant ones or homozygous recessive when they are recessive.
Heterozygous - when an organism contains alleles that are different. Heterozygous individuals
phenotypically express the dominant trait. Recessive alleles are not expressed in heterozygous
individuals.
#2
phenotypic: 3:1
genotypic: 1:2:1
#3
9:3:3:1
#4
AB
Ab
aB
ab
#5
a.
A
A
a
Aa
Aa
a
Aa
Aa
no albinism; all heterozygous (carriers)
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Answers
b.
A
a
a
Aa
aa
a
Aa
aa
½ albino; ½ heterozygous (carriers); ½ homozygous recessive
c.
A
a
A
AA
Aa
a
Aa
aa
¼ homozygous dominant; ½ heterozygous (carriers); ¼ albino
d.
A
A
A
AA
AA
a
Aa
Aa
½ homozygous dominant; ½ heterozygous (carriers)
e.
a
a
a
aa
aa
a
aa
aa
all albino
#6
50%
#7
Parents: Aa (carriers)
Child: aa (albino)
#8
Tt
#9
Parents: both Tt
Person: tt
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Answers
#10
AB only
#11
Mother: IBi
Father: IAi
Person: ii
#12
Not possible
#13
Yes; XcXc
#14
Yes, if the mother is a carrier.
XY x XcX
Xc Y
#15
No, not possible.
XY x XcX
XcX, XX
or
XY x XX
XX
#16
No, this is not possible. If all parents are colorblind, there is no chance the son will inherit an X
chromosome that does not have the colorblind trait.
#17
In the first cross, the mother must be heterozygous (a carrier). She has to be this because her father was
colorblind and passed on the one X chromosome he had to her. Her normal X chromosome came from
her mother. She marries a man "just like dear old dad" meaning he is also colorblind. So, she's a carrier
and he's colorblind. Their sons have a 50/50 chance of getting a colorblind X chromosome because the
mother has one colorblind X and one normal X chromosome. For their daughters, half will be colorblind
having inherited the colorblind X chromosome from their father and, unluckily, the same kind of X
chromosome from their mother. Daughters also have a 50% chance of being a carrier. In the second
cross, since neither parent has a colorblind X chromosome, they cannot pass that on to any of their
children, male or female.
#18
BR
BR
BBRR
Br
BBRr
bR
BbRR
br
BbRr
black with rough fur (B_R_): 9/16 * 96 = 54
white with rough fur (bbR_): 3/16 * 96 = 18
Br
bR
BBRr
BbRR
BBrr
BbRr
BbRr
bbRR
Bbrr
bbRr
black with smooth fur (B_rr): 3/16 * 96 = 18
white with smooth fur (bbrr): 1/16 * 96 = 6
Lake-Sumter State College, Leesburg Laboratory Manual for BSC 1010C
br
BbRr
Bbrr
bbRr
bbrr
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