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Unformatted text preview: © Prep101 www.prep101.com/freestuff Solutions to HMB265 Practice Midterm
1. In a diploid cell in the metaphase of mitosis there are 100 chromatids. The
haploid number of chromosomes for this organism would be:
a)
b)
c)
d)
e) 200
100
50
25
cannot accurately decide based on data given Solution: Since there are 2 sister chromatid for every chromosome, there
would be 50 chromosomes in this cell. And, given that it is a diploid cell,
there are two sets of chromosomes, or homologues, thus if 2n=50, n=25.
2. Which of the following is NOT a feature of chromosomes?
a)
b)
c)
d) Homologous chromosomes carry the same set of genes
Metacentric chromosomes are comprised of two sister chromatids
Nonhomologous chromosomes have unrelated sets of genes
The definition of karyotype is the entire chromosome complement of an
individual organism or cell, as seen during metaphase
e) All of the above are correct 3. There are two types of variation. These are continuous and discontinuous
variation. Which of the following statements is FALSE?
a) Weight is an example of continuous variation
b) Continuous variation produces a spread of variation within a population
c) Discontinuous variation is often reflective of a 1:1 relationship between
genotype:phenotype
d) An example of discontinuous variation is height
e) All of the above are correct
Solution: Discontinuous variation is variation having distinct classes of
phenotypes for a particular character. Examples of discontinuous
variation are blood type or eye colour. Page 1 of 22 © Prep101 www.prep101.com/freestuff
4. During what stage of mitosis are centromeric connections severed?
a)
b)
c)
d)
e) anaphase
metaphase
prometaphase
prophase
telophase Solution: Anaphase is characterized by centromeres dividing, 2
kinetochores splitting and sister chromatid (now daughter
chromosomes) moving to opposite poles.
5. During what stage of mitosis does the nuclear membrane reform?
a)
b)
c)
d)
e) anaphase
metaphase
prometaphase
prophase
telophase Solution: Telophase is characterized by the nuclear envelope reforming,
spindle fibers disappearing, chromosomes decondensing and nucleoli
reappearing.
6. Arabidopsis thaliana is a model organism studied by many of the world’s
plant biologists. It has a diploid chromosome number of eight. How many
chromosomes are present in the cell nucleus at the beginning of meiosis II?
a)
b)
c)
d)
e) 4
16
8
2
32 Solution: Meiosis (2N to N) takes place in 2 successive nuclear divisions
called Meiosis I (reductional division) and Meiosis II (equational
division) in special cells called meiocytes (germ cells). At the end of
meiosis I, homologous chromosomes have split and thus, the cell now
has a haploid number of chromosomes that are still comprised of two
sister chromatid, which are divided at the end of Meiosis II. Page 2 of 22 © Prep101 www.prep101.com/freestuff
7. The following is a list of mitotic events. What is the correct order?
A. chromosomes align on the midplate of the cell
B. kinetochores being attaching to spindle fibers
C. nuclear membrane reforms, chromosomes, decondense
D. chromosomes condense, centrosomes migrate to the opposite poles
E. pairs of sister chromatids separate, one of a pair moves to each pole
a)
b)
c)
d)
e) BDACE
DABEC
DBAEC
ABDCE
EDBAC Solution: The order reflects the order of mitotic events corresponding to
the details of that particular stage as outlined in the following chart.
Phases of Mitosis
Prophase Metaphase
Prometaphase Metaphase
Anaphase
Telophase Cytokinesis Important details
chromatin condense (n # of chromosomes, but 2n
chromatid)
and
become
visibleeven
though
chromosomes have replicated, because they are still
attached by the centromere, the number of
chromosomes has NOT increased!
sister chromatid attach at centromere
spindle apparatus forms outside of nucleus
centrosomes form the spindle poles
nucleoli disappear
nuclear envelope breaks down (vesiculates)
microtubules attach to kinetochore
polar microtubules span centrosomes
astral microtubules extend from centrosomes
sister chromatid attach to opposite poles
chromosomes align along the “equator”
sister chromatid face OPPOSITE polesthis is an
important difference from meiosis!!
centromeres divide; 2 kinetochores split
sister chromatid (now daughter chromosomes) move
to opposite poles
nuclear envelope reforms
spindle fibers disappear
chromosomes decondense
nucleoli reappear
cytoplasm divides at cleavage furrow via a contractile ring
produce 2 equal daughter cells Page 3 of 22 © Prep101 www.prep101.com/freestuff
8. At the beginning of meiosis II a cell has 36 chromatid, how many
chromosomes will each cell have at the anaphase II?
a) 36
b) 72
c) 18
d) 9
e) 16
Solution: At the beginning of meiosis II, homologous chromosomes
have already split, thus, there are only the haploid number of
chromosomes remaining which are each comprised of two sister
chromatid. Although sister chromatid divides and move to the opposite
poles during anaphase II. Thus, each chromatid is considered a separate
chromosome and so the number of chromosomes at the end of anaphase
II equals the number of chromosomes at the beginning of meiosis II.
9. Before the S phase of mitosis, a diploid cell has 20 chromosomes, how many
chromosomes will it have in late anaphase?
a) 20
b) 40
c) 10
d) 80
e) 30
Solution: In late anaphase centromeres divide; 2 kinetochores split,
sister chromatid (now daughter chromosomes) move to opposite poles.
Thus, the number doubles from before mitosis to the end as the 2
daughter cells need to have the exact chromosome # as the parent cell.
10. How many chromosomes does a diploid organism have at the end of
anaphase I, if n represents the haploid number of chromosomes?
a) n/2
b) n
c) 2n
d) 4n
e) 8n
Solution: Because cells go through the S phase prior to meiosis, a diploid
cell would have 2n chromosomes, but 4n chromatid. Because
homologous chromosomes split in meiosis I, the chromosome number is
the same as the original 2n number.
Page 4 of 22 © Prep101 www.prep101.com/freestuff
11. To identify the genotype of yellowseeded pea plants as either homozygous
dominant (YY) or heterozygous (Yy), you could do a testcross with plants of
genotype _______.
a)
b)
c)
d)
e) y
Y
yy
YY
Yy Solution: By definition, a testcross is a cross to an animal that is
homozygous recessive for the query genes.
12. A pea plant is heterozygous for both seed shape and seed color. S is the
allele for the dominant, spherical shape characteristic; s is the allele for the
recessive, dented shape characteristic. Y is the allele for the dominant, yellow
color characteristic; y is the allele for the recessive, green color characteristic.
What will be the distribution of these two alleles in this plant's gametes
assuming that these genes are on different chromosomes?
a) 50% of gametes are Sy; 50% of gametes are sY
b) 25% of gametes are SY; 25% of gametes are Sy;
25% of gametes are sY; 25% of gametes are sy.
c) 50% of gametes are sy; 50% of gametes are SY
d) 100% of the gametes are SsYy
e) 50% of gametes are SsYy; 50% of gametes are SSYY.
Solution: Because of independent assortment, the dominant alleles of the
different genes on different chromosomes have an equal chance of
sorting together or with the recessive alleles and likewise for the
recessive alleles. Mendel’s Second Law: Independent Assortment
Genes on different chromosomes segregate independently of each other.
So, an equal chance of SY, Sy, sY, sy is 1:1:1:1 or a 25% (equal) chance of
any of the 4 combinations. Page 5 of 22 © Prep101 www.prep101.com/freestuff 13. Which of the following genetic crosses would be predicted to give a
phenotypic ratio of 9:3:3:1?
a)
b)
c)
d)
e) SSYY x ssyy
SsYY x SSYy
SsYy x SsYy
SSyy x ssYY
ssYY x ssyy Solution: The 9:3:3:1 ratio is unique to a dihybrid cross (two
heterozygotes mating).
14. In a dihybrid cross, what fraction of the offspring will be homozygous for
both recessive traits?
a)
b)
c)
d)
e) 1/16
1/8
3/16
1/4
¾ Solution: Homozygous recessive offspring represent the “1” in the 9:3:3:1
dihybrid cross outcome. Thus, 1/(9+3+3+1=16).
15. In a trihybrid cross, how many different phenotypes would you observe in
the resulting offspring?
a)
b)
c)
d)
e) 3
8
9
12
27 Solution: Using the branching method, there are two possible
phenotypes for any of the three genes = 23 = 8. Page 6 of 22 © Prep101 www.prep101.com/freestuff
16. For the following cross, A/a; B/B; c/c; D/d; E/e x A/a; b/b; C/c; D/D;
e/e what would be the chance of having a completely heterozygous offspring
with the genotype: A/a; B/b; C/c; D/d; E/e?
a)
b)
c)
d)
e) 1/2
¼
1/8
1/16
1/32 Solution: Using the product rule, there is a ½ chance that the parental
gametes will combine to produce Aa x 1/1 for B/b (each parent only has
a B or b to give, so all offspring will have B/b) x ½ for C/c; ½ for D/d
and ½ for E/e. ½ x 1 x ½ x ½ x ½ = 1/16
17. Hemophilia in humans is recessive and is due to an Xchromosome
mutation. What will be the results of mating between a normal (noncarrier)
female and a hemophilic male?
a)
b)
c)
d)
e) half of daughters are normal and half of sons are hemophilic
all sons are normal and all daughters are carriers
half of sons are normal and half are hemophilic; all daughters are carriers
all daughters are normal and all sons are carriers
half of daughters are hemophilic and half of daughters are carriers; all
sons are normal Solution: Because the father will only contribute the disease and it is on
the X chromosome, the father will pass on the affected X chromosome to
all of the daughters and none of the sons. Because this is a sexlinked
recessive disease, the daughters will only be carriers. Page 7 of 22 © Prep101 www.prep101.com/freestuff 18. What is the inheritance pattern in this pedigree? a)
b)
c)
d)
e) Autosomal recessive
Autosomal dominant
Xlinked recessive
Xlinked dominant
Recessive lethal Solution: Because the disease is rare and there is no bias towards one
gender, it is an autosomal recessive disease. Consanguineous marriages
often results in the manifestation of disease in their offspring due to the
increased chance of rare recessive alleles combining.
19. According to the above pedigree, what is the probability that the offspring
between individuals 1 and 2 will have the trait?
a)
b)
c)
d)
e) 1/16
1/8
1/2
0
2/3 Solution: Focus on individual 1. There is a 100% chance that his dad is
A/a and so there is ½ chance that he is A/a and ½ chance that he will
pass it on to his offspring. The father of individual 2 has the trait and so
there is 100% chance that individual 2 is heterozygous and then a ½
chance that if she is a heterozygote, she will pass it on. Multiply these
chances together to give you: ½ * ½ * ½ = 1/8
Page 8 of 22 © Prep101 www.prep101.com/freestuff 20. Below is a pedigree of a human genetic disease in which stricken individuals
are solidcolored. Apply the laws of probability and calculate the probability the
offspring of the cousin marriage 1 x 4 will have the disease. a) 2/3
b) 1/2
c) 1/6
d) 1/4
e) 1/8
Solution: Because the affected individuals had unaffected parents and
there is no genderdifferences, this disease is an autosomal recessive
disease. Cousin 1 has a 100% chance that he is a carrier and ½ chance
that he will pass the recessive allele on to his offspring. Cousin 4 has the
trait and so there is a 100% chance that she will pass the trait on to her
offspring. Thus, there is a ½ chance that the offspring will inherit the
disease.
21. A man and a woman are having a baby. Both the man and woman have
siblings with sicklecell anemia, a debilitating blood disease caused by a
recessive allele, but neither the man nor the woman nor any of their parents
suffer from the disease themselves. What is the probability that the baby will
have sicklecell anemia?
a) 1/8
b) 1/16
c) 1/9
d) 2/3
e) 1/3
Solution: Because the siblings of these individuals have the disease, we
know that their parents are A/a. A/a x A/a = A/A, A/a, A/a, a/a.
Because these individuals do not have the disease, there is a 2/3 chance
that they are carriers (eliminate a/a from the possibility and you are left
with A/A, A/a, A/a = 2/3 = A/a). If they are A/a, there is a ½ chance Page 9 of 22 © Prep101 www.prep101.com/freestuff
that they will pass the disease allele on to their offspring. Thus the
chance of having a child with a/a is 2/3 * ½ * 2/3 * ½ = 1/9
22. Linked loci are loci that
a) Are on the same chromosome
b) Determine sex
c) Govern traits (such as hair texture and hair color) that are functionally
related
d) Have the same alleles residing on them
e) Govern traits that have nothing to do with one another
23. How many map units is a recombination frequency of 10% equal to?
a) 10%
b) 15%
c) 10 centimorgans
d) 90 anticentimorgans
e) Both c and d
Solution: 1% recombination = 1 map unit (m.u.) and 1m.u. = 1
centiMorgan (cM). So, 10% rf = 10 cM.
24. The pairwise map distances for four linked genes are as follows: AB = 9
m.u., BC = 4 m.u., CD = 6 m.u., AC = 5 m.u., AD= 1 m.u., BD = 10 m.u.
What is the order of these four genes?
a) ABCD
b) BDCA
c) BCAD
d) CABD
e) BCDA
Solution: Because B and D have the greatest map distance they are the
furthest apart. C is closest to B and A is closest to D. So far, you know
that the order is B, C, A, D. Check by making sure that these all add up. Page 10 of 22 © Prep101 www.prep101.com/freestuff
Use the following information for the next four questions:
An experiment was done to determine the linkage relationship between three
genes in Drosophila (d, e, and f). Homozygous females phenotypically d,f were
crossed with homozygous males phenotypically e. The F1 progeny were all wild
type. The F1 females were then crossed to a male tester and the resulting F2
offspring are listed in the following table.
Genotype Total offspring
d+f
290
d++
7
def
139
++f
81
+++
145
+ef
10
+e+
265
de+
63
25. What was the gene order and allele combination in the original parents?
a)
b)
c)
d)
e) def and +++
d+f and +e+
df+ and ++e
+df and e++
df; + and ++; e Solution:
Genotype Total
offspring
d+f
290
d++
7
def
139
++f
81
+++
145
+ef
10
+e+
265
de+
63
Total
1000 Parental or
recombinant
P
R
R
R
R
R
P
R
m.u.
(#/total*100) Page 11 of 22 de 139
81
145
63
428
42.8 ef df 7
139 7
81 145
10 10 301
30.1 63
161
16.1 © Prep101 www.prep101.com/freestuff
The offspring genotype with the largest number of offspring is the
Parental and any other combination arose due to recombinations. Go
down the list and make three columns to represent the recombinations
between the three genes based on those that deviate from the parental
combinations. Add these up and divide by the total number of progeny,
multiplied by 100 for the m.u. (recall 1% rf = 1 m.u.). From these
numbers, you will determine that d and e are the furthest apart and that f
is closer to d than e.
26. How many double recombinant offspring are there?
a)
b)
c)
d)
e) 17
34
48
144
70 Solution: Since the parentals are df+ and ++e, the double recombinants
are +fe and d++ which are 10 and 7, respectively, and 17 in total.
27. What is the distance from d to e?
a)
b)
c)
d)
e) 42.8 m.u.
46.2 m.u.
30.1 m.u.
16.1 m.u.
10.3 m.u. Solution: The distance in the de column from question 25 is 42.8, but this
is not accounting for double recombinant. Add (428 + 2 x 17)/1000 = 46.2
m.u.
28. Calculate the interference value I
a)
b)
c)
d)
e) 1
0
0.35
0.52
0.65 Page 12 of 22 © Prep101 www.prep101.com/freestuff
Solution: I = 1 – Observed number of doublerecombinant/expected
number of double recombinants. I = 1 – (17)/((161*301)/1000) = 0.65.
The following 3 questions are pertaining to the pedigree below: i
i
IB
i i
i
i
i IB
i i
i
IB
i i
i
i
i i
i i
i
i
i IBi IB
i IB
i IB
i IB
i i
i i
i
IB
i i
i IB IB i
iii i
i IB
i i
i IB
i IB
i IB
i i
i The above pedigree shows the inheritance of a dominant phenotype (D) and
blood type (type O = ii and type B = IBi). Assume that those marrying into the
family are homozygous recessive (ddii), while those with the dominant
phenotype are heterozygous (D/d).
29. How many different recombinant offspring are in this family?
a) 3
b) 6
c) 12
d) 0
e) 8
Solution: Determine the parental genotype with regards to D and blood typelook at
oldest person in the pedigree that had the phenotype (is D linked to IB or is D linked
to i)? Given that most of the people in the second generation that have the phenotype
are of blood types IBi, their mom must be DIB/di. Anytime someone has the disease
but isn’t IB (thus, Di), or someone does not have the dominant phenotype but is IB
(dIB), must have resulted from a recombination. There are 6 in total. 30. What is the RF for these two genes in this family?
a) 0.21
b) 0.19
c) 0.17
d) 0.25
e) 0.3
Page 13 of 22 © Prep101 www.prep101.com/freestuff
Solution: RF = recombinants/total = 6/24 (only include progeny, not
people who married into the family) = 0.25
31. What is the Lod score for this family?
a) 1.4
b) 1.9
c) 1.4
d) 1.9
e) none of the above
Solution: Let’s calculate the probabilities under both hypotheses, linked
(RF=0.25) and unlinked (0.5)independent assortment.
RF
P
P
R
R 0.5
0.25
0.25
0.25
0.25 0.2727
0.375
0.375
0.125
0.125 The probability of obtaining the result under independent assortment
(RF of 50%) will be equal to
0.2524=3.55 x 1015 x B (the number of possible birth orders for 18 parental
and 6 recombinant individuals)
For an RF of 0.25 the probability is
0.37518 x 0.1256 = 8.20 x 1014 x B
The ratio of the odds is 8.2 x 1014 x B/3.55 x 1015 x B (B’s cancel out) =
23.1
Log (23.1) = 1.4
32. If a striped cat from a purebreeding stock was mated to a spotted cat from
another purebreeding stock, and the resultant progeny are all both spotted and
stripped, these alleles for coat pattern are. . .
a)
b)
c)
d)
e) Recessive
Dominant
Epistatic
Codominant
Incompletely dominant Page 14 of 22 © Prep101 www.prep101.com/freestuff
Solution: The definition of Codominance is when both traits become
visible in an F1 hybrid, but not as a mix.
33. In certain breed of plant, a single gene is responsible for flower colour. If a
blue flowered plant from a purebreeding stock is crossed to a yellow flowered
plant from a purebreeding stock and the resultant progeny all have orange
flowers, these alleles for flower colour are:
a)
b)
c)
d)
e) Dominant
Recessive
Epistatic
Codominant
Incompletely dominant Solution: The definition of incomplete dominant is when the F1 hybrid
resembles neither purebred parent, but appears as a combination of the
parents.
34. Suppose you are studying eye size in a species of spider. Eye size is
controlled by two genes (each with two alleles: Q/q and R/r). If a spider carries
at least one Q allele, it has large eyes no matter what alleles it has at the R locus.
If it is qq, its phenotype depends on the R locus—if it is qqRr, it has medium
eyes; if it is qqrr, it has small eyes. What ratio of (large: medium: small)
phenotypes do you expect to see resulting from a QqRr X qqrr cross?
a) 9:3:3:1
b) 9:4:3
c) 2:1:1
d) 3:1
e) 1:2:1
Solution: Given that the first parent will contribute Q ½ of the time, ½ of
the progeny will have large eyes. The other have will have a split between
medium and small, which is determined by the fact that R locus will be
R/r half the time and r/r the other half. Thus, 2:1:1 Page 15 of 22 © Prep101 www.prep101.com/freestuff
35. Two unlinked loci effect mouse hair color. AA or Aa mice are agouti. Mice
with genotype aa are albino because all pigment production is blocked,
regardless of the phenotype at the second locus. At the second locus, the B
allele (agouti coat) is dominant to the b allele (black coat). What would be the
result of a cross between two agouti mice of genotype AaBb?
a)
b)
c)
d)
e) 4 agouti: 4 black: 8 albino
9 agouti: 3 black: 3 albino: 1 grey
9 agouti: 3 black: 4 albino
8 agouti: 4 black: 4 albino
12 agouti: 3 black: 1 albino Solution: In order to get agouti, you need A/; B/, which happens 9/16;
and black when you have A/; b/b 3/16 and a/a; / is 4/16. Thus,
albino is epistatic to black.
36. Symptoms of Marfan syndrome include skeletal, optical, and cardiovascular
abnormalities. Skeletal abnormalities include lengthening of the long bones,
scoliosis and others. Optical abnormalities can include dislocation of the lens
into the anterior chamber of the eye. Cardiovascular abnormalities are
numerous (ex. dissecting aneurysms), which attributes to the shorter life span
of Marfan syndrome patients as a group. Each person with this genetic makeup
display a range of the symptoms (from no symptoms to severe). This is an
example of
a)
b)
c)
d)
e) Pleiotropy
Expressivity
Penetrance
Incomplete dominance
Expressivity and pleiotropy Solution: Because people with the disease genotype don’t all have
symptoms, this is an example of penetrance. However, within the
population of people that have symptoms, there is a range, which is an
example of expressivity. Page 16 of 22 © Prep101 www.prep101.com/freestuff
37. Determine the order of the compounds (letters) and genes (numbers),
respectively, acting in this biosynthetic pathway given the following
information that evaluates for the presence of the end product in the pathway:
Compound added
Mutant A B C D E F
1
 +++
2
+++ +
3
++
4
+++++
5
+
a)
b)
c)
d)
e) A,E,B,D,C,F; 4,2,1,3,5
F,C,D,B,E,A; 5,3,1,2,4
F,C,D,B,E,A; 4,2,1,3,5
A,E,B,D,C,F; 5,3,1,2,4
None of the above Solution: The first enzyme in the pathway can be recognized by the fact
that it is the one that is rescued by the most number of compounds. The
compound that doesn’t rescue the mutant is before the first enzyme in
the pathway. Thus, gene 4 is first and compound A is first, then 2/E,
1/B, 3/D, 5/C, F (or to determine the first compound, look at the one
that rescues no mutants to all mutants; conversely, to ID the first
mutant, look for the enzyme that is rescued by the most compounds).
AEBDCF
4
2
1
3
5
A cannot be synthesized into E (and ultimately the product at the end of
the pathway) without a functioning enzyme 4, likewise, a mutation in
enzyme 2 requires that B be added in order to complete the pathway etc.
38. Quantitative genetics is:
a)
b)
c)
d)
e) The study of genetics of continuously varying characters
Influenced by variation in the environment
Often involves multifactorial inheritance
Uses statistical methods to make predictions about inheritance
All of the above Page 17 of 22 © Prep101 www.prep101.com/freestuff
39. The spread in a distribution refers to:
a) variation
b) dispersion of a measurement
c) central tendency
d) relation between different variables
e) Two of the above are correct
Solution: Both variation and dispersion are measures of spread. Central
tendency is the mode, while correlation describes the relation between
different variables.
40. You have two pure breeding lines of plants (average height is 10 cm and 30
cm for line 1 and 2, respectively). You make a cross of a plant from each line
and get F1 progeny with an average height of 22 cm. You cross the F1 progeny
and get an average height of 24 cm, but a much greater distribution of heights
(12 cm to 29 cm). From the F2 plants, you selectively breed two of the smallest
plants together and two of the largest plants together. What are the expected
outcomes from these crosses?
a) Means in each cross is close to the size of the F2 parents
b) An average in each cross is close to the size of the parents, but a much
larger distribution than in F2
c) An average in each cross is different from the parental heights
d) The variation in F3 distributions will be overlapping
e) Two of the above are correct
Solution: It is not likely that the distributions in each cross will be larger
than F2 rather they will likely be smaller due to the combination of
genotypes that are more similar (in comparison to the cross of the F1s
generating the F2s).
41. Which of the following examples reflects covariance that is calculated in
broadsense heritability?
a) Tall parents having tall children
b) Overweight parents that have children who are overweight
c) Siblings inheriting the same genotype
d) The survival of a butterfly with dark wings in the forest
e) Different allelic combinations in parents
Solution: Covariance in broadsense heritability is the covariance
between genotype and environment (populations variation is influenced
by the environment of individuals).
Page 18 of 22 © Prep101 www.prep101.com/freestuff
42. A heritability value of 0.8 indicates which of the following?
a)
b)
c)
d)
e) The majority of phenotypic variation is due to environmental variation
The phenotype is the result of 3 genes interacting
The majority of the phenotypic variation has a genetic basis
There is no developmental noise occurring
No answers are correct 43. You are studying the heritability of height in a given population at two
altitudes. You calculate the genotypic variance to be 33.4 and the variance of
the population to be 49.8. What is the environmental variance and heritability
(assume zero covariance) of this trait in this study?
a)
b)
c)
d)
e) 16.4, 33%
16.4, 49%
16.4, 67%
83.2, 40%
83.2, 67% Solution:
se2 = sp2  sg2
se2 = 49.8 – 33.4 se2 = 16.4
Heritability = H2 = sg2 / sp2
H2 = 33.4/49.8
H2 =67%
44. Which of the following is NOT an assumption of the HardyWeinberg
equilibrium?
a)
b)
c)
d)
e) The population in infinitely large
Individuals are mating randomly
Endogamy is taking place within subpopulations
There are no new mutations introduced into the population
There is no migration into or out of the population Solution: Endogamy is the mating of individuals within a group or
subgroup rather than at random in a population. Page 19 of 22 © Prep101 www.prep101.com/freestuff
Use the following table to answer the next 3 questions.
Gametic type
Population AB
Ab
aB
ab
1
0.25
0.3
0.15
0.3
2
0.4
0.2
0.3
0.1
45. What is the heterozygosity of the gametes for population 1?
a)
b)
c)
d)
e) 0.265
0.735
0.4875
0.2438
0.975 Solution:
Gamete heterozygosity = 1homozygosity 1 (f AB/AB + f Ab/Ab + f aB/aB + f ab/ab)
= 1 (0.25*0.25 + 0.3*0.3 + 0.15*0.15 + 0.3*0.3) = 0.735
46. What is the allelic heterozygosity for population 1?
a)
b)
c)
d)
e) 0.265
0.735
0.4875
0.2438
0.975 Solution:
Allelic heterozygosity = ( f A/a + f B/b )/2
= (2pq + 2rs)/2
= (2(0.25 + 0.3)(0.15+0.3) + 2(0.25 + 0.15)(0.3 + 0.3))/2= 0.4875 Page 20 of 22 © Prep101 www.prep101.com/freestuff
47. What is the heterozygosity of the gametes for population 2?
a)
b)
c)
d)
e) 0.7
0.3
0.45
0.9
0.225 Solution:
Gamete heterozygosity = 1homozygosity 1 (f AB/AB + f Ab/Ab + f aB/aB + f ab/ab)
= 1 (0.4*0.4 + 0.2*0.2 + 0.3*0.3 + 0.1*0.1) = 0.7
48. Humans with the genotype DD and Dd show the Rh+ blood phenotype,
whereas those with the genotype dd show the Rh blood phenotype. In a
sample of 400 Basques from Spain, 230 people were Rh+ and 170 people were
Rh. Assuming that this population is in HardyWeinberg proportions, what is
the allele frequency of the allele D?
a)
b)
c)
d)
e) 0.348
0.652
0.425
0.575
0.288 Solution: q2 = 170/400 = 0.425
q = √(0.425) = 0.6519
p = 1 – 0.6519 = 0.348 49. Which of the following statements about inbreeding is FALSE?
a) There is a 1/2N loss of heterozygosity in each generation
b) In each generation, there is an increase towards recessive
homozygosity
c) Inbreeding results in nonrandom mating which alters HardyWeinberg
frequencies
d) An inbreeding coefficient F needs to be calculated to adjust for the
proportion of shared genes between related individuals
e) Two of the above
Page 21 of 22 © Prep101 www.prep101.com/freestuff Solution: There is an increase towards either homozygous recessive or
homozygous dominant
50. If in a population of 1 million people, 200 blind (homozygous recessives,
aa) were found, how many normal (homozygous dominants, AA) individuals
would be found in the next generation under equilibrium conditions?
a)
b)
c)
d)
e) 749,850
999,800
985,858
27,888
971,916 Solution:
q2 = 200/1,000,000 = 0.0002
q = √(0.0002) = 0.01414
p = 1 – 0.01414 = 0.985858
p2 = (0.985858)2 = 0.971916 * 1,000,000
f(AA)* 1e6 =971,916 Page 22 of 22 ...
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This note was uploaded on 02/20/2012 for the course BIO 101 taught by Professor Randall during the Spring '12 term at Tunxis CC.
 Spring '12
 randall
 Mitosis

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