Download Virginia State Science Olympiad Regional Tournament

Virginia State Science Olympiad
Regional Tournament
2013 – Division B
Heredity
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You may write your name on this sheet before the event
examination has begun.
Two (2) non-programmable non-graphing calculators, one (1)
standard-size (8.5”x11”) double-sided sheet (not two one-side
sheets) of paper containing any information, and writing
implements may be used.
Any other electronic devices are not allowed for this event; please
consult the proctor about a safe location to store these devices for
the duration of the event if you happen to carry one; any team
caught with an electronic device during the event will be
immediately disqualified.
Tiebreakers will be tallied according to specified questions and
will be used in the order from the front to back; tiebreakers will
only be used to break ties and do not count towards your overall
score.
Student Names:
School:
Team #:
KEY
Heredity
Team: _______
PART A – Molecular Genetics!
1. Which of the above karyotypes represents a normal human female? [1pt]
a. A
c. C
e. E
b. B
d. D
f. F
2. Which of the above karyotypes represents a normal human male? [1pt]
a. A
c. C
e. E
b. B
d. D
f. F
3. Which of the above karyotypes represents a person with two (2) chromosomal
aneuploidies? [1pt]
a. A
c. C
e. E
b. B
d. D
f. F
4. Which of the above karyotypes represent individuals with Down ’s syndrome? [1pt]
a. A&B
d. E&F
b. B&D
e. None of the above
c. B&E
5. Most chromosomal aneuploidies are caused by nondisjunction during meiosis. What is
precisely is nondisjunction? [2pt]
Nondisjunction is the failure to separate chromosomes during meiosis.
6. Which of the following chromosomes is most likely to undergo nondisjunction? [1pt]
a. 9
d. X
b. 18
e. Y
c. 21
6
1
Points:__________
Heredity
Team: _______
7. Which of the following best demonstrates the central dogma of molecular biology? [1pt]
a. DNA  RNA  protein
d. RNA  protein  DNA
b. DNA  protein  RNA
e. protein  RNA  DNA
c. RNA  DNA  protein
8. Which of the following best describes how DNA and RNA molecules are different? [1pt]
a. DNA contain uracil residues, while RNA contain thymine residues.
b. Both molecules contain a sugar-phosphate backbone.
c. DNA is generally double-stranded while RNA is generally single-stranded.
d. DNA is read in a 5’ to 3’ direction, while RNA is read in a 3’ to 5’ direction.
e. None of the above show accurate differences between DNA and RNA molecules.
9. Which of the following terms best describes the method of replication in DNA? [1pt]
a. Conservative
d. Anti-discontinuous
b. Semi-conservative
e. Magic
c. Discontinuous
The following questions refer to the diagram to
the right:
10. Which letters represent
interphase? [2pt]
A,B,C (all letters for full credit)
11. Which letter represents the
phase where DNA
synthesis usually occurs?
[1pt]
B
12. Which letter represents the phase
where cell division actively occurs?
[1pt]
D
13. Which of the following statements best illustrates the difference between mitosis and
meiosis? [1pt]
a. Mitosis results in four haploid daughter cells, while meiosis results in two diploid
daughter cells.
b. DNA synthesis only occurs once before either mitosis or meiosis.
c. During anaphase I, homologous pairs of chromatids are separated; during regular
anaphase, homologous pairs of chromosomes are separated.
d. Recombination occurs during both mitosis and meiosis.
e. None of the above show accurate differences between mitosis and meiosis.
8
2
Points:__________
Heredity
Team: _______
PART B – Attack of the Peas!
You’ve discovered a new species of pea, with a trait for flower color
where allele “A” (red) is dominant to allele “a” (white).
One of two possible outcomes
♀\♂
A
A
a
Aa
Aa
a
Aa
Aa
14. Complete the sample parent cross to the right between truebred
red and truebred white flowers. [3pt]
15. What proportion of peas resulting from the crossing in question 2
is expected to be a hybrid plant? [1pt]
a. 0%
d. 75%
b. 25%
e. 100%
c. 50%
16. What is the expected color of the hybrid flowers if the trait is…

completely
dominant? [1pt]

red
incompletely
dominant? [1pt]

codominant? [2pt]
red and white
(stripes/variegation)
pink
17. Complete a sample cross between two of the hybrids from the
parent (P) cross to the right. [3pt]
First Filial (F1) cross
18. Which of the following best shows the expected ratio of
genotypes resulting from the cross in question 5? [1pt]
a. 50% AA; 50% Aa; 50% aa
b. 100% AA; 0% Aa; 0% aa
c. 0% AA; 100% Aa; 0% aa
d. 0% AA; 0% Aa; 100% aa
e. 25% AA; 50% Aa; 25% aa
♀\♂
A
a
A
AA
Aa
a
Aa
aa
19. What is the expected ratio of phenotypes if the flower color trait is…

completely
dominant? [1pt]
3 red : 1 white

incompletely
dominant? [1pt]
1 red : 2 pink : 1 white

codominant? [1pt]
1 red : 2 red/white:
1 white
20. After the F1 cross, what is the relative proportion of alleles (“A” and “a”) present in the
F2 pea population (the alleles present in the progeny)? [1pt]
a. 0% A; 100% a
d. 75% A; 25% a
b. 25% A; 75% a
e. 0% A; 100% a
c. 50% A; 50% a
16
3
Points:__________
Heredity
Team: _______
It is discovered sometime later that a second gene controls
the pea plants’ appetite for blood (from which the red
coloration is surprisingly derived). While preparing to
arm yourselves with herbicides, you decide to check out
the genetics of the trait, where the allele “B” (bloodthirsty)
is completely dominant to allele “b” (not-bloodthirsty).
21. Assuming you start with only dihybrid plants,
complete the Punnett square to the right. [4pt]
22. What is the probability that any progeny from the
cross in question 9 will be AaBb? [1pt]
a. 0% AaBb
d. 75% AaBb
b. 25% AaBb
e. 100% AaBb
c. 50% AaBb
This is one of many possible outcomes, which all
include the same genotypes
♀\♂
AB
aB
Ab
ab
AB
AABB
AaBB
AABb
AaBb
aB
AaBB
aaBB
AaBb
aaBb
Ab
AABb
AaBb
AAbb
Aabb
ab
AaBb
aaBb
Aabb
aabb
23. Assuming that flower color is completely dominant, what is the approximate probability
that a red flowered pea plant will also be not-bloodthirsty? [1pt]
a. 0%
d. 33%
b. 19%
e. 50%
AAbb or Aabb
c. 25%
3/12
24. What is the approximate probability that a white flowered pea plant will also be
bloodthirsty? [1pt]
aaBB or aaBb
a. 0%
d. 75%
3/4
b. 19%
e. 94%
c. 56%
25. Assuming that flower color is not completely dominant (i.e. either incomplete or
codominant), what is the approximate probability that any pea plant will be neither red
nor bloodthirsty? [1pt]
a. 0%
d. 44%
b. 6%
e. 56%
Aabb or aabb
c. 19%
3/16
26. By some fortune, you have discovered that another trait – color of pea pods (“G” is
dominant over “g”) – is actually linked to the bloodthirstiness trait. How might linkage
affect the prediction of progeny in a Punnett square? [2pt]
Linked genes cannot assort independently, so any gametes must contain the alleles that are linked together.
If B is linked to G and b is linked to g, the only possible gametes are BG and bg, not Bg or bG.
TIE-BREAKER
In the space below, draw out a Punnett square showing the crossing between dihybrid BbGg pea
plants, assuming that the dominant alleles for each trait are linked.
BG
BG
4
bg
bg
BBGG BbGg
BbGg
bbgg
alternatives acceptable,
as long as gametes and
genotypes are correct
10
Points:__________
Heredity
Team: _______
PART C - The Royal Disease
Hemophilia is a disorder that affected many of the ruling families of Europe during the 19th and
20th centuries, thanks largely to the legacy of inter-royal marriages involving children of Queen
Victoria. The following pedigree follows the descendants of Queen Victoria:
27. The first daughter of Queen Victoria – Empress Frederick (born Victoria Adelaide) of
Prussia (B) – wedded the Kaiser Frederick III and was the mother to Kaiser Wilhelm II of
Prussia (G). As far as we are aware, none of her children suffered from hemophilia (as
shown in the pedigree). Based upon this information, what is the probable genotype of
the Empress Frederick (B)? [2pt]
It is probable that Empress Frederick is not a carrier of the hemophiliac allele;
it is probable that Empress Frederick is homozygous for the non-hemophiliac allele.
28. Tsarina Alexandra (H) was the daughter of Queen Victoria’s second daughter Alice (D).
The first son of Tsarina Alexandra was born a hemophiliac. Based upon this information,
determine the genotype of Alice (D). [2pt]
Since Alice’s daughter is a carrier (and her husband is not a hemophiliac), Alice must be
heterozygous and a carrier as well.
5
4
Points:__________
Heredity
Team: _______
29. Leopold (E) was the only son of Victoria who was actually hemophiliac. Despite his
condition, he managed to live until the age of thirty and had two children – Alice,
Countess of Althona (J) and Charles Edward (I). Explain briefly why the children of
Charles Edward (J) never suffered from hemophilia, while Alice’s (J) only son was a
hemophiliac. [4pt]
Hemophilia is only transmitted to sons by the mother, so the only way Charles Edward could
have inherited hemophilia was from his mother, who was not a carrier; ergo, Charles Edward
was not hemophiliac.
Alice, Countess of Althona, on the other hand, must inherit the hemophilia gene from her father,
who was a hemophiliac. Since her mother is not hemophiliac, Alice did not suffer from
hemophilia herself. Since she has at least one hemophiliac allele, Alice is a carrier of hemophilia.
Ergo, Alice’s son has a chance to inherit the hemophilia allele.
30. The final daughter of Queen Victoria was Princess Beatrice (F), whose daughter Victoria
Eugenie (K) became consort to Alfonso XIII of Spain. Of her four sons, only two had
hemophilia while the other two were normal, including the Infante Juan, Count of
Barcelona and father to the current king of Spain. Explain briefly why half of Victoria
Eugenie’s sons were hemophiliac. [4pt]
Victoria Eugenie has two X chromosomes, each with a different allele of the hemophilia gene.
Since the father cannot donate any X alleles, he has no effect on the inheritance of hemophilia.
Any son of Victoria Eugenie has a 50% chance of inheriting the hemophilia allele and the
disease; any son of Victoria Eugenie also has a 50% chance of inheriting the normal allele and
not the disease. Since two sons had the disease while two did not, it conforms to our expectation
of possible progeny from Victoria Eugenie.
31. Based upon this information, what can you determine about the inheritance pattern of
hemophilia? [3pt]
Since hemophilia is passed down from mothers to son only, the trait is definitely X-linked. Since
daughters must have two hemophiliac alleles to have hemophilia, it is recessive.
for full credit, answer must state “X-linked recessive”
32. The present Queen Elizabeth II is the great-granddaughter of Edward VII (C). Her
husband is the great-grandson of Princess Alice (D). Base upon this information, what is
the probability that Queen Elizabeth’s son – Prince Charles – could have inherited
hemophilia [which he does not have, btw]? Explain why briefly. (4pt)
Since Queen Elizabeth is a great-great-granddaughter of Queen Victoria through a male lineage,
it is impossible for Queen Elizabeth to carry the hemophilia gene (unless her own mother had
hemophilia – which she did not). Even though Queen Elizabeth’s husband (Prince Philip) is
descended from Queen Victoria by an all-female lineage, he cannot pass the hemophiliac allele to
his son, since the gene is found on the X chromosome.
The probability Prince Charles could have inherited hemophilia is 0%.
15
6
Points:__________