Download Practise Midterm Exam

Biology 105: Midterm EXAM Spring 2014
Name ____________________________Student ID _______________________
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Make sure you have all 10 pages
You can use the back-side of the pages for scratch, but we will not grade answers
written on the back-side of the page. Unsolved fractions are acceptable answers
Part I One word answers
16 points
Part II Multiple Choice
24 points
Part III Short Answer
32 points
Part IV Extended Calculation
28 points
Total
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One word Answers and True/False (16 points)
In the space provided next to each definition, clearly write the appropriate term.
1
__nucleotides__ The building blocks of DNA and RNA
2
___genotype_ The genetic composition of an organism
3
___ haploid _ A cell with 1N chromosomes
4
___amino acids _ The 20 building blocks of proteins
5
__diploid__ A cell with 2N chromosomes
6
__transcription__ The synthesis of RNA using one strand of DNA as a
template
7
__phenotype__ An observed characteristic of an organism
8
__homozygous__ An organism with 2 identical alleles for the same gene
9
__eukaryote__ An organism with genetic material inside a nucleus
10
_heterozygous___ An organism with 2 different alleles for the same gene
11
_autosomal___ A gene that lies on any chromosome except the sex
chromosomes
12
Nucleosome A repeating structure of chromatin composed of histones
and DNA
13
You have a true breeding brown-eyed stock of flies and true breeding red-eyed
stock of flies. Brown eyed flies are crossed with red-eyed flies and the F1 generation are
self crossed. In the F2 you have a 3: 1 ratio of brown to red-eyed phenotypes for both
males and female flies.
For each statement below, circle the T preceding each statement that is true and the F
preceding each statement that is false. More than one statement may be true/false.
T
F
a. The mode of inheritance is most likely recessive
T
F
b. The mode of inheritance is most likely autosomal
T
F
c. The mode of inheritance includes two different genes for eye color
T
F
d. The red eyed flies are recessive to the brown eyed flies
Multiple Choice (24 points)
1
Gene Promoters interact with
A
DNA polymerase
B
RNA polymerase
C
Reverse transcriptase
D
Ribosomes
E
all of the above
2
Which of the following genetic elements are translated into a protein
A
exons
B
introns
C
ribosome binding site
D
stop codon
E
promoter
3
Primary transcripts in the eukaryotic nucleus have
A
exons only
B
exons and introns
C
introns only
D
none of the above
E
all of the abovese
4
A chromosome is discovered in which all of the transcripts originate from the
same DNA strand. This implies that in this chromosome
A
promoters are all oriented in the same direction
B
adjacent promoters are oriented in opposing directions
C
promoters are randomly oriented
D
none of the above
E
all of the above
5
The genetic code is a three-letter code. If the code were a four-letter code, how
many codons would exist in the genetic code?
A
4
B
16
C
64
D
256
E
none
6
Which of the following are examples of a dihybrid cross
A
Aa x Aa
B
AaBb x aaBb
C
AaBBCc x AaBBCc
D
AaBBCCDD x aabbccdd
E
none of the above
7
If you cross a Gg (G=yellow, g=green) Ww (W=round w=wrinkled) individual
with a ggWw individual what would be your expected phenotype ratio
A
3:3:1:1
B
9:3:3:1
C
7:4:3:2
D
1:1:1:1
E
9:7
8
if we cross 2 plants that are both heterozygous at four loci, what fraction of the
progeny will be homozygous for all four recessive alleles
A
1/16
B
1/64
C
6/128
D
1/256
E
12/256
9
Which of the following does not occur during meiosisI
A
recombination
B
replication of homologous chromosomes
C
separation of homologous chromosomes
D
pairing of homologous chromosomes
E
separation of sister chromatids
10
Red/green color blindness is controlled by an X linked gene in humans. A normal
man and woman marry. The fathers of both of these individuals are color blind but the
mothers of both are homozygous normal. What is the probability that their first child will
be color blind
A
1/2
B
1/3
C
1/4
D
1/8
E
0
11
Individuals heterozygous for a dominant eye mutation S have small eyes. The S
allele is also associated with recessive lethality. When two heterozygotes are crossed,
normal-eyed and small eyed individuals are expected in a
A
1:1 ratio
B
2:1 ratio
C
1:2 ratio
D
1:3 ratio
E
3:1 ratio
12
Mendels law of independent assortment has its physical basis in the
A
separation of alleles into haploid cells
B
spindle attachment in anaphaseI
C
random arrangement of chromosomes on the metaphase plate in meiosisI
D
formation of new cell walls
E
none of the above
Short answers
1
Shown below is the DNA sequence of a gene that encodes a short protein, and
also the sequence of the mRNA synthesized from this gene.
Genomic DNA sequence:
5'-AGCTCATGTGCGAGTCCTGACGCTGACGTAGG-3'
3'-TCGAGTACACGCTCAGGACTGCGACTGCATCC-5'
Mature mRNA sequence:
5'-UCAUGUGCGAACGCUGACGUAGG-3'
i)
Which strand is the template strand for the RNA?
ii)
In the genomic DNA sequence shown above, draw boxes around the exons.
iii)
Write the sequence of the peptide encoded by this gene. Indicate the NH3 and
the -COO ends of the peptide.
+
2
An X-linked dominant allele causes hypophosphatemia in humans. A man with
hypophosphatemia marries a normal woman. What proportion of their sons will have the
disease?
0%
3
The karyotype of a child is shown below. What event during meiosis can best
explain the child’s observed karyotype?
Non-dysjunction
In which parent and at what stage during meiosis did this event occur? (there are two
possibilities) Father meiosis I or mother meiosis II
4
Consider the pedigree where two first cousins have a son. A recessive X-linked
disease affects female I-1 while male I-2 does not have the disease. Individuals II-1 and
II-4 neither have nor are carriers of the disease. Individuals in generations II through IV
may or may not have the disease.
A)
100%
B)
0%
C)
100%
D)
100%
E)
50%
F)
50%
What is the probability that male II-2 will have the disease?_______________
5
Assume this is the pedigree of a family with retinitis pigmentosa.
What is the probability that female II-3 will have the disease?______________
What is the probability that II-3 will be a carrier for the disease trait?__________
What is the probability that III-1 will be a carrier for the disease trait?__________
What is the probability that male III-2 will have the disease? ____________
What is the probability that male IV-1 will have the disease?_____________
What is the most likely mode of inheritance of this trait?
Y linked inheritance.
Can you determine if this trait is dominant or recessive? Explain your answer
If you can determine that the trait is dominant or recessive, what is it?
(You cannot determine if this is dominant or recessive)
Using the nomenclature R= normal allele and r = mutant allele
What is the genotype of individual :
II-1 (X/Yr),
III-2 (X/YR)
III-5 (X/Yr)
6
Two hypothetical autosomal mouse genes XEY and STH are tightly linked and
only 2 map units apart.
Dominant alleles of XEYD cause crossed eyes.
Dominant alleles of STHD cause small toes.
A heterozygous cross-eyed, small toed female mouse mates with a true breeding normal
eyed normal toed male and they have twenty offspring. Ten of the offspring are only
cross-eyed and the other ten have only small toes.
Would you expect either of the female’s parents to be cross-eyed AND have small toes?
What is the MOST LIKELY phenotype of each of her parents?
NO. One parent is most likely cross-eyed and the other parent is most likely small
toed.
7
You have two true-breeding varieties of elongated pumpkins (strain#1 and
strain#2). True breeding wild-type pumpkin strains are round. When crossed to wild type,
strain#1 and strain#2 produce round pumpkins.
Is the phenotype of strain#1 and strain#2 dominant or recessive to the phenotype of the
wild-type strain?
Recessive
You cross true breeding strain#1 with true breeding strain#2 and find that all F1
pumpkins are elongated. After self-fertilization you find that all F2 pumpkins are
elongated.
What are the most likely genotype of the parents, the F1 and F2 generation plants?
Strain1
aa
Strain2
aa
F1
aa
F2
aa
8
In flies the character vestigial wings (v) is autosomal recessive to long wings (V).
White eyes (w) is an X-linked recessive to red eyes (W).
In an F1 the following results are obtained:
Males ½ white eyed vestigial ½ white eyed long wings
Females ½ red eyed vestigial ½ red eyed long wings
What were the genotypes of the parents for the eye color and wing shape gene
V/v and v/v
W/Y and w/w
Extended Calculation:
1
A true breeding mutant mouse strain exhibits two different traits (curly tail and no ears).
When a true breeding male with curly tail and no ears is crossed to a true breeding wildtype female, all of the female F 1 progeny exhibit both mutant traits whereas all of the
male F 1 progeny look wild type.
What is the most likely mode of inheritance of the two traits?
X-linked dominant
Write the genotypes of the parents and F1 progeny
Male=Xab/Y x female= X++/X++
F1= Xab/X++ females and X++/Y males
The male and female F1 mice described above are crossed to one another to produce F2
progeny. Of the male F2 progeny, 40% have curly tail and no ears, 40% are wild type
while 10% have only curly tail and the remainder is only earless. What fraction of the
female F2 progeny would you expect to have both traits?
40%
A----------------------B
(P) A-----------------B (40%)
A----------------------B
(R) A-----------------+ (10%)
X
+------------------------+
(R) +------------------B (10%)
+------------------------+
(P) +-------------------+(40%)
Since A and B are dominant, XabX++ will show both traits i.e. 40%
What is the map distance (in cM) between the genes for the two traits?
20CM 10+10/100=20CM
3
Eyeless, spineless and colorless are recessive traits. A zebra fish that is heterozygous for
eyeless (ey/+) spineless (sp/+) and colorless (cl/+) is test crossed to an eyeless (ey/ey),
spineless (sp/sp) colorless fish (cl/cl). The following progeny are obtained:
Eyeless
420
Spineless colorless
380
Spineless
Eyeless colorless
Colorless
Eyeless spineless
Eyeless spineless colorless
WT
Total
44
45
53
48
5
5
1000
What is the correct gene order?
Sp----ey----cl
What is the correct phase in the heterozygous?
Sp+-------ey---------Cl+
Sp--------Ey+-------cl
What is map distance between Spineless and Eyeless genes?
Sp + cl
+ ey +
53+48+5+5
1000
11%
What is map distance between Eyeless and Colorless genes?
45+44+5+5
10%
1000
Is interference occurring in this cross? Explain your answer.
No