Download Biology 120 Lab Exam 2 Review Session

Biology 120 Lab Exam 2 Review Session- ANSWER KEY
Important note: This review was written by a Biology Peer Mentor (not your professors or lab
coordinators)! It is designed to help you test yourself on topics and concepts covered in the lab
and should not in any way be considered a “preview” of the actual lab exam.
Section 1. Spot Test:
Question
Answer
1
A: Diploid
B: No
2
Telophase I
3
Diploid
4
No, because they are a different size (so would have a different number of
genes)
5
A: Archegonium/Archegonia
B: Antheridium/Antheridia
C: Rhizoid
6
A: Down Syndrome
B: Mental impairment, stunted growth, low muscle tone, large tongue,
flattened nose, almond shaped eyes, short neck. See Wikipedia.
7
A: Only one X chromosome (so missing an X or a Y)
B: Turner Syndrome
C: Short Stature, broad chest, low ears/hairline, infertile, webbed neck.
See Wikipedia.
8
A: Male
B:Klinefelter’s Syndrome
C: Sparse body hair, enlarged breasts, wide hips, small testicles, high
voice, sterile. See Wikipedia.
/16
Section 2. Fill in the blanks:
1
1. Ferns have a complex life cycle called an __alternation______ __of__
______generations______.
2. In the fern, ___sporocytes___ undergo meiosis to form haploid cells called spores.
3. Crossing over occurs in ___Prophase I___ of meiosis.
4. In the fern life cycle, the multicellular _____gametophyte____ stage is haploid, and the
_______sporophyte______ stage is diploid.
5. In animals meiosis only occurs in reproductive organs ( __testis___ or ____ovaries____),
and results in the production of gametes ( __sperm____ or ___eggs/ova/ovum____)
/10
Section 3. Definitions: 2 marks each
1. Heterozygous:
An individual or cell with two different alleles of the same gene OR both the dominant and
recessive alleles for a gene.
2. Genotype:
The genetic makeup of an organism.
3. Phenotype:
The physical appearance of an organism.
4. Allele:
The different forms of a gene OR just the form of a gene.
5. Test Cross:
A cross between a heterozygous OR unknown individual and a homozygous recessive
individual. Used to determine gene linkage or the genotype of an unknown.
6. Polygenic Traits:
Traits that are controlled/determined by the contribution of many genes.
7. Ploidy:
The number of sets/copies of chromosomes in a cell or organism
/14
Section 4. True or False:
Identify whether each of the following is True or False. If it is false, correct the statement.
__T__ The Y chromosome is smaller than the X Chromosome .
__T__ Interphase is the longest phase in the cell cycle.
__F__ Meiosis Mitosis produces two genetically identical offspring.
__T__ In animals, gametes are produced by meiosis.
__F__ S Stage of Mitosis the cell cycle is where chromosomes are duplicated by DNA
replication.
__F__ In a mouse sperm is produced by mitosis meiosis.
__F__ Phosphodiester Hydrogen bonds are formed between the complementary nitrogenous
bases of DNA. Or Phosphodiester are between adjacent nucleotides
__F__ In gel electrophoresis, sucrose salt infused water is used to conduct electricity.
__F__ DNA is collected from cells that lack with a nucleus for use in Paternity testing.
/9
2
Section 5. Short Answer:
1. Match each of the following fingerprints with the correct pattern. (3 marks)
a)
b)
c)
Loop ___C____
Whorl ___A____
Arch ___B____
2. Describe the differences between the products of mitosis and meiosis. (2 marks)
Mitosis: 2 identical cells with the same ploidy as the parent cell.
Meiosis: 4 non-identical cells with half the ploidy of the parent cell.
3. What is the difference between parental and recombinant offspring? (2 marks)
Recombinant offspring received gametes that are the result of crossing over, and are less
likely (so a lower number)
4. Briefly describe Mendel’s Laws. (4 marks)
Mendel’s First Law: Alleles of the same gene segregate from each other.
Mendel’s Second Law: Genes assort independently.
5. Describe, in detail, the structure of DNA. (4 marks)
•
•
•
•
•
•
Double Stranded
Nucleotides made of a nitrogenous base, sugar, and phosphate
Phosphodiester bonds between Nucleotides
Hydrogen bonds between strands
Negative Charge
Size, distance between turns, etc.
6. What events occur in prophase I of meiosis but NOT in prophase of mitosis? (2 marks)
Homologous chromosomes form Tetrads/Bivalents.
Crossing over occurs
7. Sally’s poodle was stolen on a Friday night. While looking inside her home for the dog
she found a piece of hair and a spatter of blood on the floor near the dog’s bed.
a. Which sample would be a better source of DNA? Why? (2 marks)
Blood, because it contains cells with nuclei. A hair does not have a nucleus unless it
has a follicle attached.
b. What three solutions would you use to extract the DNA? Briefly describe the role of
each solution. (6 marks)
Saline/Gatorade/Sports Drink: An isotonic solution that safely collects cells.
Detergent/Cell Lysis Solution: Breaks apart (lyses) the nuclear and cell membranes,
releasing the DNA in to the solution.
Ethanol/Alcohol/EtOH: Precipitates/separates the DNA from the solution.
3
c. What process can you use to create more copies of this DNA? Briefly describe each
step of this process. (4 marks)
Polymerase Chain Reaction (or PCR)
1. Heat: Denatures/seperates the two strands of the DNA helix
2. Cool: Allows primers to anneal/attach to the single stranded DNA
3. Heat: Activates taq/DNA polymerase, which extends the primers to create full
copies of the DNA.
8. List and briefly describe the steps of restriction fragment analysis. (5 marks)
Restriction Digest: Cut the DNA with a restriction enzyme
Gel Electrophoresis: Place fragments in a gel, run an electric current through the gel, and
separate DNA fragments by size
Interpret the results
9. Label the following image. (3 Marks)
Sporangium OR Sporangial Wall
Spores
Sporangial Stalk
10. Draw a chromosome diagram of a cell (2n=4) in G1 stage of the Cell Cycle. Are the
chromosomes replicated during this stage? (2 marks)
No, chromosomes
are not replicated
11. Draw a chromosome diagram of a cell (n=3) in Metaphase. (2 marks)
/41
4
Section 6. Genetics Problems:
1.
In garden peas, round peas (R) are dominant to wrinkled peas (r).
a. If you crossed a homozygous dominant pea with a homozygous recessive pea, what
would be the genotype and phenotype of the offspring? (2 marks)
Genotype: Aa
Phenotype: Round
b. Draw a chromosome diagram of an F1 plant (2n=2) in Prophase I. (2 marks)
2. After completing a cross between two lily plants with unknown genotypes, you find that 74
of the offspring have purple flowers and 26 have white. What were the possible genotypes
AND phenotypes of the parents? Use ‘P’ for the gene controlling flower colour. (2 marks)
Parents must both be Pp and have purple flowers.
We know because the 3 dominant:1 recessive ratio of offspring is a result of a Pp x Pp cross.
3.
In a grass species native to Saskatchewan, three leaves (T) are dominant to two leaves (t). If
a cross produces offspring that are 50% homozygous dominant and 50% heterozygous, what
are the genotypes of the parents? (1 mark)
The parents were TT and Tt.
4. 70% of Americans perceive a bitter taste from the chemical phenylthiocarbamide (PTC).
The ability to taste this chemical results from a dominant allele (T) and not being able to
taste PTC is the result of having two recessive alleles (t).
Albinism is also a single locus trait with normal pigment being dominant (A) and the lack
of pigment being recessive (a). A normally pigmented woman who cannot taste PTC has
a father who is an albino taster. She marries a homozygous, normally pigmented man
who is a taster but who has a mother that does not taste PTC.
a. What are the genotypes of the possible children? (1 mark)
At
at
AT
AATt
AaTt
At
AAtt
Aatt
So the children could be AATt, AAtt, AaTt, or Aatt.
b. What percentage of the children will be albino? (1 mark)
0%
c. What percentage of the children will be non-tasters of PTC? (1 mark)
50%
5
5. In a strain of soybeans, high oil (H) content in the seeds is dominant to low oil content,
and four seeds (E) in a pod is dominant to two seeds in a pod. A farmer crosses two
soybean plants, both with high oil content and four seeds per pod. The resulting F1
offspring have a phenotypic ratio of 9:3:3:1.
a. What genotype were the parent plants? (1 mark)
HhEe and HhEe. The 9:3:3:1 is another classic ratio, so no more work is
necessary.
6. The traits for fur colour and sharpness of nails are linked in rabbits. Brown fur (B) is
dominant to white (b) and sharp nails (N) are dominant to dull (n). Hannah’s mother had
brown fur and sharp nails, while her father had white hair and dull nails. If Hannah has
brown fur and sharp nails and mates with Joey, who has white fur and dull nails, what
will be the phenotypic and genotypic frequencies of the offspring? Assume no crossover.
(2 marks)
bn
BN
BbNn
Bn
Bbnn
bN
bbNn
bn
bbnn
0.5 BbNn, 0.5 bbnn
0.5 Brown fur and sharp nails, 0.5 white fur and dull nails.
7. Colour blindness (b) in humans is a sex linked recessive disorder. A carrier female
marries a normal male. (5 marks)
a. Draw Punnett square to illustrate this cross.
XB
Y
B
B B
X
X X
XBY
b
B b
X
X X
XbY
b. What proportion of the male offspring can be expected to be colorblind?
½
c. What proportion of the offspring can be expected to have normal vision?
¾
d. What proportion of the female offspring can be expected to be carriers?
½
e. What proportion of the female offspring can be expected to be phenotypically
normal?
100%
8. In spider monkeys, the allele for black fur (B) is dominant to the allele for grey fur (b),
and the allele for a long tail (T) is dominant to the allele for a short tail (t). The gene for
fur colour is 9.6 map units from the gene for tail length.
a. Draw a Punnett square to illustrate the cross between a monkey with grey fur and
a short tail and a monkey that was heterozygous for both characteristics. The
father of the heterozygous had black fur and a long tail, homozygous for both
traits. (1 mark)
bt
BT (P)
BbTt
45.2
Bt (R)
Bbtt
4.8
bT (R)
bbTt
4.8
Bt (P)
bbtt
45.2
b. Calculate the expected genotypic ratio for this cross. (1 mark)
9.42 BbTt: 1 Bbtt: 1 bbTt: 9.42 bbtt
/20
6
Section 7. Pedigrees:
1. The following pedigree illustrates the inheritance of cystic fibrosis in a population.
Use ‘F’ to represent the gene responsible.
A
Normal female
Normal male
Female with trait
C
B
Male with trait
D
a. What is most likely the mode of inheritance?
Autosomal Recessive
b. What is the genotype of A? ____Ff______
c. What is the genotype of D? ____ff_________
d. If individuals B and C have a son what is the chance that he will have cystic
fibrosis?
50%
2. The following pedigree illustrates the inheritance of polydactyly in cats.
Normal female
Normal male
Female with trait
Male with trait
a. What is the most likely mode of inheritance for this trait?
Autosomal Dominant
b. Give the genotypes of the following individuals. Use the letter “B” for this trait.
6 _____Bb______
7 _____Bb_______
8 _____bb_______
c.
If Individuals 6 and 7 have another daughter, what are the chances that she will be
affected?
75%, ¾
3. The following pedigree illustrates the inheritance of awesome looking hair.
a. What is the most likely mode of inheritance?
Autosomal Recessive
b. Using the letter “A” to represent the gene for awesome looking hair, give all
possible genotypes for the following individuals.
i. Individual 1: ____Aa or AA_____________________
ii. Individual 2: ____Aa or AA__________________________
iii. Individual 4: ____Aa only___________________________
7
4. The following pedigree illustrates the inheritance of Duchenne muscular dystrophy in a
British population. Use ‘D’ to represent the gene.
A
Normal female
C
B
D
Normal male
Female with trait
Male with trait
b. What is most likely the mode of inheritance for this disorder?
Autosomal OR Sex linked Recessive
c.
List all possible genotypes for each of the following:
i. Individual A: ____XdY or dd_____________________
ii. Individual B: _____XDXd or Dd__________________________
iii. Individual C: _____XDY or Dd________________________
d. If Individual D has a daughter with a woman who is a carrier, what is the
probability that she will have the disease?
Xd
Y
D
X
XDXd XDY
Xd
XdXd
XdY
Either way, 50% or ½
D
d
d
Dd
dd
d
Dd
dd
/18
Total:
/126
8