EXAM II Review
Questions:
1.
2.
Who was Mendel and what were his contributions to Genetics?
a. Did experimental crosses with Peas – determined the idea that we have two of each gene – on homologous chromosomes. And
that these homologous chromosomes separate during the production of egg and sperm (meiosis) to put only one of each
homologous pair in each daughter cell. So, he figured out that we’re diploid, and that chromosomes do independent assortment.
A man denies fathering a child of blood type A. He is blood type B. Could this man be the child’s biological father? Does this test prove
for certain that he is or isn’t?
a.
3.
A man with blood type A and a woman with blood type B have three children: A daughter with type AB and two sons, one with type B
and the other with type O blood. Remembering that blood type is expressed with codominant inheritance, determine the genotypes of the
parents.
a.
4.
6.
7.
8.
Man is phenotype A, so could be AA or AO. Woman is phenotype B, so could be BB or BO. The fact that one of
their children is O, and therefore OO, means that the parents have to be AO and BO.
If the hair of the field mouse was controlled by genes that were expressed with Complete dominance, what color would the hair of a “Bb”
individual be, where B is for Black and b is for the white trait. What if the genes were expressed with Incomplete Dominance? What if the
genes were expressed with Co-dominance
a.
b.
c.
5.
Child is phenotype A, so is either AA or AO. Man is phenotype B, so is either BB or BO. The man could be the
father, if his genotype is BO and the child’s genotype is AO. It doesn’t prove that he is for certain, because we don’t
know their exact genotypes, and there are lots of men with B blood.
Complete dominance: Bb = black
Incomplete dominance: Bb = gray
Codominance: Bb = white and black
What gametes (sex cells) could be produced during Meiosis in the following individual? AaBb
a. AB, Ab, aB, ab
What gametes could be produced during Meiosis in the following individual? AABb
a. AB, Ab
What does Independent Assortment have to do with the two questions above?
a. These different combinations occur because the chromosome pair that holds gene A is assorting independently from the
chromosome that holds gene B. So, all combinations of these alleles are possible.
Coat color in Labrador retrievers is governed by two genes: Bb and Ee.
B=black pigment
b=brown pigment
E=allows pigment to be deposited
e=does not allow pigment to be deposited
The B allele is dominant to the b, and E is dominant to e. The ee genotype specifies a golden lab regardless of what B/b allele combination
it carries. This is because the black or brown pigment cannot be deposited in a proper manner. If a heterozygous black lab is mated to a
black lab, and they have some golden puppies:
What are the genotypes of the parents?
What is the chance that these parents will be able to generate black lab puppies (Hint: do a dihybrid punnet square)? Brown labs? Golden
labs?
i. Black puppies – must have at least one dominant B and one dominant E – 9/16
ii. Brown puppies – must have two recessive b and at least one dominant E – 3/16
iii. Golden puppies – must have two recessive e – 4/16 or ¼
In a litter of 12 puppies, how many theoretically will be golden based on your answer in (b)? 3 puppies
9.
If a gene is “X-linked”, why does the trait usually show up in men?
a. Carrier mothers pass it to their sons (the son gets the Y from the father, and therefore always gets his X from mom).
Dad’s pass their X to their daughters, but since females have another X that could carry a dominant allele, the X
linked recessive trait is often “hidden” in the females. These females are carriers and pass it to their sons, who only
have one X.
10. The following mRNA was transcribed from a gene and has not yet been modified before Translation. The uppercase letters
represent exons. The lowercase letters represent introns.
5’–G C U A A A U G G C A G C A C A U U G A C U C G G G G U C A G A U C C G–3’
Write the double-stranded DNA segment that served as the template.
Indicate both the 5’ and the 3’ ends of both DNA strands.
3’ CGATTTACCGTtttaacggCGUGUAACUGAGCCttagctCCAGUCUAGGC 5’
5’ GGTAAATGGCAaaattgccGCACATTGACTCGGaatcgaGGTCAGATCCG 3’
Also, translate the protein that is encoded by this mRNA message.
MET-ALA-ALA-HIS
11.
Which is the correct order of events during the production of proteins?
a. A DNA sequence is transcribed into an amino acid sequence, which is used to make messenger RNA.
b. A DNA sequence is copied into messenger RNA, which is translated into an amino acid sequence
c. Messenger RNA is used to make protein, which is copied into a DNA sequence.
d. Amino acid sequences are copied into mRNA, which is used to translate protein
12. Long stems are dominant to short stems, purple flowers are dominant to white, and round seeds are dominant to wrinkled.
Each trait is determined by a single, different gene. A plant that is heterozygous at all three genes is self-crossed, and 2048
progeny are examined. How many of these plants would you expect to be long stemmed with purple flowers, producing
wrinkled seeds?
Well, this is a tri-hybrid cross – lots of potential offspring in the punnet square. .. . Possible gametes are: ABC, ABc, AbC,
Abc, aBC, abC, aBc, abc 8 gametes on top and 8 gametes on the side of a Punnett create 64 squares in between. Then
count up how many squares out of 64 have a dominant A, dominant B and recessive c phenotype. Take that percentage of
the 2048 progeny to find out how many offspring have that phenotype.
13. When and why does DNA need to replicate? How might the structure of DNA allow replication to occur with accuracy and efficiency?
DNA must replicate whenever a cell divides – during S phase of the cell cycle.
Because there are two strands that have very clear binding rules – A-T, G-C - it is clear that unzipping between the two strands
creates a template for replication that ensures the preservation of the original sequence.
14. Why shouldn’t we inbreed?
When we have children with individuals within our immediate family, we run a greater risk of mating two heterozygotes for any particular gene.
In the heterozygote, the dominant allele masks the recessive trait, but the offspring of two heterozygotes may have a homozygous recessive
genotype for a trait that may confer a disease. So, in inbred families we see a greater incidence of negative recessive traits.
15. If one strand of a double-stranded DNA chromosome has the sequence:
A T T G G C A T T A C G T A T C C G
What is the sequence of the other strand?
TAACCGTAATGCATAGGC
16. Why would a DNA structure in which each base type (A,T,G,C) could form hydrogen bonds with any of the other three bases
not produce a molecule that could be easily replicated?
17. What were the key pieces of evidence that were critical in determining the structure of DNA? Who were the scientists that
did the work in these areas?
18. What is the mode of inheritance for this trait? Autosomal Dominant
19. What is the mode of inheritance for this trait? X linked Recessive
20. What is the mode of inheritance for this trait? Autosomal Dominant
21. Why must DNA be replicated using both Leading and Lagging strands?
Because DNA strands run antiparallel, and DNA Polymerase can only add to the 3’ end of a new strand . . . . So, DNA must replicate both
strands ultimately toward each replication fork, but it must add nucleotides for one strand in the direction AWAY from the replication fork.
Therefore, at each fork, one strand has nucleotides adding toward the fork continuously, and the other strand adds nucleotides in the direction
away from the fork in fragments – this creates leading and lagging strands.
22. The following is a diagram of a DNA replication bubble from human DNA. This DNA has been unwound and unzipped at
an origin sequence (O) and DNA replication is about to commence. . Begin at the origin and draw arrows to represent the
direction of nucleotide addition for each new DNA strand created. Make the arrows long or short to denote leading
(continuous) or lagging (fragmented) strands, and label them as such. Label each end of a new strand as either 3’ or 5’.
O
5’
\
3’
3’
5’
23. The following is a diagram of DNA from fictional bacteria from the fictional planet, Vignatia. This DNA has been unwound
and unzipped at an origin sequence (O) and DNA Replication is about to commence. Note that this DNA is PARALLEL.
Begin at the origin and draw arrows to represent the direction of nucleotide addition for each new DNA strand created. Make
the arrows long or short to denote leading (continuous) or lagging (fragmented) strands, and label them as such. Label each
end of a new strand as either 3’ or 5’. Assume that Vignatian enzymes are functionally identical to those on Earth and that
Vignatian nucleotides are able to hydrogen bond A to T and G to C in a parallel fashion.
O
3’
5’
3’
5’
O