Download PC Pc pC pc PC PPCC (purple) PPCc (purple) PpCC

Name: Answer Key
Discussion Section (circle one): 1pm, 2pm, or 3pm
Group Members:
BioSci 93 Discussion Week 8
Work with 2 or 3 students sitting next to you
1) In sweet peas, the purple flower color (P) is dominant over white (p). However, there is also a
gene that controls whether the color will be able to be expressed in the petals. The C allele
allows the color to be expressed, and it is dominant over the non-expressed c.
a. Write the phenotypes for a plant that is PPCC, a plant that is ppcc, and a plant that is
Ppcc.
PPCC = purple flowers, ppcc = white flowers, and Ppcc = white flowers. Note here
that the flowers would be white regardless of the alleles for the P gene, since the
plants are homozygous recessive (cc) for the second gene.
b. What gametes can the PPCC and ppcc plants produce?
The PPCC plant can only produce gametes with PC and the ppcc plant can only
produce gametes with pc.
c. What will be the genotype and phenotype of the F1 generation of plants produced
from a cross of these parents?
This cross is PPCC x ppcc. ALL of the offspring will have the genotype of PpCc and
the phenotype of purple.
d. You self-cross the plants from the F1 generation. What are the genotypes and
phenotypes of the offspring? Draw a Punnett square.
The first thing to do here is to determine the possible gametes that these plants can
produce: PC, Pc, pC, and pc.
PC
Pc
pC
pc
PC
PPCC
(purple)
PPCc
(purple)
PpCC
(purple)
PpCc
(purple)
Pc
PPCc
(purple)
PPcc
(white)
PpCc
(purple)
Ppcc
(white)
pC
PpCC
(purple)
PpCc
(purple)
ppCC
(white)
ppCc
(white)
pc
PpCc
(purple)
Ppcc
(white)
ppCc
(white)
ppcc
(white)
* Note that the offspring in bold have the phenotype for purple pigment masked by the
recessive alleles for pigment expression
1 of 3 e. Flower color in sweet pea plants is an example of epistasis. Circle the choices that
correctly describe the relationship between the P and C genes in this example:
The P / C gene is epistatic to the P / C gene
2) The pedigree below traces the inheritance of Tay-Sachs disease, a rare disease characterized
by the progressive deterioration of nerve cells and of mental and physical abilities that could
commence as early as six months after birth. Things to note:
1. Squares represents male and circles represent female
2. Shapes without a shade filling do not show a phenotype, whereas those that are shaded do
*Quick note: I-1 is actually Tt, not TT. I couldn’t fix the image~ Sorry!!!
a. Fill in the genotypes of the individual whose genotypes you know (TT, Tt, or tt).
b. Based on the pedigree, does Tay-Sachs disease appear to be caused by a dominant or
recessive allele?
Recessive allele
c. Could you have known the genotypes for individuals II-5 and II-6 before they had a
child? Explain your answer.
II-5: Yes, because the circle is not colored in you know that the individual must be
heterozygous. The mother is homozygous recessive (aa) and there are siblings that are
also homozygous recessive. Therefore, the father must be heterozygous (Aa) for the
2 of 3 disease, and therefore, IL-5 must also be heterozygous since the phenotype is not
shown.
II-6: No, since you don’t know the parents’ genotypes (or phenotypes) so you won’t
know the second allele until after they have a child. Once the child (III-4) is born
homozygous recessive, then you know that II-6 must be a carrier for Tay-Sachs
disease.
3) The mRNA for a newly discovered gene has a lot of secondary structure in its 5’ UTR. This
means that the mRNA folds back and base pairs with itself. The RNA structures that are formed
prevent the initiation complex from scanning along the mRNA. Given this information, which of
the following is likely to be TRUE?
a. Translation initiation will be more efficient for this mRNA than for an mRNA without
secondary structure in the 5’UTR
b. Because the initiation complex cannot scan the mRNA and locate the AUG initiation
codon, the large ribosomal subunit will not be recruited
c. As the 5’ polyA tail will appear shorter due to folding, this mRNA will be unstable
d. As the ribosome must scan the mRNA to assist in splicing, the introns will not be
removed from this mRNA
e. This mRNA will only be read by the ribosome 3’ to 5’ as translation, like DNA
replication, occurs bidirectionally
4) During your Bio199, the graduate student you are working with asks you to translate a
fragment of a cytosolic mRNA they isolated from a new species of tree frog they collected in
the Amazon rain forest:
5’ CGUUUAGCAUGCUUCUCCUCUGGUUAUAAGUGC 3’
What do you tell the graduate student?
a. Rare, Amazonian frogs probably do not use the mammalian genetic code; I need a
different table.
b. Arg-Leu-Ala-Cys-Phe-Ser-Ser-Gly-Tyr-Lys-Cys
c. The protein sequence cannot be determined, I need more information about this gene.
d. Met-Leu-Leu-Leu-Trp-Leu
Remember! You do not know if the AUG is at the start or the middle of the open reading frame
or if this is part of the UTR, so you cannot determine the protein sequence.
5) All of the following regulate gene expression EXCEPT:
a. Nuclear pore function
b. Histone acetylation
c. DNA sequence around the TATA box
d. The rate at which RNA polymerase splices out introns
e. The availability of specific transcription factors
6) As a new DNA strand is synthesized
a. Covalent bonds between the nucleotides are formed by a hydrolysis reaction
b. Nucleotides are added to the 3’ end
3 of 3 c. Nucleotides are linked to each other by ester bonds
d. DNA polymerase forms hydrogen bonds linking A to T, and C to G
4 of 3