Description
1. Highschool, AP Biology, 27 questions (multiple choice + few short answers)
2. Snork Protein Synthesis
3.Mutations Practice
4.Watch the videos and take few notes. (doesn’t have to be fancy&long, but must be your own sentences. you can write your knowledge directly on it without watching the video, but must be built points and connect to the video) – Double-space,12,times new roman
* All work should be done on the document.
1/29/2021
Formative
Unit 5 Test
Name: _______________________
Date: _______________________
1
1
Which of the following is the correct use of genetic
terminology?
A heterozygous individual has two copies of the same
allele
A phenotype can be determined by two alleles in a
human
A haploid individual has two copies of a gene
Two genotypes combine to create one allele
2
1
A diploid cell in a gorilla has 48 chromosomes. How many
chromosomes will be present in a haploid gorilla cell?
24
12
96
48
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3
1
In lizards, blue is dominant to green and two-horned is
dominant to one horned. Which of the following crosses
most likely produced the following results: 56 blue, twohorned lizards, 18 blue, one-horned lizards, 20 green, twohorned lizards, and 6 green, one-horned lizards?
BBHH x bbhh
BBHh x BbHh
Bb x Bb
BbHh x BbHh
4
1
Which of the following cannot be determined by a
karyotype?
Biological sex of individual
If there were any translocations or deletions
If the individual has a heterozygous genotype
Number of chromosomes
5
1
A scientist determines that the number of dorsal (on the
back) spines of her experimental sh is determined by a
single gene. She mates a sh with one dorsal spine with a
sh of the same species with 2 dorsal spines. The mating
produces 10 progeny, each having 1 dorsal spine. Which of
the following statements could NOT be true?
2 dorsal spines is a recessive phenotype
The progeny sh are homozygous for the 1-spine allele
The 1-spine parent sh has a homozygous genotype
for the dorsal spine
The 2-spine parent sh has a homozygous genotype
for the dorsal spine
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Formative
6
1
The exchange of segments of DNA between homologous
chromosomes
produces four genetically identical gametes.
happens in prophase I.
is called pleiotropy.
produces genetic disorders.
7
1
Hemophilia is caused by an X-linked recessive allele. Only the
individuals shaded in on the pedigree below show the trait
for hemophilia. Which individuals must be carriers?

I-4 and II-2
III-2 and III-3
I-1 and II-1
II-1 only
8
What are autosomes?
1
Chromosomes #1-22
The cells produced by meiosis
The cells produced by mitosis
Chromosome #23
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9
What type of inheritance is this pedigree tracking?
1

Sex linked recessive
Autosomal dominant
Autosomal recessive
Sex linked dominant
10
1
Which of the following generates new combinations of
alleles?
I. Crossing over
II. Independent Assortment III. Random
Fertilization IV. gene linkage
I
I, III and IV
I, II and III
I and IV
I, II, III, IV
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Formative
11
1
Hunter syndrome, or mucopolysaccharidosis II (MPS II), is a
lysosomal storage disease caused by a de cient (or absent)
enzyme, iduronate-2-sulfatase (I2S). The symptoms of
Hunter syndrome (MPS II) are generally not apparent at
birth, but usually start to become noticeable after the rst
year of life. Often, the rst symptoms may include
abdominal hernias, ear infections, runny noses, and colds.
Since these symptoms are quite common among all infants,
they are not likely to lead a doctor to make a diagnosis of
Hunter syndrome right away. As the buildup of
glycosaminoglycans (GAGs) continues throughout the cells of
the body, signs of Hunter syndrome become more visible.
Physical appearances of many children with Hunter
syndrome include a distinctive coarseness in their facial
features, including a prominent forehead, a nose with a
attened bridge, and an enlarged tongue. For this reason,
unrelated children with Hunter syndrome often look alike.
They may also have a large head, as well as an enlarged
abdomen. Many continue to have frequent infections of the
ears and respiratory tract. There are only about 2,000 people
worldwide with Hunter syndrome and most of them are
males. The most likely reason for this would be
The de cient enzyme, IS2 must only cause a buildup of
GAGs in haploid individuals
The gene for Hunter syndrome is located on the y
chromosome
The allele for Hunter syndrome is inherited from the
father’s homologous chromosome during meiosis II
Hunter syndrome is X linked recessive
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12
1

Given the diagram and the chart, determine which letter
goes with which gene on the chromosome map.
A=body color, B=wing size, C=eye color
A=eye color, B=body color, C=wing size
A=body color, B=eye color, C=wing size
A=wing size, B=eye color, C=body color
13
1
Which of the following explains a signi cantly low rate of
crossing over between two genes?
The genes code for proteins that have similar functions
The genes code for proteins that have very di erent
functions
The genes are located very close together on the same
chromosome
They are located far apart on the same chromosome
14
1
Which of the following is not a possible result of
nondisjunction?
Down syndrome
A gamete has one chromosome fewer than it should
Sister chromatids divide evenly
A gamete contains two number 3 chromosomes
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15
1
In llamas, coat color is controlled by a gene that exists in two
allelic forms. If a homozygous yellow llama is crossed with a
homozygous brown llama, the o spring have grey coats. If
two of the gray-coated o spring were crossed, what
percentage of their o spring would have brown coats?
25%
100%
75%
50%
16
1
At the end of meiosis 1 the cells are ________________ while at
the end of meiosis 2 the cells are ________________.
Haploid, n, duplicated
Haploid, n unduplicated
Diploid, 2n, duplicated
Diploid, n, unduplicated
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haploid, n, unduplicated
diploid, 2n, duplicated
haploid, n, unduplicated
haploid, n, unduplicated
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Formative
17
1
You are doing an experiment with 4 degrees of freedom and
you are using a p value of 0.05. Which of the following chi
square values shows the largest di erence between
observed and expected values AND would you would reject
the null hypothesis?

a
b
c
d
18
1
What is the di erence between sister chromatids and
homologous chromosomes?
Sister chromatids are haploid while homologous
chromosomes are diploid.
Sister chromatids separate in anaphase I while
homologous chromosomes separate in anaphase II.
A homologous chromosome is one chromosome from
one from mom and one from dad while sister
chromatids are duplications of the same chromosome.
A homologous chromosome is only present in mitosis
while sister chromatids are only present in meiosis.
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19
1
Skin color isn’t controlled by just one gene with two alleles.
Scientists know of at least 34 genes that all equally
contribute to determine a person’s skin color. During the
summer, when skin is exposed to more UV rays, people
often become tan, which alters their “natural†genetic skin
tone to a darker one. This is an example of
A dihybrid cross
The environment having an e ect on the phenotype
Incomplete dominance
The Law of Segregation
20
How are phenotypes controlled in organisms?
1
A single gene with two alleles
Multiple genes that act together
One gene a ecting another gene
All of the above are possible
21
1
The o spring of Organism A that reproduces sexually are
compared to the o spring of Organism B that reproduces
asexually. How would the cells be di erent?
Organism A’s o spring will have twice the genetic
material than Organism B’s.
Organism A’s o spring will be genetically identical to it,
while Organism B’s o spring will be genetically
di erent.
Organism B’s o spring will be genetically identical to it,
while Organism A’s o spring will be genetically
di erent.
Organism B’s o spring will have twice the genetic
material than Organism A’s.
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Formative
22
1
In mice, having brown fur is dominant to having black fur.
There’s a second gene in mice that also a ects fur color. This
gene is a pigment gene. Individuals who are homozygous
recessive for the pigment gene will have no fur color (white
fur). This is an example of
Linked genes
Codominance
Incomplete dominance
Epistasis
23
3
Horses have either curly (recessive) or smooth (dominant)
hair and this trait is controlled by only one gene (much like
Mendel’s pea plants). You are trying to determine the
genotype of one of the smooth individuals. Devise an
experiment to determine the genotype and explain how you
could come to a conclusion about the individual’s genotype
by using the results.
_____________________________________________________________________
_____________________________________________________________________
______________________________________________________________
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10/13
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Formative
Tigers (Panthera tigris) with a wild-type phenotype have orange
fur (o+) and stripes (s+). Certain mutations can cause changes
to these traits. Mutant tigers may have a white body (o) and/or
no stripes (s). To study the genetics of these traits, a researcher
crossed a true-breeding wild-type male tiger with a truebreeding mutant female tiger. All of the F1 progeny displayed a
wild-type phenotype. Female tigers from the F1 generation
were crossed with homozygous mutant male tigers. Table 1
represents the researcher’s data table for the F2 generation.

24
State the genotypes of the parents of the F2 generation.
2
_____________________________________________________________________
_____________________________________________________________________
______________________________________________________________
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Formative
25
2
26
5
Calculate the predicted amount of tigers with each
phenotype according to the parent’s genotypes and a
normal dihybrid cross if there were 24 total F2 o spring. Fill
in the values in the “Number Expected†column.
State whether the amount of observed individuals of each
phenotype is due to random chance or not. Justify your
statement with evidence. (NOTE: this is worth 5 points…so
you should be explaining quite thoroughly here!)
_____________________________________________________________________
_____________________________________________________________________
______________________________________________________________
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Formative
27
Propose an explanation for the observed number of tigers.
1
_____________________________________________________________________
_____________________________________________________________________
______________________________________________________________
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DNA, RNA and Snorks
Snorks were discovered on the planet Dee Enae in a distant solar system. Snorks have only one chromosome with eight
genes on it. Based off of the snork DNA, determine the mRNA transcript, and translate the mRNA into amino acids.
Snork genes are listed in order and separated by a vertical line. To determine the amino acid, use the chart you were given
in class. You will determine what physical traits the Snork has based on the table below. Each of the samples of DNA was
taken from volunteer Snorks and each Snork gave their consent to have their DNA studied.
Genes
Amino Acid Sequence
Description
Gene 1- Body Covering
val – ser – leu
hairless
val – ser – lys
hairy
tyr – pro – glu – glu – lys
plump
val – pro – thr – glu – lys
skinny
leu – leu – leu – pro
3 legged
leu – leu – ser – ala
2 legged
ala – ala – val
round head
val – ala – ala
square head
his – ile
tail
his – his
no tail
ser – pro – val
blue pigment (hair and skin)
val – phe – tyr
red pigment (hair and skin)
asp – ile – leu – leu – pro – thr
small slanted eyes
asp – ile – pro – pro – pro – thr
large round eyes
val – asp – asp – ala
circular mouth
asp – asp – asp – ala
rectangular mouth
phe – ser – gly
pointed standing up ears
phe – phe – gly
rounded floppy ears
arg – tyr – cys – lys
long spaghetti like arms
arg – arg – asp – thr
short stumpy arms
Gene 2- Body Style
Gene 3- Legs
Gene 4- Head Shape
Gene 5- Tails
Gene 6- Body Pigment
Gene 7- Eyes
Gene 8- Mouth
Gene 9- Ears
Gene 10- Arms
Snicker Snork
DNA: CAG TCG TTT | ATG GGG CTT CTC TTT | GAG AAT TCA CGC | CGA CGA CAC |
mRNA:
aminos:
DNA: GTA GTA | CAA AAA ATG | CTA TAG AAT GAC GGG TGG | CTG CTG CTA CGG |
mRNA:
aminos:
DNA: AAA AGA CCC | TCT ATA ACA TTT
mRNA:
aminos:
Physical Description:
Snuffle Snork
DNA: CAT AGA TTT | CAA GGA TGA CTT TTC | GAA GAG GAG GGG | CAA CGC CGA |
mRNA:
aminos:
DNA: GTA GTG | CAT AAA ATA | CTA TAA GAA GAC GGG TGT | CAA CTG CTG CGT |
mRNA:
aminos:
DNA: AAG AGC CCA | TCT ATA ACA TTC
mRNA:
aminos:
Physical Description:
Snapple Snork
DNA: CAG TCG GAA | CAA GGG TGT CTT TTT | GAG AAT TCA CGC | CAA CGC CGA |
mRNA:
aminos:
DNA: GTG TAA | AGA GGG CAT | CTA TAA GGG GGG GGG TGG | CTA CTG CTG CGT |
mRNA:
aminos:
DNA: AAG AAA CCC | GCC GCC CTG TGG
mRNA:
aminos:
Physical Description:
Snoopy Snork
DNA: CAT AGG GAG | ATG GGG CTC CTT TTT | AAT AAT GAC GGG | CGA CGA CAT |
mRNA:
aminos:
DNA: GTA TAA | AGA GGG CAT | CTA TAA GAA GAC GGG TGT | CAA CTA CTA CGG |
mRNA:
aminos:
DNA: AAA AGA CCA | GCG GCA CTG TGA
mRNA:
aminos:
Physical Description:
Choose 2 of the Snorks and sketch them below/in your notebook.
Mutations Practice
Transcribe and translated the original DNA sequence. Then, do the same for each mutated sequence and determine
the consequence, if any, for each mutation. Remember to compare the mutated DNA to the original DNA. You will
need to use a codon chart to help you. Don’t forget to compare each mutated sequence to the original
sequence to help you figure out where/what the mutation is.
Original DNA
sequence:
TAC
ACC
TTG
GCG
ACG
ACT
TAC
ACT
TTG
GCG
ACG
ACT
CTT
GGC
GAC
GAC T
mRNA:
Amino acids:
Mutated DNA
sequence #1:
mRNA:
Amino acids:
Is this a point or frameshift mutation?
How did this mutation affect the amino acid sequence?
How would this mutation affect the protein that is made?
Mutated DNA
sequence #2:
TAC
GAC
mRNA:
Amino acids:
Is this a point or frameshift mutation?
How did this mutation affect the amino acid sequence?
How would this mutation affect the protein that is made?
Mutated DNA
sequence #3:
TAC
ACC
TTC
GCG
ACG
ACT
TTG
GCG
ACA
ACT
TTG
GGA
CGA
CT
mRNA:
Amino acids:
Is this a point or frameshift mutation?
How did this mutation affect the amino acid sequence?
How would this mutation affect the protein that is made?
Mutated DNA
sequence #4:
TAC
ACC
mRNA:
Amino acids:
Is this a point or frameshift mutation?
How did this mutation affect the amino acid sequence?
How would this mutation affect the protein that is made?
Mutated DNA
sequence #5:
TAC
ACC
mRNA:
Amino acids:
Is this a point or frameshift mutation?
How did this mutation affect the amino acid sequence?
How would this mutation affect the protein that is made?
Analysis Questions:
1. Examine your codon chart. Name one amino acid that has more than one codon.
2. Examine your codon chart. Name one amino acid that has only one codon.
3. Look at the following sequence: THE BADBAD RRAT ATE THE OLD BOLD BAT. Delete the first A and
regroup the letters in groups of 3. Write out the new groups of 3. Does this sentence still make sense? What
kind of mutation is this an example of?
New groups of 3:
Still make sense?
Kind of mutation:
4. Given the following 3 mRNA sequences, determine which 2 code for the same protein.
Transcript
mRNA #1
mRNA #2
mRNA #3
AGU UUA GCA ACG AGA UCA
UCG UUG GCG ACA AGU UCA
AGC CUC GCC ACU CGU AGU
Translate
Which 2 code for the same protein?
5. You have a DNA sequence that codes for a protein and is 21 nucleotides long. A frameshift mutation occurs
at the 16th base. How many amino acids will be correct in this protein? Explain your answer.
Watch the lecture about mutations and jot down some notes. Then, submit your notes to this assignment as either
pictures inserted into a google doc, or a scan.
Here’s the lecture link: https://www.youtube.com/watch?v=yAw_wY19Mcl
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