Answer all parts (A-G). Please use COMPLETE SENTENCES for any TEXT RESPONSE. (A-
ID: 300034 • Letter: A
Question
Answer all parts (A-G). Please use COMPLETE SENTENCES for any TEXT RESPONSE. (A-G)
For this problem – to obtain FULL CREDIT:
clearly diagram the crosses (full genotype and phenotypes for parents and progeny)
state the FULL hypothesis being tested
make a table showing the phenotypic classes and the observed and expected values
state the degree(s) of freedom
state your conclusion based on your calculation, and EXPLAIN how your chi-square analysis supports your conclusion
Question 3
SpongeBob SquarePants and Patrick harvest 500 sea anemone progeny (from a cross between two sea anemones, Earl and Pearl) that are
300 purple and long-spiked
75 purple and short-spiked
85 red and long-spiked
40 red and short-spiked
SpongeBob develops the hypothesis that he is observing the progeny from a dihybrid cross, with two different genes directing the color of the sea anemone and the length of its spikes.
Help Patrick EVALUATE SpongeBob’s hypothesis. In your evaluation, make sure that you: clearly diagram the crosses, state the hypothesis you are testing, make a table showing your phenotypic classes and the observed and expected values, state the degree(s) of freedom, and state your conclusion based on your calculation (see the instructions above). CLEARLY SHOW YOUR WORK FOR YOUR CALCULATION. Please respond to EACH PART of this question.
Use the letters A and a for anemone color and B and b for spike length
A. Fully state SpongeBob’s hypothesis:
B. Fill-in all possible GENOTYPES for the following sea anemone PHENOTYPES:
Use the letters A and a for anemone color and B and b for spike length
Phenotype
Genotype
purple and long
purple and short
red and long
red and short
C. Diagram the ACTUAL CROSS that resulted in the sea anemone progeny you are studying (GENOTYPE and PHENOTYPE of the PARENTS (Earl and Pearl) and PROGENY in this cross)
Use the letters A and a for anemone color and B and b for spike length
CORN
GENOTYPE
PHENOTYPE
PARENT 1
Earl (AaBb)
PARENT 2
Pearl (AaBb)
PROGENY
D. Fill in the OBSERVED and EXPECTED progeny numbers for the cross you are analyzing.
PHENOTYPIC CLASS
OBSERVED progeny
EXPECTED progeny
purple and long
purple and short
red and long
red and short
Total progeny--->
E. In your evaluation of this hypothesis, what is the “degree(s) of freedom” you are using? Show your CALCULATION.
F. SHOW YOUR CALCULATION of the c2 from the data provided. (Use the table!)
x2 = [ (observed value – expected value)2 ]/(expected value)
Fill in the table below to show your calculation of the c2 . Up to 2 decimals in answer, eg. “1.88”
Phenotypic class
observed
expected
(o – e)2
(o – e)2
e
Total Progeny -->
SUM of terms, or x2 -->
G. Based on your calculation, do the data and chi-square analysis allow you to reject or fail to reject your lab partner’s hypothesis? Please CLEARLY STATE YOUR CONCLUSION and SUPPORT YOUR CONCLUSION with your analysis. Remember to USE COMPLETE SENTENCES in your conclusion.
SpongeBob SquarePants and Patrick harvest 500 sea anemone progeny (from a cross between two sea anemones, Earl and Pearl) that are
300 purple and long-spiked
75 purple and short-spiked
85 red and long-spiked
40 red and short-spiked
Explanation / Answer
A. There would be two genes are involved for the color of the sea anemone and the length of its spikes, if four different phenotypes are observed from the cross between Earl Anemone and Pearl Anemone.
B and C. As we see that progenies with phenotypes purple and long are most number, we can say that these are dominant genes. So, ‘A’ will stand for purple, ‘a’ for red color in sea anemone. The other gene ‘B’ is showing dominant long spike over recessive ‘b’ as short spike.
Earl Anemone, genotype for parents = AaBb
Pearl Anemone, genotype for parents = AaBb
AaBb x AaBb
Gametes
AB
Ab
aB
ab
AB
AABB purple, long
AABb purple, long
AaBB purple, long
AaBb purple, long
Ab
AABb purple, long
AAbb purple, short
AaBb purple, long
Aabb purple, short
aB
AaBB purple, long
AaBb purple, long
aaBB red, long
aaBb red, long
ab
AaBb purple, long
Aabb purple, short
aaBb red, long
aabb red, short
D. The total number of progenies are= 300+75+85+40 = 500
The expected phenotypic ratio for a dihybrid cross from the Mendel’s law of inheritance is 9:3:3:1.
So the least number of progenies, red and short, would be 31 (500/16 = 31.12).
The purple and long progenies = 31 x 9 =279
The purple and short progenies = 31 x 3 =93
The red and long progenies = 31 x 3 =93
So the observed and expected progenies are given as:
Phenotype
Observed Progeny
Expected Progeny
purple and long
300
280
purple and short
75
94
red and long
85
95
red and short
40
31
Here we increased the number of progenies slightly to make the total complete It is because the decimal fraction of least number of progenies.
Gametes
AB
Ab
aB
ab
AB
AABB purple, long
AABb purple, long
AaBB purple, long
AaBb purple, long
Ab
AABb purple, long
AAbb purple, short
AaBb purple, long
Aabb purple, short
aB
AaBB purple, long
AaBb purple, long
aaBB red, long
aaBb red, long
ab
AaBb purple, long
Aabb purple, short
aaBb red, long
aabb red, short
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