Two true breeding strains of peas, one with yellow and round seeds and the other with...
Two true breeding strains of peas, one with yellow and round seeds and the other with green and wrinkled seeds, were crossed. All the F1plants had yellow and round seeds. When these F1 plants were test crossed to the green and wrinkled parental strain, the following offspring were obtained: 30 yellow and round; 20 green and round; 26 yellow and wrinkled; 28 green and wrinkled.
A) Please diagram the crosses described above, showing genotypes for each plant, using your own CLEARLY DEFINED nomenclature. Include the genotypes and proportions of each type of plant that you would expect in the F2 generation if the alleles that determine seed color assort independently from the alleles that determine seed shape.
B) Use Chi square analysis to test the hypothesis that the alleles determining seed color assort independently from the alleles determining seed shape Please show your
work along with the items listed below. (This is the part where I get confuse)
VALUE OF X^2
Degrees of freedom and conclusion regarding the hypothesis
Homework Answers
(a). Considering two factors, seed
color, and seed shape. The Y (yellow) dominates y (green) to decide
seed color, and the R allele for round dominates the r
allele for the determination of wrinkled seed. The P (Parental)
cross is between true-breeding alleles of round green peas and
wrinkled yellow peas. The F1 offspring are all round and yellow. In
the F2 generation, the genes at the two loci will separate
independently. Alternatively, the phenotypic ratio expected for
either character is 3:1.
In the F2 generation, 9:3:3:1 is the phenotypic ratio.
(b). Chi-Square Formula
Χ2 = Ʃ (observed value – expected value)2 / (expected value)
Degrees of freedom (df) = n-1 where n is the number of classes
|
OBSERVED VALUES |
EXPECTED VALUES |
|
ROUND YELLOW SEEDS (30) |
(9/16) X 30 = 16.875 |
|
ROUND GREEN SEEDS (20) |
(3/16) X 20 = 3.75 |
|
WRINKLED YELLOW SEEDS (26) |
(3/16) X 26 = 4.875 |
|
WRINKLED GREEN SEEDS (28) |
(1/16) X 28 = 1.75 |
|
TOTAL SEEDS = 104 |
TOTAL = 27.25 |
= (30-16.875)2/ 16.875 + (20- 3.75)2/3.75 + (26-4.875)2/4.875 + (28-1.75)2/1.75
= 10 + 7 + 9 + 3.9
= 29.9 = 30
Number of classes (n) = 4
Degree of freedom = n-1 + 4-1
Chi square value = 32
The probability of the chi square is greater than 0.05. If the value is less than the 0.05, the hypothesis can be accepted and if the value is greater than the 0.05, the hypothesis will be rejected.
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Two true breeding strains of peas, one with yellow and round
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A true breeding plant for round and yellow seeds is crossed to a true breeding wrinkled and green seeds. All progeny are round and yellow seeds. A plant from the F1 is crossed to a true breeding wrinkled and green seed's plant, producing the following offspring: 32 yellow and round seeds, 36 wrinkled and green seeds, 15 yellow and wrinkled seeds and 17 green and round seeds. What is the map distance between the color gene and the size gene?...
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In this tutorial you will examine dihybrid crosses: crosses where alleles at separate loci assort independently...
In this tutorial you will examine dihybrid crosses: crosses
where alleles at separate loci assort independently into gametes at
meiosis. You will also use logic to determine unknown genotypes,
phenotypes, and genetic ratios from given data.
Part A - Deducing phenotypes and genotypes of selfed parents
Mendel studied pea plants dihybrid for seed shape (round versus
wrinkled) and seed color (yellow versus green). Recall that
the round allele (R) is dominant to the wrinkled allele
(r) and
the yellow allele...
12) In peas, the yellow seed color is dominant (YY or Yy) over the green seed...
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3. Yellow seeds is a dominant characteristic in pea plants. Pure breeding pea plants for yellow...
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Part B - Deducing genotypes of crossed parents A plant grown from a [round, yellow] seed...
Part B - Deducing genotypes of crossed parents A plant grown from a [round, yellow] seed is crossed with a plant grown from a [wrinkled, yellow] seed. This cross produces four progeny types in the F [round, yellow], [wrinkled, yellow], [round, green], and [wrinkled, green]. Use this information to deduce the genotypes of the parent plants. Indicate the genotypes by dragging the correct label to the appropriate location.
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In this tutorial you will examine dihybrid crosses: crosses
where alleles at separate loci assort independently into gametes at
meiosis. You will also use logic to determine unknown genotypes,
phenotypes, and genetic ratios from given data.
Part A - Deducing phenotypes and genotypes of selfed parents
Mendel studied pea plants dihybrid for seed shape (round versus
wrinkled) and seed color (yellow versus green). Recall that
the round allele (R) is dominant to the wrinkled allele
(r) and
the yellow allele...
12) In peas, the yellow seed color is dominant (YY or Yy) over the green seed color (vy). You have been given a plant that only produces yellow seeds and need to determine if the plant is YY or Yy. Which of the following crosses would be most effective for determining the genotype of your plant? A)your plant x a plant that produces only green seeds B)your plant x another plant with known genotype YY C)your plant x another plant...
3. Yellow seeds is a dominant characteristic in pea plants. Pure breeding pea plants for yellow seeds were crossed with pure breeding pea plants for green seeds. The F1 generation all exhibited yellow seeds. An F1 plant was crossed to a green plant. Of 800 F2 plants, how many are expected to exhibit green seeds? Show work for full credit.
Part B - Deducing genotypes of crossed parents A plant grown from a [round, yellow] seed is crossed with a plant grown from a [wrinkled, yellow] seed. This cross produces four progeny types in the F [round, yellow], [wrinkled, yellow], [round, green], and [wrinkled, green]. Use this information to deduce the genotypes of the parent plants. Indicate the genotypes by dragging the correct label to the appropriate location.
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