2 SEM ACC W/RAVEN CARDED
12th Edition
ISBN: 9781264439218
Author: Raven
Publisher: MCGRAW-HILL HIGHER EDUCATION
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Textbook Question
Chapter 12, Problem 1S
z1. Create a Punnett square for the following crosses and use this to predict
a. A monohybrid cross between individuals with the genotype Aa and Aa
b. A dihybrid cross between two individuals with the genotype AaBb
c. A dihybrid cross between individuals with the genotype AaBb and aabb
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. An allele A that is not lethal when homozygous causes rats to have yellow coats. The allele R of a separate gene that assorts independently produces a black coat. Together, A and R produce a grayish coat, whereas a and r produce a white coat. A gray male is crossed with a yel-low female, and the F1 is 3 8 yellow, 3 8 gray, 1 8 black, and 1 8 white. Determine the genotypes of the parents.
A. Theoretical: The Punnett square will be used to examine the theoretical outcome of possible monohybrid
crosses.
1. The first cross is between a Homozygous dominant parent (PP), and a Homozygous recessive or just simply
say recessive parent (pp):
a. Fill in the Punnett square. Each box represents a genotype possibility for an offspring.
b. Place the allele donated by each parent in the corresponding box. Now list the possible genotypes and
their corresponding phenotype.
c. If an individual's genotype is heterozygous, the dominant trait will be expressed in the phenotype.
Give the percent possible for the phenotypes.
P
Genotype:
p
Phenotype:
Phenotype % probable:
1| Page
2. Cross between a Homozygous dominant parent (PP) and a Heterozygous parent (Pp). Fill in as in step one.
3. Cross between a Heterozygous parent (Pp), and another Heterozygous parent (Pp). Fill-in as before.
4. Test cross: A test cross is between a recessive parent (pp), and a Heterozygous parent (Pp). Again, fill-in…
1. Genes A and E are independent. Two individuals with the genotypes AaEe x aaEe are
crossed (upper case = dominant, lowercase = recessive).
a. Draw your Punnett square(s) in the space below.
b. On each line, show your calculations and the final answer. ie 1/2 x 1/2 = 1/4
What is the probability of these two individuals having the following:
an offspring with the phenotype aE ?
an offspring that is heterozygous for both traits?
an male offspring with the genotype AaEE ?
Chapter 12 Solutions
2 SEM ACC W/RAVEN CARDED
Ch. 12.1 - Prob. 1LOCh. 12.1 - Explain the advantages of Mendels experimental...Ch. 12.2 - Evaluate the outcome of a monohybrid cross.Ch. 12.2 - Explain Mendels Principle of Segregation.Ch. 12.2 - Compare the segregation of alleles with the...Ch. 12.3 - Evaluate the outcome of a dihybrid cross.Ch. 12.3 - Explain Mendels Principle of Independent...Ch. 12.3 - Compare the segregation of alleles for different...Ch. 12.4 - Prob. 1LOCh. 12.4 - Prob. 2LO
Ch. 12.5 - Interpret data from testcrosses to infer unknown...Ch. 12.6 - Describe how assumptions in Mendels model result...Ch. 12.6 - Prob. 2LOCh. 12.6 - Explain the genetic basis for observed alterations...Ch. 12 - Inquiry question What confounding problems could...Ch. 12 - Prob. 2IQCh. 12 - Prob. 1DACh. 12 - Prob. 2DACh. 12 - Prob. 3DACh. 12 - What property distinguished Mendels investigation...Ch. 12 - The F1 generation of the monohybrid cross purple...Ch. 12 - The F1 plants from the previous question are...Ch. 12 - In a cross of Aa Bb cc X Aa Bb Cc, what is the...Ch. 12 - An organisms __________ is/are determined by its...Ch. 12 - Phenotypes like height in humans, which show a...Ch. 12 - Japanese four oclocks that are red and tall are...Ch. 12 - If the two genes in the previous question showed...Ch. 12 - What is the probability of obtaining an individual...Ch. 12 - Prob. 4ACh. 12 - Prob. 5ACh. 12 - Mendels model assumes that each trait is...Ch. 12 - z1. Create a Punnett square for the following...Ch. 12 - Explain how the events of meiosis can explain both...Ch. 12 - Prob. 3SCh. 12 - In mammals, a variety of genes affect coat color....
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- a. 1 dominant allele will contribute 120/10 = 12 cm to the base height of the plant.b. The height of the parent plant 1 Genotype of the parent plant 1 – D1D1D2D2D3D3d4d4d5d5 The height of the parent plant 2 Genotype of the parent plant 2 – d1d1d2d2d3d3D4D4D5D5Contributing alleles – D4D4D5D5. The height of the plant without any contributing alleles would be 80 cm. The plant with genotype d1d1d2d2d3d3D4D4D5D5 has 4 contributing allele each of which contributes 12 cm to the base. Hence, the height of the plant with genotype d1d1d2d2d3d3D4D4D5D5 would be 80 + 12 + 12 + 12 + 12 = 128 cm. c. Parents – D1D1D2D2D3D3d4d4d5d5 × d1d1d2d2d3d3D4D4D5D5 Gametes – D1D2D3d4d5 × d1d2d3D4D5 F1 generation – D1d1D2d2D3d3D4d4D5d5 The height of the plants of F1 generation = 80 + 12 + 12 + 12 + 12 + 12 = 140 cm Hence, Genotype of the F1 = D1d1D2d2D3d3D4d4D5d5 Phenotype of…arrow_forwardPurple flowers are dominant to white flowers. Identify the phenotypefor the following genotype Ff, FF, ff and determine if the genotype is heterozygous or homozygous. * For each row, you should select two columns. Purple flowers White flowers Heterozygous Homozygous Ff FF ff Brown eyes are dominant to blue eyes. Identify the phenotype for the following genotype BB, Bb, bb and determine if the genotype is heterozygous or homozygous. * 口 ロ口arrow_forward3. Short-tailed pup distribution. In mice, the dominant T allele results in a short tail. Homozygous T/T genotype is lethal, which means mouse embryos with this genotype die before they are born. Homozygous t/t is normal, with normal tail. A cross between two short-tailed mice produces a litter of 5 pups. a). What are the genotype(s) of the two short-tailed mice used in the cross? b). What are the possible genotypes and phenotypes of the 5 pups? What are their perspective ratio?arrow_forward
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