Concept explainers
Still referring to Problem 1, what will be the possible genotypes of offspring from the following matings? With what frequency will each genotype show up?
- a. AABB × aaBB
- b. AaBB × AABb
- c. AaBb × aabb
- d. AaBb × AaBb
a.
To predict: The predicted genotype frequencies among the offspring for the mating AABB × aaBB.
Introduction: The law of independent assortment and law of segregation are the laws proposed by Gregor Johann Mendel, a geneticist. Mendel proposed the law of independent assortment based on the results of monohybrid cross that “two alleles at any locus tend to separate from each other during meiosis, so they end up in different gametes”. Mendel proposed the law of independent assortment based on the results of dihybrid cross that “alleles at one locus tend to assort into gametes independently of alleles at other loci”. The number of possible allele combination for the given genotype is given by the formula 2n where “n” is the number of heterozygous alleles that are present.
Explanation of Solution
A monohybrid cross is constructed with Punnett square for the parents having the genotype AABB and aaBB.
All the offspring have the same genotype AaBB. Therefore, there is no genotype frequency.
b.
To predict: The predicted genotype frequencies among the offspring for the mating AaBB × AABb.
Introduction: The law of independent assortment and law of segregation are the laws proposed by Gregor Johann Mendel, a geneticist. Mendel proposed the law of independent assortment based on the results of monohybrid cross that “two alleles at any locus tend to separate from each other during meiosis, so they end up in different gametes”. Mendel proposed the law of independent assortment based on the results of dihybrid cross that “alleles at one locus tend to assort into gametes independently of alleles at other loci”. The number of possible allele combination for the given genotype is given by the formula 2n where “n” is the number of heterozygous alleles that are present.
Explanation of Solution
A monohybrid cross is constructed with Punnett square for the parents having the genotype AaBB and AABb.
The following is the frequency of genotypes.
Total number of possible genotypes = 4.
AABB=
AABb=
AaBB=
AaBb =
The frequency of all four genotypes is 25%.
c.
To predict: The predicted genotype frequencies among the offspring for the mating AaBb × aabb.
Introduction: The law of independent assortment and law of segregation are the laws proposed by Gregor Johann Mendel, a geneticist. Mendel proposed the law of independent assortment based on the results of monohybrid cross that “two alleles at any locus tend to separate from each other during meiosis, so they end up in different gametes”. Mendel proposed the law of independent assortment based on the results of dihybrid cross that “alleles at one locus tend to assort into gametes independently of alleles at other loci”. The number of possible allele combination for the given genotype is given by the formula 2n where “n” is the number of heterozygous alleles that are present.
Explanation of Solution
A monohybrid cross is constructed with Punnett square for the parents having the genotype AaBb and aabb
The following is the frequency of genotypes.
Total number of possible genotypes = 4.
AABB=
aaBB=
Aabb=
aabb =
The frequency of all four genotypes is 25%.
d.
To predict: The predicted genotype frequencies among the offspring for the mating AaBb × AaBb.
Introduction: The law of independent assortment and law of segregation are the laws proposed by Gregor Johann Mendel, a geneticist. Mendel proposed the law of independent assortment based on the results of monohybrid cross that “two alleles at any locus tend to separate from each other during meiosis, so they end up in different gametes”. Mendel proposed the law of independent assortment based on the results of dihybrid cross that “alleles at one locus tend to assort into gametes independently of alleles at other loci”. The number of possible allele combination for the given genotype is given by the formula 2n where “n” is the number of heterozygous alleles that are present.
Explanation of Solution
A monohybrid cross is constructed with Punnett square for the parents having the genotype AaBb and AaBb.
The following is the frequency of genotypes.
Total number of possible genotypes = 16.
AABB=
AABb=
AaBB=
AaBb=
AAbb=
Aabb=
aaBB=
aaBb=
aabb=
The frequency of genotypes AABB, aaBB , AAbb and aabb is 6.25%.
The frequency of genotypes AaBB, AABb, aaBb and Aabb is 12.5%.
The frequency of genotypes AaBb is 25%.
Want to see more full solutions like this?
Chapter 19 Solutions
Human Biology (MindTap Course List)
- 3) A wild type fruit fly (dihybrid for gray body color and normal wings) is mated with a black fly with vestigial wings. Determine the expected phenotypic ratio resulting from this cross if we initially assume the alleles for these traits are unlinked and exhibit complete dominance. The offspring of the mating described above actually exhibited the following phenotypic ratio: Wild type: 778 Black vestigial: 785 Black normal: 158 Gray vestigial: 162 a) What does the observed phenotypic ratio suggest regarding the alleles for body color and wing shape? b) What could account for the existence of the observed recombinant phenotypes in the ratios shown above? c) Calculate the recombination frequency between the genes for body color and wing shape. What is the relative distance between these alleles on the same chromosome?arrow_forward4) The round pea seed allele (R) is dominant, while the wrinkled pea seed allele (r) isrecessive. A heterozygous round-seeded pea plant is crossed with a wrinkle-seeded peaplant. Use a Punnett Square to solve the following: a. Determine the predicted genotype ratio of the offspring. b. Determine the predicted phenotype ratio of the offspring. c. If this cross produced 50 plants, how many plants would you predict would bewrinkle-seeded pea plants? please if punnet square needed write down the P G and F1 also thanksarrow_forwardIn a diploid plant species, an F1 with the genotype Mm Rr Ss is test crossed to a pure breeding recessive plant with the genotype mm rr ss. The offspring genotypes are as follows: Genotype Number Mm Rr Ss 687 Mm Rr ss 5 Mm rr Ss 68 Mm rr ss 196 mm Rr Ss 185 mm Rr ss 72 mm rr Ss 8 mm rr ss 679 Total 1900 1. What is the interference value for this data set?arrow_forward
- You have determined that the gene order for three linked genes being studied is CBA. The number of recombinants resulting from crossover between genes A and B alone totals 40 and 42, while the double-crossover progeny total 4 and 6. What is the recombination frequency between genes A and B if the total number of progeny from the cross is 1000? O 0.102% O 0.092% O 10.2% O 0.082% O 8.2% O 1% O 9.2%arrow_forwardThx!arrow_forwardPlease answer a and b and explain how you came up with the ratio?arrow_forward
- CC DD and cc dd individuals were crossed to eachother, and the F1 generation was backcrossed to thecc dd parent. 997 Cc Dd, 999 cc dd, 1 Cc dd, and 3 ccDd offspring resulted.a. How far apart are the c and d loci?b. What progeny and in what frequencies would youexpect to result from testcrossing the F1 generationfrom a CC dd × cc DD cross to cc dd?c. In a typical meiosis, how many crossovers occurbetween genes C and D?d. Assume that the C and D loci are on the samechromosome, but the offspring from the testcrossdescribed in part (b) were 498 Cc Dd, 502 cc dd,504 Cc dd, and 496 cc Dd. How would your answer to part (c) change?arrow_forwardA is dominant over a and B is dominant over b. Genes A/a and B/b assort independently. The parental cross (P) is between true- breeding (homozygous) strains (AA bb x aa BB). The resulting F1 offspring would be [Select ] A F1 x F1 cross gives the F2 generation. The ratio of phenotypes in the F2 generation can be described as [ Select ] In the phenotypic ratio, what number refers to individuals that are: dominant for both traits? [ Select ] recessive for both traits? [Select ] To what phenotypes do the "3"s in the ratio refer? [ Select ] > >arrow_forwardAnswer the following with a short solution if needed: a. The gametes of a worm's genotype SsYy should produce what genotypes? b. A genetic cross between two F1-hybrid rose plants having yellow petals will yield what percent green-petal plants in the F2 generation? Yellow petals are dominant to green. c. Brown fur is dominant over light-colored fur. What is the phenotype of the resulting offspring if you cross a heterozygous brown fur and a light-colored fur?arrow_forward
- Human Biology (MindTap Course List)BiologyISBN:9781305112100Author:Cecie Starr, Beverly McMillanPublisher:Cengage LearningBiology (MindTap Course List)BiologyISBN:9781337392938Author:Eldra Solomon, Charles Martin, Diana W. Martin, Linda R. BergPublisher:Cengage Learning