Concept explainers
a.
To determine:
The probability of one offspring having normal
Introduction:
Gametes carry genes from parents. Each parent passes on a haploid set of genes to the progeny during the sexual reproduction. The fusion of gametes carrying the haploid gene set from each of the parents gives rise to a diploid progeny.
Explanation of Solution
In the given case, dominant characters are normal wings (T) and normal/oval eyes (N), whereas the recessive characters have tiny wings (t) and narrow eyes (n). Cross between genotypes Ttnn (male) and TtNn (female) can be shown as given below.
TN | Tn | tN | tn | |
Tn | TTNn | TTnn | TtNn | Ttnn |
tn | TtNn | Ttnn | ttNn | ttnn |
Based on the above table, the probability of each of the genotypes is found to be:
Normal wings =3/4
Tiny wings = 1/4
Oval eyes = 1/2
Narrow eyes = 1/2
Hence, the total probability of an offspring having normal characters can be calculated by multiplying the respective probabilities of both the normal characters.
Hence, 3/8 of the total offspring is having normal genotype (TTNN, TtNn, TTNn, or TtNN).
b.
To determine:
The expected phenotype in the offspring from the above cross along with the number of offsprings in each phenotypic class if there are a total of 200 progeny obtained.
Introduction:
During sexual reproduction, the offspring somewhat resembles the parents because they carry the genes that are passed on from the parents to them through the gametes. Sexual reproduction brings genes from both the parents into the progeny.
Explanation of Solution
Based on the cross between genotypes Ttnn (male) and TtNn (female), probabilities are found to be as follows:
Normal wings = 3/4
Tiny wings = 1/4
Oval eyes = 1/2
Narrow eyes = 1/2
The expected phenotype in the offspring from the above cross can be calculated as follows:
The total number of offsprings is given as 200. Based on this, the total number of offspring with each of the character can be calculated as follows:
This gives the total number of offsprings as follows:
Normal wings and oval eyes: 75
Normal wings and narrow eyes: 75
Tiny wings and oval eyes: 25
Tiny wings and narrow eyes: 25
Want to see more full solutions like this?
Chapter 2 Solutions
Connect 1-semester Access Card For Genetics
- Assume that in a series of experiments, plants with round seeds were crossed with plants with wrinkled seeds and the following offspring were obtained: 220 round and 180 wrinkled. (a) What is the most probable genotype of each parent?(b) What genotypic and phenotypic ratios are expected?(c) Based on the information provided in part (b) above, what are the expected (theoretical) numbers of progeny (400 total) of each phenotypic class?arrow_forwardIn barley, a self-fertilizing species that can be cross-fertilized, two true-breeding strains with virescent leaves occur. In strain A, the trait is caused by a cytoplasmic gene while in strain B it is by a recessive chromosomal gene. What phenotypes would you expect among the progeny, and in what proportions in each of the following? Illustrate your crosses below, indicate and the female and male parent for each cross, and write the phenotype of all the parents and offspring(s). a. reciprocal crosses between A and Bb. crossing of each F1 in (a) to each of the paternal strainsc. self-fertilization of the F1’s in (a)d. reciprocal crosses between F1’s in (a) Use the following gene assignments: Strain A (trait is in Cytoplasm) A – virescent a – not virescent Strain B (recessive chromosomal gene) B – not virescent b - virescentarrow_forwardIn barley, a self-fertilizing species that can be cross-fertilized, two true-breeding strains with virescent leaves occur. In strain A, the trait is caused by a cytoplasmic gene while in strain B it is by a recessive chromosomal gene. What phenotypes would you expect among the progeny, and in what proportions in each of the following? Illustrate your crosses below, indicate and the female and male parent for each cross, and write the phenotype of all the parents and offspring(s). a. reciprocal crosses between A and Barrow_forward
- Two pure-breeding strains of flies are mated, and the F1 are intercrossed. The first strain has curled wings and black bodies. The second strain has straight wings and brown bodies. The F2 progeny are 271 straight wings with brown bodies, 31 curled wings with black bodies, 94 curled wings with brown bodies and 90 straight wings with black bodies. If the F1 were backcrossed to the straight, wing brown bodied parent, what phenotypes would be produced among the progeny? What would be the proportion of each phenotype?arrow_forwardThree autosomal recessive mutations in yeast, all producing the same phenotype (m1, m2, and m3), are subjected to complementation analysis. Of the results shown below, which, if any, are alleles of one another? Predict the results of the cross that is not shown—that is, m2 * m3. Cross 1: m1 * m24 F1: all wild-type progeny Cross 2: m1 * m34 F1: all mutant progenyarrow_forwardGiven the distance between the orange eye color locus and the STRs, how frequent will you expect to get recombinant progeny with orange eyes? What would the probability of this marriage be producing children with orange eyes?arrow_forward
- A female of genotype a b c + + + produces 100 meiotic tetrads. Of these, 68 show no crossover events. Of the remaining 32, 20 show a crossover between a and b, 10 show a crossover between b and c, and 2 show a double crossover between a and b and between b and c. Of the 400 gametes produced, how many of each of the 8 different genotypes will be produced? Assuming the order a–b–c and the allele arrangement previously shown, what is the map distance between these loci?arrow_forwardThe genotype of F1 individuals in a tetrahybrid cross is AaBbCcDd X AaBbCcDd. Assuming independent assortment of these four genes, what are the probabilities that F2 offspring will have the following genotypes? Calculate the probability of each using multiplication and addition rules. Show your work. a. aabbccdd b. AaBbCcDd c. AaBBCCdd d. Any of the 3 genotypes abovearrow_forwardThe phenotype of crooked wings (cw) in Drosophila melanogaster is caused by a recessive mutant gene that independently assorts with a recessive mutant gene for hairy (h) body. Assume that a cross is made between a fly with normal wings and a hairy body and a fly with crooked wings and normal body hair. All F1 flies from this cross were wild-type, and these flies were crossed among each other to produce 288 F2 offspring. Which phenotypes would you expect among the offspring in the F2 generation, and how many of each phenotype would you expect?arrow_forward
- In barley, a self-fertilizing species that can be cross-fertilized, two true-breeding strains withvirescent leaves occur. In strain A, the trait is caused by a cytoplasmic gene while in strainB it is by a recessive chromosomal gene. Illustrate the crossing of each F1 in (a) to each of the paternal strains What phenotypes would you expect among theprogeny, and in what proportions in each of the following? Illustrate your crosses below,indicate and the female and male parent for each cross, and write the phenotype of allthe parents and offspring(s).arrow_forwardErminette fowls have mostly light colored feathers with an occasional black one, giving a flecked appearance. A cross of two erminettes produced a total of 48 progeny, consisting of 22 erminettes, 14 blacks and 12 pure whites. What genetic basis of the erminette pattern is suggested? How would you test your hypotheses?arrow_forwardIn Drosophila, ebony body colour is produced by a recessive gene a and wild-type (gray) body colour by its dominant allele a+. Vestigial wings are governed by a recessive gene vg, and normal wing size (wild type) by its dominant allele vg+. If wild-type dihybrid flies are crossed and produce 256 progeny, how many of these progeny flies are expected in each phenotypic class?arrow_forward
- Biology (MindTap Course List)BiologyISBN:9781337392938Author:Eldra Solomon, Charles Martin, Diana W. Martin, Linda R. BergPublisher:Cengage Learning