Concepts of Genetics (12th Edition)
12th Edition
ISBN: 9780134604718
Author: William S. Klug, Michael R. Cummings, Charlotte A. Spencer, Michael A. Palladino, Darrell Killian
Publisher: PEARSON
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Chapter 5, Problem 29ESP
A female of genotype
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?
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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?
A fruit fly of genotype B R/b r is testcrossed with b r/b r.In 84 percent of the meioses, there are no chiasmata between the linked genes; in 16 percent of the meioses,there is one chiasma between the genes. What proportion of the progeny will be B r/b r?
In the following cross, imagine that you have a female fly that has two Xs and one Y
due to a nondisjunction event in her mother's germ cells. Draw out what the
possible gametes are for both the female and the male and also a Punnett square
showing the genotypes, phenotypes, and sex of the possible flies as a result of this
cross. You do not need to provide the probabilities of each of these.
Red-eyed wi
C Ở Red-eyed
wt
XX Y
X Y
Meiosis
Chapter 5 Solutions
Concepts of Genetics (12th Edition)
Ch. 5 - In a family with one autistic child the risk for...Ch. 5 - Given that the prenatal test can provide only a...Ch. 5 - Consider two hypothetical recessive autosomal...Ch. 5 - With two pairs of genes involved (P/p and Z/z), a...Ch. 5 - In Drosophila, a heterozygous female for the...Ch. 5 - HOW DO WE KNOW? In this chapter, we focused on...Ch. 5 - Review the Chapter Concepts list on page 94. Most...Ch. 5 - Describe the cytological observation that suggests...Ch. 5 - Why does more crossing over occur between two...Ch. 5 - Explain why a 50 percent recovery of...
Ch. 5 - Why are double-crossover events expected less...Ch. 5 - What is the proposed basis for positive...Ch. 5 - What two essential criteria must be met in order...Ch. 5 - The genes dumpy (dp), clot (cl), and apterous (ap)...Ch. 5 - Colored aleurone in the kernels of com is due to...Ch. 5 - In the cross shown here, involving two linked...Ch. 5 - In a series of two-point mapping crosses involving...Ch. 5 - Two different female Drosophila were isolated,...Ch. 5 - In Drosophila, a cross was made between femalesall...Ch. 5 - Another cross in Drosophila involved the...Ch. 5 - In Drosophila, Dichaete (D) is a mutation on...Ch. 5 - Drosophila females homozygous for the third...Ch. 5 - In Drosophila, two mutations, Stubble (Sb) and...Ch. 5 - If the cross described in Problem 18 were made,...Ch. 5 - Are mitotic recombinations and sister chromatid...Ch. 5 - What possible conclusions can be drawn from the...Ch. 5 - An organism of the genotype AaBbCc was testcrossed...Ch. 5 - Based on our discussion of the potential...Ch. 5 - Traditional gene mapping has been applied...Ch. 5 - DNA markers have greatly enhanced the mapping of...Ch. 5 - In a certain plant, fruit is either red or yellow,...Ch. 5 - Two plants in a cross were each heterozygous for...Ch. 5 - A number of humanmouse somatic cell hybrid clones...Ch. 5 - A female of genotype produces 100 meiotic tetrads....Ch. 5 - In laboratory class, a genetics student was...Ch. 5 - Drosophila melanogaster has one pair of sex...Ch. 5 - In Drosophila, a female fly is heterozygous for...Ch. 5 - The gene controlling the Xg blood group alleles...Ch. 5 - Prob. 34ESP
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- Figure 8.10 In pea plants, purple flowers (P) are dominant to white (p), and yellow peas (Y) are dominant to green (y). What are the possible genotypes and phenotypes for a cross between PpYY and ppYy pea plants? How many squares would you need to complete a Punnett square analysis of this cross?arrow_forwardHuman females have two X chromosomes XX; males have one X and one Y chromosome XY. a. With respect to X-linked alleles, how many different types of gametes can a male produce? b. A female homozygous for an X-linked allele can produce how many types of gametes with respect to that allele? c. A female heterozygous for an X-linked allele can produce how many types of gametes with respect to that allele?arrow_forwardTwo plants in a cross were each heterozygous for two gene pairs (AB /ab) whose loci are linked and 30 map units (mu) apart. (Recall that 1 mu is equal to 1% recombination between two genes.) Assuming that crossing over occurs during the formation of both male and female gametes and that the A and B alleles are dominant, determine the phenotypic ratio of their offspring. Part E: What proportion of the offspring of two plants (both (AB/ab ) will be A - B- if the genes are 30 mu apart? Part F: What proportion of the offspring of two plants (both (AB/ab)) will be A - bb if the genes are 30 mu apart? Part G: What proportion of the offspring of two plants (both (AB/ab)) will be aaB- If the genes are 30 mu apart? Part H: What proportion of the offspring of two plants (both (AB/ab)) will be aabb if the genes are 30 mu apart?arrow_forward
- Two plants in a cross were each heterozygous for two gene pairs (AB/ab) whose loci are linked and 10 map units (mu) apart. (Recall that 1 mu is equal to 1% recombination between two genes.) Assuming that crossing over occurs during the formation of both male and female gametes and that the A and B alleles are dominant, determine the phenotypic ratio of their offspring. Part D If the two genes are 15 mu apart and the plant is (Ab/aB), what proportion of gametes from a signal plant will be ab? Part E What proportion of the offspring of two plants ( both (Ab/aB)) will be A_B_ if the genes are 15 mu apart? Part F What proportion of the offspring of two plants ( both (Ab/aB)) will be A_bb if the genes are 15 mu apart? Part G What proportion of the offspring of two plants ( both (Ab/aB)) will be aaB_ if the genes are 15 mu apart? Part H What proportion of the offspring of two plants ( both (Ab/aB)) will be aabb if the genes are 15 mu apart? How would I solve these?arrow_forwardAn individual that is homozygous AABBcc is crossed to an individual that is homozygous aabbCC producing F1 individuals with genotype AaBbCc. Assume that the A, B, and C loci are all on the same chromosome. One chromatid at the start of meiosis (before crossing over) has the alleles A, B, and c. What alleles are on its sister chromatid? What alleles are on the homolog? Group of answer choices A. Sister chromatid ABc, Homolog no way to tell B. Sister chromatid ABC, Homolog abc C. Sister chromatid ABc, Homolog abC D. Sister chromatid abC, Homolog ABcarrow_forwardWhat is the correct table in these two given? 1. Take two coins and assume that heads represent the dominantallele (A) and tails represents the recessive allele (a). The genotype for each coin isheterozygous (Aa).2. Assume that each coin represents one parent. When a single coin is flipped, one gameteis formed (through the process of meiosis). If the flipped coin is on heads, then thegamete has the dominant allele (A). When both coins are flipped simultaneously, therewill be two possible gametes that can combine through fertilization to form a zygote. Eachtime you flip both coins, you will record the “genotype” of the offspring.3. Flip the coins 100 times and record your results in the chart belowarrow_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_forwardAssume that there is complete interference in the Lf-W region. The cross between the triple heterozygote (Lf J W/ lf j w) and a triple homozygous recessive produced 350 progenies. Give the frequency of each specific progeny type:a. with crossover in the Lf-J and J-W regionsb. with crossover in the Lf-J regionc. with crossover in the J-W regiond. without crossover in the Lf-W regionarrow_forwardAssume that D, E, F, G, H, and I are autosomal genes on different chromosomes. From the mating (parent A) Dd ee Ff GG Hh li x (parent B) Dd EE FF Gg Hh ii. A. What is the probability that one of the offspring will have the genotype Dd Ee FF Gg hh li? B. What is the probability that one of the offspring will be heterozygous for each allele?arrow_forward
- Fruitflies of genotype AABB are bred to fruitflies with genotype aabb. The F1 offspring are then bred to fruitflies of genotype aabb. The F2 offspring have the following genotypes: Aabb 14 flies AaBb 413 flies aabb 419 flies aaBb 15 flies a. Are genes A and B on the same chromosome? Yes No b. Show how you would calculate the genetic distance between genes A and B.arrow_forwardBased on the information you discovered in the previous problem, answer the following:a. A female fruit fly with genotype Tt nn is mated toa male of genotype Tt Nn. What is the probabilitythat any one of their offspring will have normalphenotypes for both characters?b. What phenotypes would you expect among the offspring of this cross? If you obtained 200 progeny, howmany of each phenotypic class would you expect?arrow_forwardIn onion, male sterility is produced when the nuclear genotype is aa and the mitochondrial gene S (sterile) are present. Any other combination of nuclear genotype and mitochondrial gene (including gene F for fertile) will result in a male fertile plant. Give the genotypic ratio and the phenotypic ratio or the percentage of male sterile and male fertile offspring that will be produced in the following crosses. 1. Aa + S male x aa + F female 2. Reciprocal cross of number 1. (Note that when we do reciprocal cross, we interchange/swap the genotypes of the parents (if there is a nuclear gene involved, you interchange the nuclear genotype as well). 3. Aa + S female x Aa + F male 4. Reciprocal cross of number 3.arrow_forward
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