BackGenetics Exam Practice: Linkage, Mapping, Bacterial Genetics, and DNA Replication
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Q36. In Drosophila, the two genes v and sn are X-linked and 25 map units apart. A female fly of genotype v+ sn/v sn+ is crossed to a male from a wild-type line. What percent of male progeny will be v sn?
Background
Topic: X-linked gene mapping and recombination frequency in Drosophila.
This question tests your understanding of X-linked inheritance, recombination frequency (map units), and how to predict progeny phenotypes from a testcross.
Key Terms and Formulas:
Recombination frequency (in map units): 1 map unit = 1% recombination.
Testcross: Crossing a heterozygote with a homozygous recessive (or wild-type male for X-linked genes in Drosophila).
Gamete types: Parental and recombinant types, with frequencies determined by recombination percentage.
Step-by-Step Guidance
Draw the female's genotype: v+ sn / v sn+ (cis configuration).
Recall that male Drosophila inherit their X chromosome from their mother and Y from their father.
Determine the possible gametes produced by the female, and their expected frequencies based on the recombination rate (25%).
Identify which gamete will produce v sn males when combined with the Y chromosome from the wild-type male.
Set up the calculation for the expected percentage of v sn male progeny, but do not compute the final value yet.
Try solving on your own before revealing the answer!
Final Answer: 12.5%
Since the recombination frequency is 25%, each recombinant gamete (v sn and v+ sn+) will be produced at 12.5% frequency. Only male progeny receiving the v sn recombinant X chromosome from the mother and a Y from the father will be v sn males.
Q37. In maize, two plants heterozygous for the recessive alleles a and b are crossed. If a and b are 7.2 map units apart and both parents carry a and b in trans, what frequency of double-mutant progeny will appear?
Background
Topic: Linkage and recombination in plants, specifically calculating recombinant frequencies for genes in trans configuration.
This question tests your ability to use recombination frequencies to predict the proportion of double-mutant offspring from a cross involving linked genes.
Key Terms and Formulas:
Trans configuration: Each chromosome carries one dominant and one recessive allele (Ab/aB).
Recombination frequency: 7.2 map units = 7.2% recombination.
Double-mutant progeny: Individuals with genotype ab.
Probability of double-mutant progeny = (recombination frequency / 2) × (recombination frequency / 2).
Step-by-Step Guidance
Write the parental genotypes: Ab/aB × Ab/aB.
Determine the gametes produced by each parent and their frequencies, considering the recombination rate.
Identify which gamete combinations will produce ab/ab double-mutant progeny.
Set up the calculation for the expected frequency of double-mutant progeny, but do not compute the final value yet.
Try solving on your own before revealing the answer!
Final Answer: 0.001296
The frequency of double-mutant progeny (ab/ab) is calculated as (0.072/2) × (0.072/2) = 0.001296, since each parent produces ab gametes at a frequency of 0.036.
Q39. In corn, the following recessive alleles are found: g (glossy leaf) and r (red aleurone). A hybrid of unknown origin is testcrossed, resulting in 1,000 progeny: red, 127; glossy, 153; wild-type, 352; and glossy + red, 368. Which of the following could have been a parent of the dihybrid, assuming the parents were homozygous?
Background
Topic: Testcross analysis and deduction of parental genotypes in dihybrid crosses.
This question tests your ability to interpret testcross results and deduce possible parental genotypes based on progeny phenotypes and numbers.
Key Terms and Formulas:
Testcross: Crossing a heterozygote with a homozygous recessive individual.
Parental and recombinant types: The most frequent progeny classes are usually parental types.
Step-by-Step Guidance
List the phenotypes and their frequencies: red (127), glossy (153), wild-type (352), glossy + red (368).
Identify the two most frequent classes (wild-type and glossy + red) as likely parental types.
Determine the possible homozygous parental genotypes that could produce these F1 dihybrids.
Set up the logic for which parental combinations could yield the observed F1 and testcross results, but do not state the final answer yet.
Try solving on your own before revealing the answer!
Final Answer: a. glossy, red
The most frequent progeny classes (wild-type and glossy + red) indicate that the parents were likely glossy, red (gg rr) and wild-type (GG RR).
Q40. In corn, the following recessive alleles are found: g (glossy leaf) and r (red aleurone). A hybrid is testcrossed, resulting in 1,000 progeny: red, 127; glossy, 153; wild-type, 352; and glossy + red, 368. What is your best estimate of the map distance between glossy and red, in map units?
Background
Topic: Genetic mapping and calculation of map distance from testcross data.
This question tests your ability to calculate recombination frequency (map units) from observed progeny numbers in a testcross.
Key Terms and Formulas:
Map distance (in map units or centimorgans, cM):
Recombinant progeny: The less frequent classes (red and glossy).
Step-by-Step Guidance
Identify the recombinant classes (red and glossy) and sum their numbers.
Calculate the total number of progeny.
Set up the formula for map distance using the recombinant and total progeny numbers, but do not compute the final value yet.
Try solving on your own before revealing the answer!
Final Answer: 28.0 map units
Map distance = ((127 + 153) / 1000) × 100 = 28.0 map units.
Q41. Genes a and b are 10 map units apart, b and c are 20 map units apart, and a and c are 30 map units apart. If a triple heterozygote is testcrossed, among 1,000 progeny, how many are expected to result from double crossovers if there is no interference?
Background
Topic: Three-point testcross and calculation of expected double crossovers.
This question tests your understanding of genetic mapping, recombination frequencies, and the concept of interference.
Key Terms and Formulas:
Double crossover frequency = (frequency between a and b) × (frequency between b and c)
No interference means the expected frequency is simply the product of the two single crossover frequencies.
Step-by-Step Guidance
Convert map units to recombination frequencies: 10% (0.10) and 20% (0.20).
Calculate the expected double crossover frequency: 0.10 × 0.20.
Multiply this frequency by the total number of progeny (1,000) to get the expected number of double crossovers, but do not compute the final value yet.
Try solving on your own before revealing the answer!
Final Answer: 20
Expected double crossovers = 0.10 × 0.20 × 1,000 = 20.
Q42. Genes a and b are 10 map units apart, b and c are 20 map units apart, and a and c are 30 map units apart. If a triple heterozygote is testcrossed, among 1,000 progeny, how many are expected to result from double crossovers if interference = 0.5?
Background
Topic: Interference in genetic mapping and its effect on double crossover frequency.
This question tests your ability to adjust expected double crossover numbers based on the coefficient of coincidence (1 - interference).
Key Terms and Formulas:
Interference (I) = 1 - coefficient of coincidence (CoC)
Expected double crossovers (with interference) = expected (no interference) × CoC
CoC = 1 - interference
Step-by-Step Guidance
Recall from Q41 the expected number of double crossovers without interference.
Calculate the coefficient of coincidence: CoC = 1 - 0.5 = 0.5.
Multiply the expected number of double crossovers (from Q41) by the coefficient of coincidence, but do not compute the final value yet.
Try solving on your own before revealing the answer!
Final Answer: 10
With interference = 0.5, expected double crossovers = 20 × 0.5 = 10.
Q43. Genes a and b are 10 map units apart, b and c are 20 map units apart, and a and c are 30 map units apart. If a triple heterozygote is testcrossed, among 1,000 progeny, how many are expected to result from single crossovers between a and b if there is no interference?
Background
Topic: Calculation of single crossover progeny in a three-point testcross.
This question tests your ability to use recombination frequencies to predict the number of single crossover progeny between two genes.
Key Terms and Formulas:
Single crossover frequency between a and b = 10% (0.10).
Number of single crossover progeny = frequency × total progeny.
Step-by-Step Guidance
Identify the recombination frequency between a and b (0.10).
Multiply this frequency by the total number of progeny (1,000) to get the expected number of single crossovers, but do not compute the final value yet.
Try solving on your own before revealing the answer!
Final Answer: 100
Expected single crossovers between a and b = 0.10 × 1,000 = 100.
Q44. Three recessive genes in the tomato plant produce an absence of anthocyanin pigment (a), hairless plants (h), and jointless fruit stems (j). Among 3,000 progeny from a trihybrid F1 that was testcrossed, the phenotypes in figure 5 below were observed. What is the linkage arrangement of these three loci?
Background
Topic: Three-point testcross and gene order determination.
This question tests your ability to analyze progeny phenotypes from a three-point testcross to deduce the order of three linked genes.
Key Terms and Formulas:
Parental, single crossover, and double crossover classes: Parental types are most frequent, double crossovers are least frequent.
Gene order is determined by comparing double crossover phenotypes to parental types.
Step-by-Step Guidance
List the phenotypes and their frequencies from the table.
Identify the parental (most frequent) and double crossover (least frequent) classes.
Compare the double crossover phenotypes to the parental types to deduce which gene is in the middle.
Set up the possible gene orders and determine which matches the observed data, but do not state the final answer yet.
Try solving on your own before revealing the answer!
Final Answer: a-h-j
The gene order is a-h-j, based on the comparison of double crossover and parental phenotypes.
Q45. Three recessive genes in the tomato plant produce an absence of anthocyanin pigment (a), hairless plants (h), and jointless fruit stems (j). Among 3,000 progeny from a trihybrid F1 that was testcrossed, the phenotypes in figure 5 above were observed. If the parents (P1) were homozygous, which of the following could have been a parent?
Background
Topic: Parental genotype deduction from three-point testcross data.
This question tests your ability to deduce possible parental genotypes based on observed progeny classes and gene order.
Key Terms and Formulas:
Parental types: The most frequent progeny classes indicate the original parental genotypes.
Homozygous parents: One parent is homozygous dominant, the other is homozygous recessive for all three genes.
Step-by-Step Guidance
Identify the most frequent progeny classes from the data (parental types).
Determine the phenotypes corresponding to the homozygous recessive and dominant parents.
Match the options to the possible parental phenotypes, but do not state the final answer yet.
Try solving on your own before revealing the answer!
Final Answer: a. anthocyaninless, hairless
The most frequent class corresponds to the homozygous recessive parent: anthocyaninless, hairless (a/a h/h j/j).
Q46. Three recessive genes in the tomato plant produce an absence of anthocyanin pigment (a), hairless plants (h), and jointless fruit stems (j). Among 3,000 progeny from a trihybrid F1 that was testcrossed, the phenotypes in figure 5 above were observed. What is your best estimate of the anthocyaninless-hairless distance?
Background
Topic: Calculation of map distance between two genes using three-point testcross data.
This question tests your ability to identify recombinant classes and calculate map distance between two genes.
Key Terms and Formulas:
Map distance = (Number of recombinant progeny between two genes / Total progeny) × 100
Include both single and double crossover classes in the recombinant count.
Step-by-Step Guidance
Identify the recombinant classes for anthocyaninless and hairless genes.
Sum the numbers of these recombinant progeny (including double crossovers).
Set up the formula for map distance using the recombinant and total progeny numbers, but do not compute the final value yet.
Try solving on your own before revealing the answer!
Final Answer: 20.0 map units
Map distance = (number of single crossovers + double crossovers) / 3,000 × 100 = 20.0 map units.
Q47. Three recessive genes in the tomato plant produce an absence of anthocyanin pigment (a), hairless plants (h), and jointless fruit stems (j). Among 3,000 progeny from a trihybrid F1 that was testcrossed, the phenotypes in figure 5 above were observed. What is your best estimate of the anthocyaninless-jointless distance?
Background
Topic: Calculation of map distance between two genes using three-point testcross data.
This question tests your ability to identify recombinant classes and calculate map distance between two genes.
Key Terms and Formulas:
Map distance = (Number of recombinant progeny between two genes / Total progeny) × 100
Include both single and double crossover classes in the recombinant count.
Step-by-Step Guidance
Identify the recombinant classes for anthocyaninless and jointless genes.
Sum the numbers of these recombinant progeny (including double crossovers).
Set up the formula for map distance using the recombinant and total progeny numbers, but do not compute the final value yet.
Try solving on your own before revealing the answer!
Final Answer: 35.2 map units
Map distance = (number of single crossovers + double crossovers) / 3,000 × 100 = 35.2 map units.
Q48. Three recessive genes in the tomato plant produce an absence of anthocyanin pigment (a), hairless plants (h), and jointless fruit stems (j). Among 3,000 progeny from a trihybrid F1 that was testcrossed, the phenotypes in figure 5 above were observed. What is your best estimate of the hairless-jointless distance?
Background
Topic: Calculation of map distance between two genes using three-point testcross data.
This question tests your ability to identify recombinant classes and calculate map distance between two genes.
Key Terms and Formulas:
Map distance = (Number of recombinant progeny between two genes / Total progeny) × 100
Include both single and double crossover classes in the recombinant count.
Step-by-Step Guidance
Identify the recombinant classes for hairless and jointless genes.
Sum the numbers of these recombinant progeny (including double crossovers).
Set up the formula for map distance using the recombinant and total progeny numbers, but do not compute the final value yet.
Try solving on your own before revealing the answer!
Final Answer: 15.3 map units
Map distance = (number of single crossovers + double crossovers) / 3,000 × 100 = 15.3 map units.
Q49. Three recessive genes in the tomato plant produce an absence of anthocyanin pigment (a), hairless plants (h), and jointless fruit stems (j). Among 3,000 progeny from a trihybrid F1 that was testcrossed, the phenotypes in figure 5 above were observed. Approximately what is the coefficient of coincidence?
Background
Topic: Coefficient of coincidence and interference in genetic mapping.
This question tests your ability to calculate the coefficient of coincidence from observed and expected double crossover numbers.
Key Terms and Formulas:
Coefficient of coincidence (CoC) = Observed double crossovers / Expected double crossovers
Expected double crossovers = (recombination frequency between a-h) × (recombination frequency between h-j) × total progeny
Step-by-Step Guidance
Determine the observed number of double crossovers from the data (sum the two least frequent classes).
Calculate the expected number of double crossovers using the product of the two single crossover frequencies and the total progeny.
Set up the formula for coefficient of coincidence, but do not compute the final value yet.
Try solving on your own before revealing the answer!
Final Answer: 0.6
Coefficient of coincidence = observed double crossovers / expected double crossovers = 0.6.
