44. 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?
259 hairless 268 anthocyaninless, jointless, hairless
40 jointless, hairless 941 anthocyaninless, hairless
931 jointless 32 anthocyaninless
260 normal 269 anthocyaninless, jointless
a. h-a-j. b. a-h-j. c. a-j-h. d. j-h-a. e. two of these choices.
45. 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 parentals (P1) were
homozygous, which of the following could have been a parental?
a. anthocyaninless. b. jointless. c. anthocyaninless, jointless.
d. hairless, jointless. e. hairless.
46. 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?
a. 20.0. b. 35.2. c. 37.6. d. 40.0. e. 60.0.
47. 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?
a. 20.0. b. 35.2. c. 37.6. d. 40.0. e. 60.0.
48. 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 were observed. What is your best estimate of the
hairless-jointless distance?
a. 20.0. b. 35.2. c. 37.6. d. 40.0. e. 60.0.
49. 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 were observed. Approximately what is the coefficient
of coincidence?
a. 0.4 b. 0.5 c. 0.6 d. 1.8 e. irrelevant
how do you solve these problems explain please
To solve these problems, we need to analyze the observed phenotypes of the progeny and deduce the linkage arrangements and distances between the genes.
44. To determine the linkage arrangement of the three loci, we need to look at the phenotypes observed in the progeny:
- 259 hairless plants were observed, which means the hairless trait (h) is most likely not linked with the other two traits since it appears more frequently.
- 268 plants were anthocyaninless, jointless, and hairless.
- 40 plants were jointless and hairless.
- 941 plants were anthocyaninless and hairless.
- 931 plants were jointless.
- 32 plants were anthocyaninless.
From these observations, we can conclude that the anthocyaninless trait (a) and the jointless trait (j) are linked. The hairless trait (h) appears to be unlinked. Thus, the linkage arrangement is a-j, with h unlinked.
The correct answer is (a) h-a-j.
45. If the parental (P1) plants were homozygous, we need to look for a phenotype that appears in large numbers. From the observations, we see that 941 progeny were anthocyaninless and hairless. This is the largest number observed, so it is most likely the parental phenotype.
The correct answer is (c) anthocyaninless, hairless.
46. To estimate the distance between the anthocyaninless (a) and hairless (h) genes, we need to identify the progeny that have both traits and calculate their percentage among all progeny. From the observations, 268 progeny were anthocyaninless, jointless, and hairless.
% recombinant = (268 / 3000) * 100 = 8.93%
The correct answer is approximately (a) 20.0.
47. To estimate the distance between the anthocyaninless (a) and jointless (j) genes, we need to identify the progeny that have both traits and calculate their percentage among all progeny. From the observations, 268 progeny were anthocyaninless, jointless, and hairless.
% recombinant = (268 / 3000) * 100 = 8.93%
The correct answer is approximately (a) 20.0.
48. To estimate the distance between the hairless (h) and jointless (j) genes, we need to identify the progeny that have both traits and calculate their percentage among all progeny. From the observations, 40 progeny were jointless and hairless.
% recombinant = (40 / 3000) * 100 = 1.33%
The correct answer is approximately (a) 20.0.
49. To calculate the coefficient of coincidence, we need to divide the observed double recombinant progeny by the expected double recombinant progeny.
Observed double recombinant progeny: 268 (found in 44)
Expected double recombinant progeny: (268 / 3000) * 32 = 2.845
Coefficient of coincidence = observed double recombinants / expected double recombinants
Coefficient of coincidence = 268 / 2.845 = 93.93
The correct answer is (e) irrelevant (the options provided do not include the correct value).
Note: The calculations above assume that all observed progeny are derived from single crossover events.
To solve these problems, we need to analyze the information provided and apply the principles of genetic linkage and recombination.
Problem 44: The first question asks about the linkage arrangement of the three loci (genes) involved: anthocyanin pigment (a), hairless plants (h), and jointless fruit stems (j). To determine the linkage arrangement, we need to calculate the map distance between each pair of genes and determine if they are linked or on separate chromosomes.
From the phenotypes observed in the progeny, we can count the number of recombinant offspring for each pair of genes. The pairings are as follows:
- Hairless (h) and Anthocyaninless (a): 268 recombinant offspring observed
- Hairless (h) and Jointless (j): 260 recombinant offspring observed
- Jointless (j) and Anthocyaninless (a): 931 recombinant offspring observed
Based on these numbers, we can see that there is a higher number of recombinant offspring for the pair Jointless (j) and Anthocyaninless (a) compared to the other two pairs. This suggests that the genes for Jointless (j) and Anthocyaninless (a) are furthest apart and are likely on different chromosomes, while the genes for Hairless (h) and Jointless (j) are closest together and thus more likely to be linked.
Therefore, the best estimate for the linkage arrangement of these three loci is: a-j-h. So, the correct choice is (c) a-j-h.
Problem 45: This question asks which phenotypes could be present in the parental generation (P1) if they were homozygous for the three traits.
We can look at the phenotypes observed in the progeny to determine the possible phenotypes in the parental generation. From the given data, we can see that the following phenotypes were observed in the progeny: anthocyaninless, hairless, and jointless.
If the parental generation were homozygous for the three traits, then all the offspring would exhibit the same phenotypes as the parents. So, the correct answer is (c) anthocyaninless, jointless.
Problem 46 to 48: These questions ask for the best estimate of the distances between pairs of genes: anthocyaninless/hairless (a-h), anthocyaninless/jointless (a-j), and hairless/jointless (h-j).
To estimate the distances, we can use the formula:
Map distance (%) = (Number of recombinant offspring / Total number of offspring) x 100
Using the provided data, we can calculate the following map distances:
- a-h distance: (269 / 3000) x 100 ≈ 8.97% (approximately 9%)
- a-j distance: (931 / 3000) x 100 ≈ 31.03% (approximately 31%)
- h-j distance: (260 / 3000) x 100 ≈ 8.67% (approximately 9%)
So, the best estimate for the distances between pairs of genes are: a-h ≈ 9%, a-j ≈ 31%, and h-j ≈ 9%. The closest estimate available is 9%, so the correct answers are:
- Problem 46: (a) 20.0
- Problem 47: (a) 20.0
- Problem 48: (a) 20.0
Problem 49: This question asks for the coefficient of coincidence, which is a measure of the interference between two crossing-over events in a double crossover.
The coefficient of coincidence is calculated by dividing the observed double crossovers by the expected double crossovers.
From the given data, we have:
- Observed double crossovers: 32
- Expected double crossovers can be calculated by multiplying the recombinant frequencies for each pair of genes: (269 / 3000) x (931 / 3000) = 0.00008383333
Coefficient of coincidence = Observed double crossovers / Expected double crossovers = 32 / 0.00008383333 ≈ 381.70
Since the options provided do not match the calculated value, the correct answer is (e) irrelevant.
To solve these problems, we need to analyze the observed phenotypes and use the principles of linkage and recombination. Let's go through each question step by step:
44. The first question asks for the linkage arrangement of the three loci (genes) based on the observed phenotypes. From the given data, we can see that 259 progeny are hairless, and among them, 268 are also anthocyaninless and jointless. This indicates that the hairless and anthocyaninless genes are linked together, while the jointless gene is independent. Therefore, the correct answer is e. two of these choices: a-h-j or j-h-a.
45. The second question asks which of the phenotypes could be a parental type when the parents (P1) are homozygous. By looking at the options, we see that the parental types are those without any recombinations. In this case, the option c. anthocyaninless, jointless can be a parental type since both genes are independent of each other.
46. The third question asks for the best estimate of the distance between the anthocyaninless and hairless genes. To calculate the distance, we need to determine the frequency of recombination events between these two genes. From the data, we observe that 268 progeny are both anthocyaninless and jointless, while 259 are hairless. The difference between these two values (268-259 = 9) represents the number of recombination events. To calculate the recombination percentage, we use the formula: (recombinant/total progeny) x 100 = (9/3000) x 100 = 0.3%. Therefore, the best estimate is 0.3 units or 0.3 centimorgans. Unfortunately, none of the given options provide this exact value, so we cannot choose a specific answer.
47. The fourth question asks for the best estimate of the distance between the anthocyaninless and jointless genes. To calculate the distance, we follow the same procedure as in question 46. From the data, we observe that 268 progeny are both anthocyaninless and jointless, while 931 are jointless. The difference between these two values (931-268 = 663) represents the number of recombination events. To calculate the recombination percentage, we use the formula: (recombinant/total progeny) x 100 = (663/3000) x 100 = 22.1%. Therefore, the best estimate is 22.1 units or 22.1 centimorgans. Unfortunately, none of the given options include this value, so we cannot choose a specific answer.
48. The fifth question asks for the best estimate of the distance between the hairless and jointless genes. Again, we follow the same procedure as in questions 46 and 47. From the data, we observe that 941 progeny are hairless, while 931 are jointless. The difference between these two values (941-931 = 10) represents the number of recombination events. To calculate the recombination percentage, we use the formula: (recombinant/total progeny) x 100 = (10/3000) x 100 = 0.3%. Therefore, the best estimate is 0.3 units or 0.3 centimorgans. Unfortunately, none of the given options provide this exact value, so we cannot choose a specific answer.
49. The final question asks for the approximate coefficient of coincidence. The coefficient of coincidence measures the degree to which recombination events are suppressed. To calculate it, we need to determine the expected number of double recombinants (DR) and compare it to the observed number of double recombinants (ObsDR). In this case, we are not given the ObsDR value, so it is impossible to calculate the coefficient of coincidence. Therefore, the answer is e. irrelevant.
In summary, while we can solve some aspects of these problems, such as the presence of linkage and the estimation of recombination distances, there are no exact answers provided for the distances or the coefficient of coincidence.