CHAPTER 7
Application and Experimental Questions (Includes Most Mapping Questions)
E1.Figure 7.1 shows the first experimental results that indicated linkage between two different genes. Conduct a chi square analysis to confirm that the genes are really linked and the data could not be explained by independent assortment.
Answer: If we hypothesize two genes independently assorting, the predicted ratio is 9:3:3:1. The total number of offspring is 427. The expected numbers of offspring are
9/16 427 = 240 purple flowers, long pollen
3/16 427 = 80 purple flowers, round pollen
3/16 427 = 80 red flowers, long pollen
1/16 427 = 27 red flowers, round pollen
Plugging these values into our chi square formula, along with the observed numbers, we get:
Looking up this value in the chi square table under 3 degrees of freedom, we find that such a large value is expected by chance less than 1% of the time. Therefore, we reject the hypothesis that the genes assort independently and conclude that the genes are linked.
E2.In the experiment of Figure 7.6, the researchers followed the inheritance pattern of chromosomes that were abnormal at both ends to correlate genetic recombination with the physical exchange of chromosome pieces. Is it necessary to use a chromosome that is abnormal at both ends, or could the researchers have used a strain with two abnormal versions of chromosome 9, one with a knob at one end and its homolog with a translocation at the other end?
Answer: They could have used a strain with two abnormal chromosomes. In this case, the recombinant chromosomes would either look normal or have abnormalities at both ends.
E3.How would you determine that genes in mammals are located on the Y chromosome linkage group? Is it possible to conduct crosses (let’s say in mice) to map the distances between genes along the Y chromosome? Explain.
Answer: A gene on the Y chromosome in mammals would only be transmitted from father to son. It would be difficult to genetically map Y-linked genes because a normal male has only one copy of the Y chromosome, so you do not get any crossing over between two Y chromosomes. Occasionally, abnormal males (XYY) are born with two Y chromosomes. If such males were heterozygous for alleles of Y-linked genes, one could examine the normal male offspring of XYY fathers and determine if crossing over has occurred.
E4.Explain the rationale behind a testcross. Is it necessary for one of the parents to be homozygous recessive for the genes of interest? In the heterozygous parent of a testcross, must all of the dominant alleles be linked on the same chromosome and all of the recessive alleles be linked on the homolog?
Answer: The rationale behind a testcross is to determine if recombination has occurred during meiosis in the heterozygous parent. The other parent is usually homozygous recessive, so we cannot tell if crossing over has occurred in the recessive parent. It is easier to interpret the data if a testcross does use a completely homozygous recessive parent. However, in the other parent, it is not necessary for all of the dominant alleles to be on one chromosome and all of the recessive alleles on the other. The parental generation provides us with information concerning the original linkage pattern between the dominant and recessive alleles.
E5.In your own words, explain why a testcross cannot produce more than 50% recombinant offspring. When a testcross does produce 50% recombinant offspring, what do these results mean?
Answer: The answer is explained in solved problem S5. We cannot get more than 50% recombinant offspring because the pattern of multiple crossovers can yield an average maximum value of only 50%. When a testcross does yield a value of 50% recombinant offspring, it can mean two different things. Either the two genes are on different chromosomes or the two genes are on the same chromosome but at least 50 mu apart.
E6.Explain why the percentage of recombinant offspring in a testcross is a more accurate measure of map distance when two genes are close together. When two genes are far apart, is the percentage of recombinant offspring an underestimate or overestimate of the actual map distance?
Answer: The reason why the percentage of recombinant offspring is more accurate when the genes are close together is because fewer double crossovers occur. The inability to detect double crossover causes the map distance to be underestimated. If two genes are very close together, very few double crossovers occur, so that underestimation due to double crossovers is minimized.
E7.If two genes are more than 50 mu apart, how would you ever be able to show experimentally that they are located on the same chromosome?
Answer: If two genes are at least 50 mu apart, you would need to map genes between them to show that the two genes were actually in the same linkage group. For example, if gene A was 55 mu from gene B, there might be a third gene (e.g., gene C) that was 20 mu from A and 35 mu from B. These results would indicate that A and B are 55 mu apart, assuming dihybrid testcrosses between genes A and B yielded 50% recombinant offspring.
E8.In Morgan’s trihybrid testcross of Figure 7.3, he realized that crossing over was more frequent between the eye color and wing length genes than between the body color and eye color genes. Explain how he determined this.
Answer: Morgan determined this by analyzing the data in gene pairs. This analysis revealed that there were fewer recombinants between certain gene pairs (e.g., body color and eye color) than between other gene pairs (e.g., eye color and wing length). From this comparison, he hypothesized that genes that are close together on the same chromosome will produce fewer recombinants than genes that are farther apart.
E9.In the experiment of Figure 7.9, list the gene pairs from the particular dihybrid crosses that Sturtevant used to construct his genetic map.
Answer: Sturtevant used the data involving the following pairs: y and w,w and v,v and r, and v and m.
E10.In the tomato, red fruit (R) is dominant over yellow fruit (r), and yellow flowers (Wf) are dominant over white flowers (wf). A cross was made between true-breeding plants with red fruit and yellow flowers, and plants with yellow fruit and white flowers. The F1 generation plants were then crossed to plants with yellow fruit and white flowers. The following results were obtained:
333 red fruit, yellow flowers
64 red fruit, white flowers
58 yellow fruit, yellow flowers
350 yellow fruit, white flowers
Calculate the map distance between the two genes.
Answer:
Map distance:
E11.Two genes are located on the same chromosome and are known to be 12 mu apart. An AABB individual was crossed to an aabb individual to produce AaBb offspring. The AaBb offspring were then crossed to aabb individuals.
A.If this cross produces 1,000 offspring, what are the predicted numbers of offspring with each of the four genotypes: AaBb,Aabb,aaBb, and aabb?
B.What would be the predicted numbers of offspring with these four genotypes if the parental generation had been AAbb and aaBB instead of AABB and aabb?
Answer:
A.Because they are 12 mu apart, we expect 12% (or 120) recombinant offspring. This would be approximately 60 Aabb and 60 aaBb plus 440 AaBb and 440 aabb.
B.We would expect 60 AaBb, 60 aabb, 440 Aabb, and 440 aaBb.
E12.Two genes, designated A and B, are located 10 mu from each other. A third gene, designated C, is located 15 mu from B and 5 mu from A. A parental generation consisting of AA bb CC and aa BB cc individuals were crossed to each other. The F1 heterozygotes were then testcrossed to aa bb cc individuals. If we assume no double crossovers occur in this region, what percentage of offspring would you expect with the following genotypes?
A.Aa Bb Cc
B.aa Bb Cc
C.Aa bb cc
Answer: We consider the genes in pairs: there should be 10% offspring due to crossing over between genes A and B, and 5% due to crossing over between A and C.
A.This is due to a crossover between B and A. The parentals are Aa bb Cc and aa Bb cc. The 10% recombinants are Aa Bb Cc and aa bb cc. If we assume an equal number of both types of recombinants, 5% are Aa Bb Cc.
B.This is due to a crossover between A and C. The parentals are Aa bb Cc and aa Bb cc. The 5% recombinants are aa Bb Cc and Aa bb cc. If we assume an equal number of both types of recombinants, 2.5% are aa Bb Cc.
C.This is also due to a crossover between A and C. The parentals are Aa bb Cc and aa Bb cc. The 5% recombinants are aa Bb Cc and Aa bb cc. If we assume an equal number of both types of recombinants, 2.5% are Aa bb cc.
E13.Two genes in tomatoes are 61 mu apart; normal fruit (F) is dominant to fasciated fruit (f), and normal numbers of leaves (Lf) is dominant to leafy (lf). A true-breeding plant with normal leaves and fruit was crossed to a leafy plant with fasciated fruit. The F1 offspring were then crossed to leafy plants with fasciated fruit. If this cross produced 600 offspring, what are the expected numbers of plants in each of the four possible categories: normal leaves, normal fruit; normal leaves, fasciated fruit; leafy, normal fruit; and leafy, fasciated fruit?
Answer: Due to the large distance between the two genes, they will assort independently even though they are actually on the same chromosome. According to independent assortment, we expect 50% parental and 50% recombinant offspring. Therefore, this cross will produce 150 offspring in each of the four phenotypic categories.
E14.In the tomato, three genes are linked on the same chromosome. Tall is dominant to dwarf, skin that is smooth is dominant to skin that is peachy, and fruit with a normal tomato shape is dominant to oblate shape. A plant that is true-breeding for the dominant traits was crossed to a dwarf plant with peachy skin and oblate fruit. The F1 plants were then testcrossed to dwarf plants with peachy skin and oblate fruit. The following results were obtained:
151 tall, smooth, normal
33 tall, smooth, oblate
11 tall, peach, oblate
2 tall, peach, normal
155 dwarf, peach, oblate
29 dwarf, peach, normal
12 dwarf, smooth, normal
0 dwarf, smooth, oblate
Construct a genetic map that describes the order of these three genes and the distances between them.
Answer:
A.One basic strategy to solve this problem is to divide the data up into gene pairs and determine the map distance between two genes.
184 tall, smooth
13 tall, peach
184 dwarf, peach
12 dwarf, smooth
153 tall, normal
44 tall, oblate
155 dwarf, oblate
41 dwarf, normal
163 smooth, normal
33 smooth, oblate
31 peach, normal
166 peach, oblate
Use the two shortest distances to compute the map:
Tall,dwarf 6.4 Smooth,peach 16.3 Normal,oblate
E15.A trait in garden peas involves the curling of leaves. A dihybrid cross was made involving a plant with yellow pods and curling leaves to a wild-type plant with green pods and normal leaves. All F1 offspring had green pods and normal leaves. The F1 plants were then crossed to plants with yellow pods and curling leaves. The following results were obtained:
117 green pods, normal leaves
115 yellow pods, curling leaves
78 green pods, curling leaves
80 yellow pods, normal leaves
A.Conduct a chi square analysis to determine if these two genes are linked.
B.If they are linked, calculate the map distance between the two genes. How accurate do you think this distance is?
Answer:
A.If we hypothesize the two genes are independently assorting, then the predicted ratio is 1:1:1:1. There are a total of 390 offspring. The expected number of offspring in each category is about 98. Plugging the figures into our chi square formula,
Looking up this value in the chi square table under 3 degrees of freedom, we reject our hypothesis, because the chi square value is above 7.815.
B.Map distance:
Because the value is relatively close to 50 mu, it is probably a significant underestimate of the true distance between these two genes.
E16.In mice, the gene that encodes the enzyme inosine triphosphatase is 12 mu from the gene that encodes the enzyme ornithine decarboxylase. Let’s suppose you have identified a strain of mice homozygous for a defective inosine triphosphatase gene that does not produce any of this enzyme and is also homozygous for a defective ornithine decarboxylase gene. In other words, this strain of mice cannot make either enzyme. You crossed this homozygous recessive strain to a normal strain of mice to produce heterozygotes. The heterozygotes were then backcrossed to the strain that cannot produce either enzyme. What is the probability of obtaining a mouse that cannot make either enzyme?
Answer: In the backcross, the two parental types would be the homozygotes that cannot make either enzyme, and the heterozygotes that can make both enzymes. The recombinants would make one enzyme but not both. Because the two genes are 12 mu apart, 12% would be recombinants and 88% would be parental types. Because there are two parental types are produced in equal numbers, we would expect 44% of the mice to be unable to make either enzyme.
E17.In the garden pea, several different genes affect pod characteristics. A gene affecting pod color (green is dominant to yellow) is approximately 7 mu away from a gene affecting pod width (wide is dominant to narrow). Both genes are located on chromosome 5. A third gene, located on chromosome 4, affects pod length (long is dominant to short). A true-breeding wild-type plant (green, wide, long pods) was crossed to a plant with yellow, narrow, short pods. The F1 offspring were then testcrossed to plants with yellow, narrow, short pods. If the testcross produced 800 offspring, what are the expected numbers of the eight possible phenotypic combinations?
Answer: The percentage of recombinants for the green, yellow and wide, narrow is 7%, or 0.07; there will be 3.5% of the green, narrow and 3.5% of the yellow, wide. The remaining 93% parentals will be 46.5% green, wide and 46.5% yellow, narrow. The third gene assorts independently. There will be 50% long and 50% short with respect to each of the other two genes. To calculate the number of offspring out of a total of 800, we multiply 800 by the percentages in each category.
(0.465 green, wide)(0.5 long)(800) = 186 green, wide, long
(0.465 yellow, narrow)(0.5 long)(800) = 186 yellow, narrow, long
(0.465 green, wide)(0.5 short)(800) = 186 green, wide, short
(0.465 yellow, narrow)(0.5 short)(800) = 186 yellow, narrow, short
(0.035 green, narrow)(0.5 long)(800) = 14 green, narrow, long
(0.035 yellow, wide)(0.5 long)(800) = 14 yellow, wide, long
(0.035 green, narrow)(0.5 short)(800) = 14 green, narrow, short
(0.035 yellow, wide)(0.5 short)(800) = 14 yellow, wide, short
E18.A sex-influenced trait is dominant in males and causes bushy tails. The same trait is recessive in females and results in a normal tail. Fur color is not sex influenced. Yellow fur is dominant to white fur. A true-breeding female with a bushy tail and yellow fur was crossed to a white male without a bushy tail. The F1 females were then crossed to white males without bushy tails. The following results were obtained:
MalesFemales
28 normal tails, yellow102 normal tails, yellow
72 normal tails, white96 normal tails, white
68 bushy tails, yellow0 bushy tails, yellow
29 bushy tails, white0 bushy tails, white
A.Conduct a chi square analysis to determine if these two genes are linked.
B.If the genes are linked, calculate the map distance between them. Explain which data you used in your calculation.
Answer:
A.If we represent B (bushy tail) and b (normal tail) for one gene, and Y (yellow) and y (white) for the second gene:
Parent generation: BBYYbbyy
F1 generation:All BbYy(NOTE: if the two genes are linked, B would be linked to Y and b would be linked to y.)
Testcross: F1BbYybbyy
Nonrecombinant offspring from testcross: BbYy and bbyy
BbYy males—bushy tails, yellow
bbyy males—normal tails, white
BbYy females—normal tails, yellow
bbyy females—normal tails, white
Recombinant offspring from testcross: Bbyy and bbYy
Bbyy males—bushy tails, white
bbYy males—normal tails, yellow
Bbyy females—normal tails, white
bbYy females—normal tails, yellow
We cannot use the data regarding female offspring, because we cannot tell if females are recombinant or nonrecombinant, because all females have normal tails. However, we can tell if male offspring are recombinant.
If we use the data on males to conduct a chi-square analysis, we expect a 1:1:1:1 ratio among the male offspring. Because there are 197 male offspring total, we expect 1/4, or 49 (rounded to the nearest whole number), of the four possible phenotypes. To compute the chi square:
If we look up the value of 35.4 in our chi square table, with 3 degrees of freedom, the value lies far beyond the 0.01 probability level. Therefore, it is very unlikely to get such a large deviation if our hypothesis of independent assortment is correct. Therefore, we reject our hypothesis and conclude that the genes are linked.
B.To compute map distance:
E19.Three recessive traits in garden pea plants are as follows: yellow pods are recessive to green pods, bluish-green seedlings are recessive to green seedlings, creeper (a plant that cannot stand up) is recessive to normal. A true-breeding normal plant with green pods and green seedlings was crossed to a creeper with yellow pods and bluish green seedlings. The F1 plants were then crossed to creepers with yellow pods and bluish green seedlings. The following results were obtained:
2,059 green pods, green seedlings, normal