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Chapter 7 Application and Experimental Questions


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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:



Males Females

28 normal tails, yellow 102 normal tails, yellow

72 normal tails, white   96 normal tails, white

68 bushy tails, yellow 0 bushy tails, yellow

29 bushy tails, white 0 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: BBYYbbyy

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: F1 BbYybbyy

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

151 green pods, green seedlings, creeper

281 green pods, bluish green seedlings, normal

15 green pods, bluish green seedlings, creeper

2,041 yellow pods, bluish green seedlings, creeper

157 yellow pods, bluish green seedlings, normal

282 yellow pods, green seedlings, creeper

11 yellow pods, green seedlings, normal

Construct a genetic map that describes the map distance between these three genes.

Answer: Let’s use the following symbols: G for green pods, g for yellow pods, S for green seedlings, s for bluish green seedlings, C for normal plants, c for creepers. The parental cross is GG SS CC crossed to gg ss cc.

The F1 plants would all be Gg Ss Cc. If the genes are linked, the alleles G, S, and C would be linked on one chromosome, and the alleles g, s, and c would be linked on the homologous chromosome.

The testcross is F1 plants, which are Gg Ss Cc, crossed to gg ss cc.

To measure the distances between the genes, we can separate the data into gene pairs.

Pod color, seedling color

2,210 green pods, green seedlings—nonrecombinant

296 green pods, bluish green seedlings—recombinant

2,198 yellow pods, bluish green seedlings—nonrecombinant

293 yellow pods, green seedlings—recombinant

Pod color, plant stature

2,340 green pods, normal—nonrecombinant

166 green pods, creeper—recombinant

2,323 yellow pods, creeper—nonrecombinant

168 yellow pods, normal—recombinant



Seedling color, plant stature

2,070 green seedlings, normal—nonrecombinant

433 green seedlings, creeper—recombinant

2,056 bluish green seedlings, creeper—nonrecombinant

438 bluish green seedlings, normal—recombinant



The order of the genes is seedling color, pod color, and plant stature (or you could say the opposite order). Pod color is in the middle. If we use the two shortest distances to construct our map:



S 11.8 G 6.7 C

E20. In mice, a trait called snubnose is recessive to a wild-type nose, a trait called pintail is dominant to a normal tail, and a trait called jerker (a defect in motor skills) is recessive to a normal gait. Jerker mice with a snubnose and pintail were crossed to normal mice, and then the F1 mice were crossed to jerker mice that have a snubnose and normal tail. The outcome of this cross was as follows:

560 jerker, snubnose, pintail

548 normal gait, normal nose, normal tail

102 jerker, snubnose, normal tail

104 normal gait, normal nose, pintail

77 jerker, normal nose, normal tail

71 normal gait, snubnose, pintail

11 jerker, normal nose, pintail

9 normal gait, snubnose, normal tail

Construct a genetic map that describes the order and distance between these genes.

Answer: Let’s use the following symbols: S for normal nose, s for snubnose, p for normal tail, P for pintail, J for normal gait, j for jerker.

The parental cross is ss PP jj crossed to SS pp JJ.

The F1 offspring would all be Ss Pp Jj. If the genes are linked, the alleles s, P, and j would be linked on one chromosome, and the alleles S, p, and J would be linked on the homologous chromosome.

The testcross is F1 mice, which are Ss Pp Jj, crossed to ss pp jj mice.

To measure the distances between the genes, we can separate the data into gene pairs.

Nose shape, tail length

631 snubnose, pintail—nonrecombinant

111 snubnose, normal tail—recombinant

625 normal nose, normal tail—nonrecombinant

115 normal nose, pintail—recombinant



Nose shape, normal gait

662 snubnose, jerker—nonrecombinant

80 snubnose, normal gait—recombinant

652 normal nose, normal gait—nonrecombinant

88 normal nose, jerker—recombinant



Tail length, normal gait

571 pintail, jerker—nonrecombinant

175 pintail, normal gait—recombinant

557 normal tail, normal gait—nonrecombinant

179 normal tail, jerker—recombinant



The order of the genes is tail length, nose shape, and normal gait (or you could say the opposite order). Nose shape is in the middle.

If we use the two shortest distances to construct our map:

P 15.2 S 11.3 J

E21. In Drosophila, an allele causing vestigial wings is 12.5 mu away from another gene that causes purple eyes. A third gene that affects body color has an allele that causes black body color. This third gene is 18.5 mu away from the vestigial wings gene and 6 mu away from the gene causing purple eyes. The alleles causing vestigial wings, purple eyes, and black body are all recessive. The dominant (wild-type) traits are long wings, red eyes, and gray body. A researcher crossed wild-type flies to flies with vestigial wings, purple eyes, and black bodies. All F1 flies were wild type. F1 female flies were then crossed to male flies with vestigial wings, purple eyes, and black bodies. If 1,000 offspring were observed, what are the expected numbers of the following types of flies?

Long wings, red eyes, gray body

Long wings, purple eyes, gray body

Long wings, red eyes, black body

Long wings, purple eyes, black body

Short wings, red eyes, gray body

Short wings, purple eyes, gray body

Short wings, red eyes, black body

Short wings, purple eyes, black body

Which kinds of flies can be produced only by a double crossover event?

Answer: To answer this question, we can consider genes in pairs. Let’s consider the two gene pairs that are closest together. The distance between the wing length and eye color genes is 12.5 mu. From this cross, we expect 87.5% to have long wings and red eyes or short wings and purple eyes, and 12.5% to have long wings and purple eyes or short wings and red eyes. Therefore, we expect 43.75% to have long wings and red eyes, 43.75% to have short wings and purple eyes, 6.25% to have long wings and purple eyes, and 6.25% to have short wings and red eyes. If we have 1,000 flies, we expect 438 to have long wings and red eyes, 438 to have short wings and purple eyes, 62 to have long wings and purple eyes, and 62 to have short wings and red eyes (rounding to the nearest whole number).

The distance between the eye color and body color genes is 6 mu. From this cross, we expect 94% to have a parental combination (red eyes and gray body or purple eyes and black body) and 6% to have a nonparental combination (red eyes and black body or purple eyes and gray body). Therefore, of our 438 flies with long wings and red eyes, we expect 94% of them (or about 412) to have long wings, red eyes, and gray body, and 6% of them (or about 26) to have long wings, red eyes, and black bodies. Of our 438 flies with short wings and purple eyes, we expect about 412 to have short wings, purple eyes, and black bodies, and 26 to have short wings, purple eyes, and gray bodies.

Of the 62 flies with long wings and purple eyes, we expect 94% of them (or about 58) to have long wings, purple eyes, and black bodies, and 6% of them (or about 4) to have long wings, purple eyes, and gray bodies. Of the 62 flies with short wings and red eyes, we expect 94% (or about 58) to have short wings, red eyes, and gray bodies, and 6% (or about 4) to have short wings, red eyes, and black bodies.

In summary:

Long wings, red eyes, gray body 412

Long wings, purple eyes, gray body 4

Long wings, red eyes, black body 26

Long wings, purple eyes, black body 58

Short wings, red eyes, gray body 58

Short wings, purple eyes, gray body 26

Short wings, red eyes, black body 4

Short wings, purple eyes, black body 412

The flies with long wings, purple eyes, and gray bodies, or short wings, red eyes, and black bodies, are produced by a double crossover event.

E22. Three autosomal genes are linked along the same chromosome. The distance between gene A and B is 7 mu, the distance between B and C is 11 mu, and the distance between A and C is 4 mu. An individual who is AA bb CC was crossed to an individual who is aa BB cc to produce heterozygous F1 offspring. The F1 offspring were then crossed to homozygous aa bb cc individuals to produce F2 ­offspring.

A. Draw the arrangement of alleles on the chromosomes in the parents and in the F1 offspring.

B. Where would a crossover have to occur to produce an F2 offspring that was heterozygous for all three genes?

C. If we assume that no double crossovers occur in this region, what percentage of F2 offspring is likely to be homozygous for all three genes?

Answer:

A.


Parent Parent

b   7   A  4  CB   7  a  4  c

b   7   A  4  C B   7  a  4  c


b   7   A  4  C   Offspring



B   7   a  4  c
B. A heterozygous F2 offspring would have to inherit a chromosome carrying all of the dominant alleles. In the F1 parent (of the F2 offspring), a crossover in the 7 mu region between genes b and A (and between B and a) would yield a chromosome that was B A C and b a c. If an F2 offspring inherited the BAC chromosome from its F1 parent and the b a c chromosome from the homozygous parent, it would be heterozygous for all three genes.

C. If you look at the answer to part B, a crossover between genes b and A (and between B and a) would yield B A C and b a c chromosomes. If an offspring inherited the b a c chromosome from its F1 parent and the b a c chromosome from its homozygous parent, it would be homozygous for all three genes. The chances of a crossover in this region are 7%. However, half of this 7% crossover event yields chromosomes that are B A C and the other half yields chromosomes that are bac. Therefore, the chances are 3.5% of getting homozygous F2 offspring.

E23. Let’s suppose that two different X-linked genes exist in mice, designated with the letters N and L. Gene N exists in a dominant, normal allele, and in a recessive allele, n, that is lethal. Similarly, gene L exists in a dominant, normal allele, and in a recessive allele, l, that is lethal. Heterozygous females are normal, while males that carry either recessive allele are born dead. Explain whether or not it would be possible to map the distance between these two genes by making crosses and analyzing the number of living and dead offspring. You may assume that you have strains of mice in which females are heterozygous for one or both genes.

Answer: Yes. Begin with females that have one X chromosome that is XNl and the other X chromosome that is XnL. These females have to be mated to XNLY males because a living male cannot carry the n or l allele. In the absence of crossing over, a mating between XNlXnL females to XNLY males should not produce any surviving male offspring. However, during oogenesis in these heterozygous female mice, there could be a crossover in the region between the two genes, which would produce an XNL chromosome and an Xnl chromosome. Male offspring inheriting these recombinant chromosomes will be either XNLY or XnlY (whereas nonrecombinant males will be XnLY or XNlY). Only the male mice that inherit XNLY will live. The living males represent only half of the recombinant offspring. (The other half are XnlY, which are born dead.)

To compute map distance:





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