Genetics Linkage Recombination PDF

8 Linkage, Recombination and Gene Mapping In Mendel’s dihybrid experiments the F1 double heterozygote always showed independent assortment of the two pairs of genes at the time of gamete formation. In fact the law was justified only because the t...

8 Linkage, Recombination and Gene Mapping In Mendel’s dihybrid experiments the F1 double heterozygote always showed independent assortment of the two pairs of genes at the time of gamete formation. In fact the law was justified only because the two genes were not linked to each other. This was true because each gene was located on a different chromosome. Mendel studied seven characters and the pea plant has only seven pairs of chromosomes. It followed therefore that in Mendelian inheritance one character must be located on one chromosome. Sutton in 1903, working on the Chromosome Theory of Heredity pointed out that the number of characters which obeyed Mendel’s laws when studied singly in monohybrid crosses, was much more than the number of chromosome pairs to which they could be assigned. This means that there must be many genes located on the same chromosome. If we focus attention on any two genes located on the same chromosome and perform a dihybrid cross, we cannot expect the two genes to assort independently as in Mendelism. On the contrary the two genes must be linked and show a tendency to be inherited together. If this is true, then the two phenotypes controlled by these genes must also appear together in an individual more often than if there was independent assortment. Stating it in another way, genes on the same chromosome are linked and tend to be transmitted together in a single unit. Obviously then the 9 : 3 : 3 : 1 ratio typical for a dihybrid Mendelian cross would be expected to become modified for two genes that are not assorting independently due to linkage. Indeed Bateson, Saunders and Punnett (1905, 1906) found results of a dihybrid cross in sweat peas different from those expected in independent assortment. The experiment involved two characters, flower colour (purple vs. red) and pollen shape (long vs. round). When two varieties of sweet peas, one purple long the other red round were crossed, the F1 progeny did not appear in the expected 9 : 3 : 3 : 1 ratio. Instead, plants with purple long and those with red round combinations were more frequent than expected. It will be noticed that these are the same combinations that were present in the parents. The new combinations purple round and red long which were not present in the parents were observed in F2 with lesser frequency (Fig. 8.1). The fact emerges that when genes are linked, parental combinations occur more frequently than recombinations. 66  LINKAGE, RECOMBINATION AND GENE MAPPING 67 P: Purple long × Red Round PPLL ppll F1: Purple Long × self PpLl F2: Purple long Purple Round Red long Red Round Observed 1,528 106 117 381 Expected 1,199 400 400 133 (9) (3) (3) (1) Fig. 8.1 Dihybrid cross in sweet peas showing linkage. The same cross was repeated by Bateson and Punnett in a different way. This time the parents had a different combination of characters namely Purple Round and Red Long. The F1 plants were all Purple Long. The F2 again showed the parental combinations (Purple Round and Red Long) in a higher frequency, and the recombinations (Purple Long and Red Round) in a lower frequency than expected. Bateson and Punnett applied the testcross method to the above mentioned crosses in sweet peas. They crossed the F1 double heterozygote (Purple Long PpLl) with the double recessive parent (red round ppll). In Mendelian inheritance such a cross indicates that F1 plants are producing four types of gametes in equal frequency, which combine with the single type of gamete from the recessive parent to produce F2 progeny of four types in the ratio of 1 : 1 : 1 : 1. Instead, the actual ratio observed by Bateson and Punnett was 7 : 1 : 1 : 7. The above experiment clearly indicates that the two genes for flower colour and pollen shape are located on the same chromosome. The genes are said to be linked. During meiosis linked genes tend to pass en bloc to the same gamete and are responsible for the appearance of parental combinations in the resulting progeny. The small number of recombinations in the progeny are due to the fact that some amount of recombination does take place between the linked genes at the pachytene stage of meiosis. More about this will be discussed later. Incidentally, none of the dihybrid crosses performed by Mendel had linked genes. Genes for all seven characters studied by him showed random assortment. It was later demonstrated cytologically that sweet pea has seven pairs of chromosomes. It was also shown genetically that the gene for each of the seven characters studied by Mendel was located on a separate chromosome. Had Mendel studied some more characters that were linked, he would not have been able to interpret his results and his Principle of Independent Assortment may not have been formulated. Bateson coined the term coupling for referring to the situation where two dominant alleles of a gene are both present in one parent and the two recessive alleles in the other. Thus both dominant genes pass together into one gamete in one parent and both recessives together in the gamete of the other parent as in the cross purple long (PPLL) and red round (ppll). The term repulsion was applied when the parents were heterozygous such as in the cross PpLl and PpLl. In this case the two dominant genes come from the different parents so that they are said to be in repulsion. In other words in the case of two linked loci in a double heterozygote if the two dominant alleles are on one chromosome and the two recessives on the other (PL/pl) the linkage relationship is said to be in the coupling phase. But when the dominant allele of  68 GENETICS one locus and the recessive allele of the other occupy the same chromosome (pl/pL) linkage is said to be in repulsion phase. The terms coupling and repulsion are now of historical interest only. MORGAN’S WORK ON DROSOPHILA In the early decades of the twentieth century T.H. Morgan and his associates A.H. Sturtevant, H.J. Muller, C.B. Bridges and a few others at Columbia University, New York were actively engaged in studies on Chromosome Theory of Heredity with a very suitable material Drosophila. From their studies on mutants in Drosophila they could assign several genes to chromosomes. For their extensive researches on Drosophila, T.H. Morgan became the first geneticist to be awarded Nobel Prize in Medicine in 1934. Some of the experiments performed in Morgan’s laboratory indicated linkage because the genes did not assort independently. In Drosophila the normal fly has grey body and long wings. There are some recessive mutations due to which the body colour of the fly is black and the wings are underdeveloped or vestigeal. It was proposed by Bridges that in linkage experiments the wild normal type allele be represented by the sign +, and mutants recessive to wild type by abbreviated initials in small letters such as b for black, and v for vestigial. Since linked genes are present on the same chromosome, they are represented by their symbols above and below a horizontal line, genes on one homologue being above the line, those on the other below the line. Bridges crossed a wild type female fly with grey body and long wings with a male having black body and vestigial wings. The F1 was wild type with grey body and long wings. The females from F1 were testcrossed with double recessive males i.e. black vestigial. Now if the genes are assorting independently, the test cross would yield all four combinations in the ratio of 1 : 1 : 1 : 1. But the actual results showed that the parental types (grey long and black vestigial) were more and the recombinations (grey vestigial and black long) were less than expected (Fig. 8.2). Obviously the genes were linked. One more striking feature was revealed by these experiments, that is, the absence of crossover in the Drosophila male. Due to this when a male fly was used as a double recessive parent in a testcross, there were recombinants present in the F2 progeny. Reciprocal testcrosses where female flies were employed as double recessives, recombinants were absent in the F2 progeny. The crosses provide excellent illustrations of linkage and absence of crossing over in male Drosophila. The female silk moth (Bombyx) is another example where crossing over does not take place. The Drosophila school performed numerous experiments in their efforts to understand the chromosome theory of heredity. They had accumulated a large number of mutants for various characters. It was possible to isolate stocks of flies carrying two mutant genes for different characters on the same chromosome. Interestingly, they also found some hereditary characters which were linked to the sex chromosome X in females. Characters such as white eyes, yellow body, miniature wings and a few others were therefore said to be sex-linked.  LINKAGE, RECOMBINATION AND GENE MAPPING 69 P: Grey long + + b b Black vestigial ++ bv ++ bv + + v v + b Gametes + v + b b b Black vestigial Grey long F1: hybrid bv bv ++ + v v v bv during meiosis + + b b b + v + v v Sperm + b + b b + + v v v Parental Recombinants Parental type egg in eggs type egg + b b b + b b b F2: + v + v v v v v Grey long Black long Grey vestigial Black vestigial ++ b+ +v bv 41% 9% 9% 41% bv bv bv bv Fig. 8.2 Cross made by Bridges demonstrating linkage in Drosophila. In all D. melanogaster was found to have four linked groups of genes. Drosophila has four pairs of chromosomes (Fig. 8.3). Each linkage group in Drosophila was associated with one pair of chromosomes and the linkage groups were numbered 1, 2, 3 and 4. Further, the number of genes in each linkage group was found to be proportional to the size of chromosomes.  70 GENETICS Thus, the tiniest dot-like chromosomes are found to carry the smallest linkage group of only about 12 genes whereas the larger chromosomes each had more than 150 linked genes. 4 4 2 3 2 3 X X X Y 1 1 Fig. 8.3 The chromosome complement of the cell of Drosophila. CROSSING OVER If linkage is complete, there should be all parental combinations only and no recombination. But actually there is no absolute linkage, thus allowing for some recombination. How does recombination take place ? At the beginning of the 20th century, Janssens had made cytological observations on meiotic chromosomes in salamanders. He found that chromosomes showed cross-shaped configurations and suggested that they represented a break and exchange of chromosome segments. A few years later, Morgan supplemented his genetical studies on Drosophila with cytological observations and explained linkage on the basis of breakage and exchange in synapsed chromosomes. He could thus account for the greater frequency of parental combinations and also why linkage was not absolute so that recombinant types occurred in F2 progeny. Morgan termed the cross-shaped configuration observed by Janssens as chiasma. The term crossing over referred to the actual exchange of segments between homologous chromosomes and could take place due to breakage and reunion in the paired homologues. Recombination is a genetic outcome of breakage and exchange of segments. It cannot be observed cytologically, but can by inferred genetically from experiments. Crossing over is the process of exchange of genetic segments which cannot be observed cytologically but can be estimated genetically from the frequency of recombinants in the F2 progeny. Chiasma Frequency Every pair of homologous chromosomes usually forms at least one chiasma somewhere along its length. There is a characteristic average number of chiasmata for each type of chromosome. In general, the longer the chromosome the greater the number of chiasmata. Moreover, the further apart two genes are located on a chromosome, the greater the chance for a chiasma to occur between them. The percentage of crossover (recombinant) gametes formed by a given genotype indicates the frequency of chiasma formation between the genes in question. When chiasma forms in one cell between two gene loci, only half of meiotic products will be of crossover type. Therefore, chiasma frequency is twice the frequency of crossover products. Chiasma % = 2 × crossover % Crossover % = 1/2 Chiasma % In other words, if chiasma forms between the loci of genes A and B in 30 % of the tetrads (paired homologous chromosomes) of an individual of genotype AB/ab, then 15% of the gametes  LINKAGE, RECOMBINATION AND GENE MAPPING 71 will be recombinant (Ab or aB), and 85% will be parental (AB or ab). Further, the map distance between A and B would be 15 units apart. It is noteworthy that the Drosophila male shows complete linkage due to absence of crossing over. In the cross between a normal red-eyed long-winged fly and purple vestigial, the F1 hybrids are all red-eyed and long-winged. If heterozygotes from the F1 progeny are used as male parents and backcrossed with purple vestigial females, only two phenotypes appear in the progeny: the homozygous purple vestigial and the heterozygous red-eyed long-winged. The recombinations are absent demonstrating absence of cross over and presence of complete linkage in Drosophila male. If a reciprocal testcross is performed using F1 heterozygotes as females and purple vestigial as male, recombinants appear in the progeny. Frequency of Crossing Over In experiments on linkage, the proportions of parental phenotypes and the new combinations can be counted. From the percentage of recombinants the amount of crossing over can be calculated. In the cross between purple long and red round sweet peas described earlier, the sum of the new combinations (106 + 117) = 223 when divided by total progeny (1528 + 106 + 117 + 381) = 2132 and multiplied by 100 indicates 10.4% recombination or frequency of crossing over in F1 gametes. In this cross the parental types equal 89.6%. The fact that recombinants occurred in F2 indicates that the distance between genes for flower colour and pollen shape allowed crossing over to take place between one parental chromosome and its homologue from the other parent. Now if distance between two linked genes is more, there are greater chances of chiasma formation between them resulting in higher percentage of recombinants in the progeny. Contrarily, if distance between linked genes is less, chiasma formation between them will be less, and corresponding reduction in the number of recombinants in the progeny. Thus it is possible to locate genes on chromosomes on the basis of their crossover frequency. The distance between genes is measured in map units. According to Sturtevant one map unit is equal to 1% crossing over. In other words, if one gamete out of 100 gametes carries a crossover chromosome for two linked genes, we say that the two genes are one map unit apart. The correlation between crossing over and distances between genes may not be true for all genes on a chromosome. This is because chiasma formation does not occur at random throughout the length of a chromosome. It occurs with different frequency in different parts of the chromosome. In the two large chromosomes of Drosophila there is less crossing over near the centromere and towards the ends, but more in the middle of the two arms. This would suggest that genes near the centromere are closer when actually they are further apart than the crossover percentage indicates. After determining positions of several linked genes in Drosophila, Morgan hypothesised that genes occurred in linear order along the length of the chromosome. The position of each gene was called the locus. THE THREE-POINT CROSS When two genes are mapped by performing a cross, it is called a two-point cross. The three- point cross involves three pairs of linked genes, is a valuable method for determining positions of genes in relation to each other and for mapping distances between genes. In 1926, Bridges and Olbrycht used this method for mapping three recessive sex-linked genes in Drosophila:  72 GENETICS scute sc (without bristles), echinus ec (rough eyes), and crossveinless wings cv (absence of transverse veins). The cross involved mating of a scute crossveinless fly with an echinus fly, and then testcrossing female F1 heterozygotes with the recessive hemizygous male showing all three recessive phenotypes (Fig. 8.4). ( + + ec ) In the data of Bridges and Olbrycht clearly the first group of progeny ( sc cv + ) ( + cv + ) ( + cv ec ) represents the parental combinations, and the second two groups and ( sc + ec ) ( sc + + ) are recombinations. sc cv ec P: × + + ec sc cv + + + ec × sc cv ec ec cv + Backcross Progeny: + + ec 810 Parentals = 1638 sc cv + 828 + cv + 88 Recombinations = 150 sc + ec 62 + cv ec 89 Recombinations = 192 sc + + 103 Fig. 8.4 The three point cross in Drosophila. Data from Bridges and Olbrycht (1926). Since chiasma formation takes place between linked genes, in order to determine crossover percentage in a three point cross, the genes must be analysed two at a time, ignoring the third gene each time. Thus we can examine three combinations: cv – ec, cv – sc, and ec – sc in each group of progeny. Considering genes cv – ec, these are found to be present in progeny of the first group in the same way as in the parent. That is cv is present in its normal wild form (+) on one homologue along with ec; in the other homologue cv is present along with the wild form of ec. Thus in the first group of progeny cv and ec are present as cv + and ec +. Since this is the way they are present in the parent, we can infer that there is no recombination between cv and ec. The second group of progeny again shows cv + on one homologue and ec + on the other as in the parents. Therefore, the second group of progeny also does not represent recombination between genes cv and ec. In the third group of progeny both cv and ec are present on one homologous chromosome whereas the other homologue has wild forms (+ +) of both cv and ec. This arrangement of cv and ec i.e., + and + is different from that in the parental combination and has arisen due to crossing over between the loci for cv and ec. The data in Fig. 8.4 shows that there were 192 flies in the third group of progeny. We can calculate the percentage recombination between cv and ec by dividing 192 by the total progeny (1638 + 150 + 192 = 1980) and multiplying by 100. The amount of crossing over between cv and ec is thus found to be 9.7%.  LINKAGE, RECOMBINATION AND GENE MAPPING 73 Let us now consider genes sc and cv. In the first group of progeny they are present as in the parental combinations, that is ++ on one chromosome and as sc cv on the other. In the second group of progeny their arrangements are different, sc + being together in one chromosome and cv + together on the other homologue. Their changed positions indicate recombination between sc and cv. In the third group of progeny also sc and cv are present differently than in parental combinations. Thus all the 150 members of second group of progeny and 192 of the third group represent recombination between sc and cv. The recombination percentage between sc and cv is calculated as described earlier, i.e., 150 + 192 × 100 = 17.3% 1980 Analysing the genes ec – sc in the same way we find that in the first group of progeny they are present as in the parental combinations. In the second group they are present as ++ on one chromosome, and as sc ec on the other i.e. they represent recombination. In the third group of progeny they are present as in the parental combinations. Therefore, percentage 150 × 100 recombination between sc – ec is = 7.6%. We have thus found that percentage 1980 recombination between scute and crossveinless is 17.3, between scute and echinus 7.6 and between crossveinless and echinus 9.7. The recombination percentage also represents crossover percentage and map distance between the genes. With the data available, it is now possible to map the genes. The chromosome is drawn as a line and the two genes showing lowest recombination frequency are marked first (Fig. 8.5a). In this case they are scute and echinus 7.6 map units apart. Next mark cv and ec which are 9.7 units apart by indicating cv either on the left or right of ec. If we mark cv 9.7 map units on the left of ec, then as seen in Fig. 8.5b, the distance between cv and sc would equal 9.7 minus 7.6 equal to 2.1 units. This does not agree with the data on recombination percentage found experimentally. But if we mark cv on the right of ec 9.7 units apart, then it indicates a map distance between sc and ev equal to 9.7 plus 7.6 i.e., 17.3 which is identical to the experimental data. 7.6 (a) sc ec 9.7 2.1 7.6 9.7 (b) cv sc ec cv 17.3 7.6 9.7 (c) sc ec cv Fig. 8.5 Mapping genes involved in a three-point cross on a chromosome.  74 GENETICS DETECTION OF LINKAGE Whether there is linkage or independent assortment between genes can be detected by backcrossing the dihybrid or heterozygote with the recessive parent and different results would be obtained. This method allows expression of all recessive genes in the heterozygote. The number of parentals and recombinations (noncrossovers and crossovers) in the resulting progeny determines the presence of linkage (Fig. 8.6). P: AaBb × aabb Gametes: AB, Ab, aB, ab ab F1: 1/4 AaBb: 1/4 Aabb: 1/4 aaBb: 1/4 aabb Above cross indicates independent assortment P: AB/ab × ab/ab Gametes: AB, ab ab F1: 1/2 AB/ab : 1/2 ab/ab Same cross as above but with linked genes. Deviations from 1 : 1 : 1 : 1 in testcross progeny indicate linkage. Fig. 8.6 Cross illustrating the backcross method of detecting linkage. DOUBLE CROSSING OVER In a three-point cross involving three genes A, B and C there are eight possible combinations of genes, namely ABC, AbC, ABc, abC, aBc, Abc, abc. Sometimes one or more of the expected combinations do not appear in the progeny. This is due to two crossovers occurring simultaneously in two regions (Fig. 8.7). A b C A b C a B c a B c Fig. 8.7 The result of double crossovers between genes A and B and C and D. When a single crossover occurs, two genes are exchanged resulting in the formation of a crossover gamete. But if at the same time a second crossover also takes place between the next two genes, the original combination of genes is restored on each chromatid resulting in a parental combination. It happens then that in such a cross a double crossover is represented as a noncrossover, giving a recombination frequency lower than the actual. This also reveals that Sturtevant’s statement that 1% crossing over equals one map unit is not always justified. Clearly, crossover percentage is not always equal to recombination percentage. When there are double crossovers between same two chromatids, the number of recombinants in the progeny is less than the number of crossover gametes. The crossover percentages are important for mapping  LINKAGE, RECOMBINATION AND GENE MAPPING 75 genes accurately. Some geneticists prefer to use morgan for map units, one morgan being equal to one per cent recombination frequency and one centimorgan equal to 0.01 morgan. Double crossovers cannot occur between genes that are located close to each other. In Drosophila it has been found that double crossovers cannot occur between genes closer than 10 or 15 map units apart. Moreover, the class of progeny that occurs least frequently represents the double crossovers. It also indicates greater map distance between two genes. Maximum frequency of double crossovers can occur between gene loci at each end of the chromosome. In any case, more than 50% recombination cannot be expected between two genes because only two of the four chromatids in a paired meiotic bivalent are involved in a crossover. Interference Sturtevant pointed out that certain parts of chromosomes were more liable to exchange segments than others. Thus if we consider three hypothetical genes A, B and C the probability of a crossover occurring between A and B may be 10%, and between B and C may be 15%. But what is the probability that two crossovers between A and B and between B and C should occur simultaneously ? We know that the probability of two chance events occurring simultaneously is equal to the product of the individual probabilities. In the hypothetical cross stated above, the probability that two crossovers occur between A and B, and B and C would be 10% × 15% or 0.1 × 0.15 = 0.015 = 1.5%. It has also been found experimentally that the actual percentage of double crossovers is a little less than that expected theoretically. This is due to interference, a term coined by Muller. Accordingly, the occurrence of one crossover reduces the chance of a second crossover in its neighbourhood. Although some explanations have been put forward for interference, both at the cytological and molecular levels, none is considered satisfactory. Muller further proposed the terms coefficient of coincidence to describe the strength or degree of interference. The coefficient of coincidence is equal to the ratio of the observed percentage of double crossovers to the expected percentage of double crossovers. The extent of interference is different between different pairs of genes. The value of coincidence falls and the value of interference rises when the distance between genes decreases. Based on the coefficient of coincidence interference can be described to range from absolute (no double crossovers) to partial (doubles less frequent than expected), none (doubles equal to expected frequency) or negative (doubles more frequent than expected). CYTOLOGICAL BASIS FOR CROSSING OVER At the zygotene stage of meiosis homologous chromosomes come together and start pairing. By pachytene pairing is stabilised, and each ribbon-like chromosome actually consists of two homologues paired (synapsed) close to each other called bivalents. Each homologue in a bivalent consists of two identical sister chromatids. Chromatids belonging to two different homologues in a bivalent are called nonsister chromatids. Due to presence of four chromatids, the pachytene bivalent is sometimes called a tetrad. Crossing over takes place between nonsister chromatids and involves breakage and reunion of only two of the four chromatids at a given point on the chromosomes. Figure 8.8 illustrates how one homologous pair of chromosomes goes through meiosis to form four gametes. Two of the gametes receive a chromosome with genes linked in the same way as in the parental chromosomes (ABC and abc). These gametes represent noncrossovers or parental types and are produced from chromatids that were not involved in  76 GENETICS crossing over. The other two gametes, (ABc, abC) represent the recombinant or crossover types and were produced after crossing over and recombination between the originally linked genes. A B C chromosome I 1 chromosome II a b c A B C chromatid 1 sister chromatids chromatid 2 of chromosome I 2 chromatid 1 sister chromatids chromatid 2 of chromosome II a b c A B C 3 diplotene-diakinesis configuration a b c A B C A B c arrangement of exchanged 4 a B C segments at diakinesis a b c Metaphase I Anaphase I Telophase I Metaphase II A A a a B B b b C C c c 1 2 3 4 Anaphase II Telophase II The product gametes Fig. 8.8 Diagrams illustrating the cytological basis of crossing over.  LINKAGE, RECOMBINATION AND GENE MAPPING 77 SEX LINKAGE When genes are carried on sex chromosomes they are said to be sex-linked. Inheritance of eye colour in Drosophila demonstrates this phenomenon. There is a white eye gene which behaves as a recessive to the normal red eye gene. In a cross between a red-eyed fly and a white-eyed fly, eye colour in the progeny depends upon which parent has red eyes and which one white. If a red-eyed female fly is crossed with a white eyed male, the F1 consists of all red-eyed male and female flies (Fig. 8.9A). The F1 flies on inbreeding gave in F2 both red and white-eyed males, and only red-eyed females. There were no white-eyed females in the progeny. But when the F1 red-eyed flies were backcrossed with white-eyed males, the progeny consisted of both red and white-eyed males, as well as red and white eyed females (Fig. 8.9). In the reciprocal cross in which a female fly with white eyes is crossed with a red-eyed male (Fig. 8.9B), the F1 progeny shows both red and white-eyed flies in equal proportions. But the striking feature is that all the red-eyed flies are females, all white-eyed ones males ! The F1 red-eyed flies on inbreeding gave the same results as F1 reds of the above cross. But when F1 white-eyed males were crossed with white females, they gave only white-eyed males and females. Morgan explained his results by assuming that the gene for eye colour is carried on the X chromosome, i.e., the gene is sex-linked. It was shown cytologically that in Drosophila and some other animals the male is heterogametic (XY) and female homogametic (XX). Since the Y carries no gene, the single X in the male expresses even the single recessive gene it carries as there is no dominant allele of the same gene on the Y chromosome to mask it. Males are therefore said to be hemizygous. In explanation of his cross Morgan proposed that the F1 red- eyed females were heterozygous for white eye character. The white gene was recessive so that the flies were red-eyed. White eyes could appear in female flies that carried both genes for white (homozygous). The male flies, due to hemizygous condition could be red-eyed if the single X chromosome carried one red gene, or white-eyed due to a single allele for white eyes on the X chromosome. The results of Morgan’s other crosses between red-eyed and white-eyed flies agreed with his explanations, thus providing support for his hypothesis. Sex-Linkage in Poultry Spillman in 1909 found sex-linked inheritance of feather colour in poultry. In the Plymoth Rock strain of chickens one variety called barred has black and white pigments in feathers alternating to give a barred appearance. The other variety of chicken was black or nonbarred. When a nonbarred female was crossed to a barred male, the F1 chickens were all barred. The F1 males and females were mated with nonbarred chickens to give a testcross progeny consisting of 4 classes: barred males, nonbarred males, nonbarred females, barred females. When barred females were crossed to nonbarred males, the testcross progeny yielded only barred males and nonbarred females (Fig. 8.10). In poultry male is homogametic (XX) and females heterogametic (X0). 78 GENETICS Red-eyed White-eyed P: R R r X X X Y Gametes: X X Y F1: R r R X X X Y Red-eyed Red-eyed 50% 50% after inbreeding F2: Red-eyed s White-eyed s Red-eyed s A White-eyed Red-eyed P: r r R X X X Y Gametes: r R X X Y F1: R r r X X X Y Red-eyed White-eyed 50% 50% after inbreeding F2: Red-eyed s White-eyed s Red-eyed s B Fig. 8.9 A, B Inheritance of eye colour in Drosophila demonstrating absence of crossing over in male and sex-linkage.  LINKAGE, RECOMBINATION AND GENE MAPPING 79 P: Nonbarred female Barred male b BB F1: Barred male and Barred female Bb B Testcross Progeny: Barred (Bb) and Barred (Bb) males nonbarred (bb) and nonbarred (b) males: barred females (B) and non- barred (b) females Fig. 8.10 Cross illustrating sex-linkage in poultry. RECOMBINATION IN NEUROSPORA The fungus Neurospora crassa has some advantages for study of crossing over. In the vegetative phase of its life cycle, there is growth of filamentous, coenocytic hyphae to form the haploid mycelium. Sexual reproduction is achieved by coming together of two hyphae of proper mating type and their nuclei fuse to form a diploid zygote (Fig. 8.11). The zygote starts enlarging into an elongated structure called ascus and the zygote nucleus starts dividing meiotically. The four products of meiosis are arranged in a single linear row inside the ascus. Each one of these undergoes a mitotic division so that 8 nuclei are formed which develop a wall and become ascospores. 4 4 3 3 2 2 1 1 Fig. 8.11 Mating hyphae in Neurospora and stages in formation of ascus. What makes Neurospora interesting for a geneticist is the fact that the ascospores are linearly ordered in the same sequence as the chromatids were on the meiotic metaphase plate. It is therefore possible to recover all the four products of a single meiosis and analyse them, as also each chromatid from a tetrad. This is called tetrad analysis. Neurospora thus presents a direct way of demonstrating recombination which in higher organisms can be inferred on a  80 GENETICS statistical basis from progeny counts. A still further advantage is the use of centromere as a marker for determining map distances as explained later in this chapter. + al al normal albino + + al al al al zygote + + + al al al al al al + + + + al al al al al al al al + + + + al al al al al al al al + + + + al + + + + al al al al al al al al al al al al al al al + + + 4al : 4al 2al : 2al : 2al : 2al First division segregation Second division segregation Fig. 8.12 Results of first and second division segregation in Neurospora.  LINKAGE, RECOMBINATION AND GENE MAPPING 81 When the segregation of a single pair of alleles is analysed, the ascus of Neurospora shows two arrangements of the eight ascospores: 4 : 4 ratio resulting from first division segregation; and 2 : 2 : 2 : 2 ratio resulting from second division segregation (Fig. 8.12). First Division Segregation: The results of a cross between a normal (al+) and an albino (al) strain of Neurospora are shown in Fig. 8.12. If the 8 ascospores are removed from the ascus and each is grown separately, it is found that 4 ascospores produce mycelium of normal type and 4 of albino type. This is due to first division segregation explained as follows: during anaphase I of meiosis in the zygote, one homologue carrying al+al+ segregates from the second homologue which has the other allele alal and both move to opposite poles. Their products are recovered in a 4 : 4 ratio after meiosis II (and after mitosis) because no crossing over has occurred between the gene and the centromere. This situation results when the gene is located close to the centromere. Therefore, if a large number of asci are found to exhibit 4 : 4 ratio, it indicates that the gene locus in question is close to the centromere. Second Division Segregation: As shown in Fig. 8.12 if crossing over occurs between the said gene and the centromere, then one chromatid of each homologue will carry al+ and the other al. Therefore al+ and al will not be able to segregate from each other at anaphase of first meiosis when the two homologues separate to the two poles. It is only at anaphase of second meiosis when centromeres divide and chromatids separate from each other to the two poles that al+ and al will segregate from each other. This is called second division segregation and produces a ratio of 2al+ : 2al : 2al+ : 2al, which also indicates that crossing over has occurred between the gene locus and centromere. GENE MAPPING IN FUNGI In Neurospora the centromere is a marker for determining map distances. For detecting linkage and map distances, the frequency of crossing over is determined from the number of asci showing second division segregation. If there is one crossover, the resulting ascus shows 50% of ascospores with parental combinations and 50% with recombinations. Suppose in a cross involving a pair of alleles 30% of asci show second division segregation. This shows that 30% of zygotes had crossing over during meiosis and 70% did not. Since there are four chromatids in each tetrad, the 50% asci have resulted from 30 × 4 = 120 original chromatids in meiosis. When there is crossing over only two of the four chromatids are involved in an exchange. Therefore only half of the 120 chromatids i.e., 60 are crossover chromatids, the remaining 60 chromatids are noncrossover chromatids. It was also stated above that 70% of zygotes did not have crossing over, which means that 70 × 4 = 280 are noncrossover chromatids. The actual number of noncrossover chromatids is larger because the 30% asci showing second division segregation also have 60 noncrossover chromatids. The exact number is therefore 280 + 60 = 340. Therefore, of the original 100 tetrads or asci 340 are noncrossover chromatids and 60 are crossover chromatids. Since 100 tetrads or asci also mean 400 chromatids, the percentage of crossover 60 chromatids is × 100 = 15%. From this we can conclude that there was 15% crossing over 400 between the gene and the centromere. We can also say that the gene in question is 15 map units apart from the centromere. Because the centromere itself serves as a marker, in Neurospora it is possible to map a single gene pair. It is also called a two-point cross.  82 GENETICS It follows that the method of detecting gene linkage in fungi is basically similar to that for diploids. The main feature in all cases consists in comparing the frequency of parental types to recombinant types. If there is a significant reduction in the frequency of recombinant types from the frequencies expected on the basis of independent assortment, we can consider linkage. Whereas in Neurospora the meiotic products occur in an ascus in a linear order (ordered tetrads) this is not true for other fungi in Ascomycetes which have unordered tetrads. Let us examine a cross involving two linked genes + +/ab in a fungus having unordered tetrads. The zygote (++/ab) undergoes meiosis, and depending upon the occurrence of crossover, the tetrads are of following three types: (a) Parental ditype (PD) (Fig. 8.13a): + + + + + + + + [50% ++ ; 50% ab] a b a b a b a b Fig. 8.13(a) If a crossover does not occur between these two loci, or if a two strand double crossover occurs between them, the resulting meiotic products will be of two kinds, both resembling parental combinations, and appear in equal frequency (1 + +; 1 ab). Such a tetrad is called parental ditype (PD). (b) Nonparental ditype (NPD) (Fig. 8.13b) + + + + 50% + b; 50% + a a b a b Fig. 8.13(b) If a four strand double crossover occurs between the two genes, two kinds of products are formed, both being recombinations. Such a tetrad is called a nonparental ditype (NPD). (c) Tetratype (TT) (Fig. 8.13 c): This is produced either by a single crossover or a three- strand double crossover (of two types) between the two genes. Whenever the number of parental ditypes and nonparental ditypes are significantly unequal, linkage between the two genes must be considered. For determining the amount of recombination between the two genes, the following formula is used: NPD + 1/ 2 TT Recombination Frequency = Total number of tetrads  LINKAGE, RECOMBINATION AND GENE MAPPING 83 + + + + a b a b + + + + 25% ++; 25% + b; 25% a +; 25% ab a b a b + + + + a b a b Fig. 8.13(c) Cytological Proof of Crossing Over In 1917 Goldschmidt proposed that recombination takes place due to exchange of alleles with- out exchange of chromosome segments. He assumed that at metaphase genes get detached from the chromosome. Later during meiosis the genes get reabsorbed on the chromosome either in the same or in a different place. In 1930 Winkler put forth his “gene conversion” hypothesis. Accordingly, if gene replication occurs in closely synapsed homologous chromosomes, the wrong allele may get replicated. When this occurs at only one locus it appears like a crossover. Both the above theories presented difficulties in understanding. Bellings hypothesis: In 1931 John Belling proposed a theory of crossing over based on exchange of chromosome segments in lily plants. Belling studied the morphology of bead-like chromomeres which are arranged linearly on the chromonema. Since the structure and arrangement of chromomeres is identical in a pair of homologous chromosomes, Belling thought they might represent genes. He explained recombination by assuming that the chromomeres were synthesised first, and the chromonemata which were synthesised later, became connected to the chromomeres. Wherever the strands of chromonemata did not get connected with chromomeres in the original linear order, they crossed over and passed through another chromomere resulting in recombination and genetic exchange. The name copy choice was later given to Belling’s mechanism for recombination. However, the idea was disproved due to lack of evidence from genetic tests (see Chapter 22).  84 GENETICS Significance of Crossing Over Crossing over occurs in living organisms ranging from viruses to man. It constitutes evidence for sexual reproduction in an organism. Its widespread occurrence in organisms ensures exchange of genes and production of new types which increase genetic diversity. This increases phenotypic diversity, which at the species level is responsible for genetic polymorphism. The occurrence of polymorphism is of advantage to a species because it leads to groups of individuals becoming adapted to a wider range of habitats. This increases the potential for evolutionary success. Factors Affecting Crossing Over The following external factors can affect the frequency of crossing over: 1. Bridges showed that in Drosophila, as maternal age increases, crossing over decreases. 2. H.H. Plough, a student of Morgan found that both low and high temperatures changed the frequency of crossing over. 3. The existence of crossing over factors has been shown in the cytoplasm of females. Consequently, females with reduced recombination frequencies can pass on this trait to their daughters. 4. Calcium and magnesium ions affect crossover frequency. Antibiotics such as mitomycin-C and actinomycin-D increase crossing over. Similarly X-ray irradiation can increase crossover frequency in Drosophila females and induce it in males. 5. Heat and shock treatments also change rates of crossing over. Genotypic effects were observed in Drosophila by R.P. and E.E. Levine by which a gene on the 3rd chromosome completely suppresses crossing over in homozygous females. Genotypic Control of Recombination In some cases at least recombination itself is under the control of genes. In Neurospora crassa several recombination genes control frequency of recombination either between genes or within genes. Thus a recessive gene rec-1 controls frequency of recombination at the his-1 locus. Both rec-1 and his-1 are on the same chromosome. In maize chiasma formation can be entirely suppressed by a recessive gene as present on chromosome 1. There are quite a few examples known of single genes as well as polygenes which can cause variation in frequency and distribution of chiasma in one or more chromosomes. QUESTIONS 1. A plant heterozygous for three alleles AaBbCc is cross pollinated by pollen from a homozygous recessive plant aabbcc. Out of the 100 progeny plants raised 45 showed the phenotypes of alleles A, B and C. Does this indicate linkage ? Explain. 2. If all the genes in an organism are mapped, how many linkage groups would be expected to be found in (a) man; (b) E. coli (main chromosome only):(c) the haploid mycelium of a fungus having 11 chromosomes in each cell; (d) the bacteriophage. LINKAGE, RECOMBINATION AND GENE MAPPING 85 3. In Drosophila genes for body colour and wing size are linked. In performing testcrosses with F1 heterozygotes would the results be different if the double recessive parent was a female instead of male? Explain. 4. If two linked loci C and G are 40 map units apart, whereas C and E have 8 map units between them, what proportions of the gametes of a heterozygous individual would be expected to have CG? With what frequency would CE gametes be formed? 5. Explain how double crossovers can increase the frequency of parental combinations. How do they affect map distances? 6. A trihybrid PpQqRr is crossed. The F1 progeny indicate that the trihybrid parent produced the following gametes. PQR—25 pqr—20 Pqr—240 pqR—230 PQr—201 pQR—250 PqR—20 pQr—28 (a) which loci are linked and which show independent assortment? (b) what is the genotype of the other parent? (c) calculate the map distances between the linked genes. 7. If an attached-X female of Drosophila is heterozygous for both genes A and B, what kind of gametes would form if (a) there was one crossover between A and B, (b) if there was no cross- over. 8. In chickens females are XO and males XX and the gene for barred feathers is carried on the X chromosome. What would be the results of the following crosses: (a) barred male x barred female; (b) nonbarred male x nonbarred female; (c) nonbarred male x barred female. 9. You are asked to perform a cross between a pink strain of Neurospora and a white strain and then grow all the 8 ascospores in an ascus. What results would you expect if (a) first division segregation had occurred; (b) second division segregation had occurred? 10. If you are given the percentage of Neurospora asci showing second division segregation, how can you calculate the map distance of the gene from the centromere. 11. Given the following map distances between individual pairs of genes construct a map of the following chromosome. o–p—10 units m–p—15 units p–n—13 units m–o—8 units n–o—3 units SELECTED READINGS Barratt, R.W. et al., 1954. Map Construction in Neurospora crassa. Adv. Genetics 6 : 1. Bridges, C.B. and Olbrycht, T.M. 1926. The Multiple Stock “Xple” and its Use. Genetics 11 : 41. Creighton, H.S. and McClintock, B. 1931. A Correlation of Cytological and Genetic Crossing over in Zea Mays. Science 17 : 492. Fincham, J.R.S. 1970. Fungal Genetics. Annu. Rev. Genetics 4 : 347. Fincham, J.R.S. and Day, P.R. 1971. Fungal Genetics. Blackwell Scientific Publ. Ltd. Fu, T.K. and Sears, E.R. 1973. The Relationship between Chiasmata and Crossing-over in Triticum aestivum. Genetics 75 : 231. 86 GENETICS Henderson, S.A. 1970. The Time and Place of Meiotic Crossing Over. Annu. Rev. Genetics 4 : 295. Mather, K. 1951. Measurement of Linkage in Heredity. Wiley, New York. Morgan, T.H. 1911. Random Segregation Versus Coupling in Mendelian Inheritance. Science 34 : 384. Muller, H.J. 1916. The Mechanism of Crossing Over. Amer. Nat. 50 : 193. Sears, E.R. 1976. Genetic Control of Chromosome Pairing in Wheat. Annu. Rev. Genetics 10 : 31. Stern, C. 1931. Zytologisch—genetische Untersuchungen an Beweise für die morganische Theorie des Faktorenaustauchs. Biol. Zentralblatt 51 : 547. Sturtevant, A.H. 1913. The Linear Arrangement of Six-linked Factors in Drosophila, as shown by their mode of association. J. Exptl. Zool. 14 : 43. Whitehouse, H.L.K. 1973. Towards an Understanding of the Mechanism of Heredity. St. Martin’s Press, New York.  9 Extranuclear Transmission of Traits In Mendelian inheritance the transmission of hereditary traits depends almost entirely on the chromosomes contained in the nucleus and their behaviour during meiosis. One of the exceptions to this regular pattern is the transmission of genetic information from parent to offspring through the cytoplasm and is called extranuclear or cytoplasmic inheritance. The role of cytoplasm in heredity is determined from results of reciprocal crosses in which the sources of male and female gametes are reversed. In chromosomal inheritance it makes no difference in the transmission of a gene whether it comes from male parent, and identical phenotypes are obtained in reciprocal crosses. But if there is extranuclear inheritance, the resulting phenotypes in a reciprocal cross are nonidentical and indicate uniparental transmission, in most cases from the maternal parent. The bigger role of the maternal parent results from the unequal cytoplasmic contributions of the male and female parents as most of cytoplasm of the zygote comes only from the egg. However, a few cases of paternal transmission are also known. Sometimes confusion can arise in recognising a true case of cytoplasmic inheritance due to certain other modes of transmission which deviate from the regular pattern of nuclear inheritance. In practice, therefore, breeding tests for extranuclear inheritance are so designed as to exclude step by step all possible chromosomal explanations of the observed behaviour as mentioned below: (a) Reciprocal crosses in which identical phenotypes are produced. (b) Sex-linked inheritance, it follows a definite pattern and is easily recognised. It is rare in plants. (c) Dauer (persistent) modifications: Sometimes maternal influence is exerted on the offspring due to nutrition, environment or certain treatments given to the female parent which affect the cytoplasm of the egg. The effects shown by the progeny and their inheritance are transient. They will either disappear in the next generation or will appear with a diminishing effect in succeeding generations. In a true case of cytoplasmic inheritance a trait must persist undiminished through many generations. (d) Maternal effect due to a cytoplasmic trait under the control of nuclear genes. Inheritance of some traits is influenced by cytoplasm in the egg, but the expression of the trait is modified by chromosomal genes. Coiling in snails (dextral and sinistral), inheritance of some intranuclear symbionts and of eye colour in moth Limnaea are good examples. Maternal effects can also result 87  88 GENETICS in a developing organism from mRNA that was included in the egg cytoplasm prior to fertilisation. The stored RNA originated from chromosomal genes. Such maternal effects usually last in the offspring for one generation only. The term extranuclear inheritance is restricted to cytoplasmic factors that exhibit replication and independent transmission. These factors have their own genetic material and the traits controlled by their genes are not transmitted in Mendelian fashion. Because of their location outside the nucleus, these genetic factors are referred to as plasmagenes or cytogenes. They are most commonly detected from differences in the results of reciprocal crosses. These differences persist as long as the extra-chromosomal factor can perpetuate itself. They are mostly transmitted through the egg cytoplasm of the female parent because the amount of cytoplasm contributed to the zygote by the male parent is negligible. However, a few cases of male cytoplasmic inheritance are known and will be described later. KILLER TRAIT IN PARAMECIUM Sonneborn in 1938 discovered certain strains in Paramecium which showed a killer trait due to the presence of a cytoplasmic factor called kappa. The killer strain can destroy the sensitive strains growing in culture which do not have kappa by liberating a toxic substance paramecin. Killer strains are not killed by their own paramecin. Paramecium has two kinds of nuclei, a small micronucleus and a very large macronucleus which is highly polyploid and irregular in shape and behaviour during cell division. Only the micronucleus behaves according to Mendelian principles. Paramecium has three modes of reproduction. The first is a simple mitotic division called binary fission. The second method is conjugation. Here two protozoans divide meiotically to form four micronuclei in each cell; out of these, three nuclei degenerate and only one remains which divides by mitosis to produce two genetically identical haploid nuclei in each Paramecium. During conjugation only one of the two haploid nuclei is exchanged through a cytoplasmic bridge formed between the two ciliates. The cells then separate as two exconjugants. The third method of reproduction is called autogamy. Here a single Paramecium divides meiotically and by the same process that occurs in conjugation, two identical haploid nuclei are formed which fuse to form a diploid organism. As there was no genetic exchange, the diploid Paramecium is homozygous. One noteworthy feature of the sensitive strains is that they are not killed by paramecin while they are in the process of conjugation. This has an advantage because it allows the investigator to perform crosses between the killer and sensitive strains. The two strains can be distinguished morphologically as killers have granular cytoplasm and sensitives are clear. When a cross is made between a killer and a sensitive Paramecium (each made homozygous by autogamy), there is exchange of genetic material through conjugation. This is followed by separation of the two genetically identical exconjugants. It is found that killer exconjugants produce only killer Paramecia and the sensitive exconjugants only the sensitive Paramecia. Obviously, the killer and sensitive traits are not controlled by Mendelian genes (Fig. 9.1). If the heterozygous (KK) killer exconjugant is inbred to another heterozygous killer, it produces three-quarter killer (1 KK and 2 kk) and one quarter sensitives (kk). But if the sensitive exconjugant (Kk) is crossed to another heterozygous sensitive, it results in all sensitive progeny  EXTRANUCLEAR TRANSMISSION OF TRAITS 89 even though their genotypes are in the ratio of 1 KK : 2Kk : lkk. The results suggest nonchromosomal inheritance of killer trait. Killer Sensitive KK kk Killer Sensitive Kk Kk Exconjugants Produces Produces killers sensitives Fig. 9.1 Conjugation in Paramecium demonstrating extranuclear inheritance of the Killer trait. The final proof regarding inheritance of killer trait was obtained by modifying the experiment in the following way. The cross between killer and sensitive was prolonged, allowing enough time for exchange of cytoplasm to take place. In this way, some of the kappa particles could move from killer into the sensitive strain. All the resulting progeny of such a cross consisted of killers thus confirming the cytoplasmic inheritance of kappa particles. The above experiment can be performed in such a way that Paramecia divide very rapidly by controlling the nutrient conditions. Under such conditions, a homozygous killer strain (KK) containing kappa particles can produce a few individuals that are sensitive and without kappa particles. The explanation is that kappa particles cannot multiply as rapidly as the cells, and become fewer in number in comparison with the number of Paramecium cells. Due to their reduced number kappa particles are not passed on to some members of the progeny at all. It was found that although kappa particles are transmitted cytoplasmically, yet they require a dominant K gene for maintenance. The K gene cannot initiate the presence of kappa particles. Kappa are virus-like particles about 0.2 micron in diameter and have ability to reproduce independent of the nucleus. They have their own DNA, can multiply and produce the substance paramecin. CO2 SENSITIVITY IN DROSOPHILA Carbon dioxide is used as an anaesthetic for Drosophila, and normal flies can withstand high CO2 concentrations without any adverse effect. But some strains of Drosophila are sensitive to CO2 so  90 GENETICS that within a few minutes they become paralysed and die. Reciprocal crosses between CO2 sensitive and resistant flies gave differing results. If a CO2 sensitive female is crossed with a normal (resistant) male, almost all the offspring are sensitive. But the reciprocal cross of a sensitive male with a normal female yields almost all normal offspring indicating extranuclear inheritance of this trait. There is mostly maternal transmission although some transmission through sperm may also take place. Moreover, normal flies can be induced to become sensitive by injecting a cell- free filtrate of haemolymph obtained from sensitive flies. The causal agent for CO2 sensitivity has been observed in electron micrographs to be a virus-like particle called sigma. In flies that have been induced to become sensitive, sigma enters the egg cytoplasm and is transmitted maternally. In the naturally occurring sensitive strains, sigma particles become incorporated into the nucleus (without associating with a specific chromosome) of the egg or some of the sperms and are transmitted in a non-Mendelian manner. They are self-reproducing independent bodies and can mutate. PLASTID INHERITANCE Plastids are cytoplasmic organelles which contain DNA and duplicate themselves independently of nuclear genes, and are distributed more or less equally to daughter cells during mitosis. While studying leaf pigmentation in Mirabilis jalapa (four O’clock plant), Correns (1909) found for the first time that plastids could be transmitted to the offspring through the egg cytoplasm. In a variegated strain of Mirabilis the cytoplasm of the zygote contains both green and colourless plastids. During cell division in the zygote to form embryo, these plastids are unequally distributed. Such an embryo on germination produces a variegated plant with three types of branches—those bearing green foliage, colourless foliage and variegated (mixture of green and white) foliage. No matter what the colour of the branch from which the pollen is used for fertilisation, it was found that seeds produced by green branches gave rise only to green plants; and seeds from colourless branches produced colourless seedlings which did not survive due to lack of chlorophyll. But seeds from the variegated branch could produce three types of progeny: plants with green, colourless or variegated branches. This is because the egg cell of a variegated plant will have both kinds of plastids—green as well as white. At the time of cell division some cells will receive only green plastids, some only white, and others will receive both green and white plastids. Likewise the progeny of such a branch could be green, white or variegated (Fig. 9.2). Green × White Variegated × Green All green Variegated × White Variegated Fig. 9.2 Crosses in M. jalapa showing no influence of pollen. All the progeny inherit maternal characteristics of foliage through egg cytoplasm.  EXTRANUCLEAR TRANSMISSION OF TRAITS 91 MALE TRANSMISSION IN GERANIUM (PELARGONIUM) As in Mirabilis, plants of geranium could be green, white or variegated. But cytoplasmic inheritance in this case follows a unique pattern. If the egg cell from a white plant is fertilised with pollen from green plants, the progeny consists of green, white and variegated plants. The question arises—when the egg cytoplasm has only white plastids, what is the source of the green plastids in the progeny? The answer revealed an unusual feature of Pelargonium, that is the male parent contributes plastids through the cytoplasm of the pollen grain. In Oenothera also male transmission of cytoplasmic traits is known. IOJAP STRAIN OF MAIZE The iojap strain of maize is characterised by green and white stripes on the leaf. The name iojap originates from two parental strains Iowa which is green and Japonica, a striped variety. In crosses between iojap and green varieties when iojap is used as the male parent, the trait is inherited according to Mendelian pattern with F1 progeny all green, and F2 segregating into 3/4 green and 1/4 iojap. But in the reciprocal cross where iojap is used as the female parent, the F1 plants showed all three phenotypes viz. green, white and striped. P: Iojap × Green iijj IIjj F1: Iijj Phenotypes: Green, White and Striped F1 Striped Iijj × IIjj Green F2 Genotypes: lijj IIjj Phenotypes Green, Green, White, and White, and Striped Striped It was found that the iojap gene in the homozygous recessive condition (ii) causes some of the plastids to mutate giving rise to colourless plastids. The mixture of green and colourless plastids accounts for the origin of striped plants. Once created, further transmission of the striped character takes place maternally through the egg cytoplasm as evident from the phenotypes of the F2 progeny of the cross above. MALE STERILITY IN PLANTS The failure of pollen formation results in male sterile plants. In some crop plants male sterile mutants occur in which inheritance of male sterility follows one of the following two patterns: those in which the trait is inherited through a single recessive chromosomal gene segregating in Mendelian ratios; secondly, those that show maternal transmission. Rhoades in 1933 described maternal inheritance of male sterility in Zea mays.  92 GENETICS P: Male Sterile line × Normal Pollen F1: All progeny male sterile F1 females backcross any number of times to normal pollen. Progeny of all succeeding generations are male sterile. In the cross shown above, the process of repeated backcrossing results in substitution of all the chromosomes of the fertile line for those of the male sterile line. The experiment therefore, not only shows cytoplasmic maternal transmission, but also confirms that chromosomal genes are not responsible for male sterility. INHERITANCE THROUGH MITOCHONDRIA Mitochondria are cytoplasmic organelles containing their own DNA and some enzymes that catalyse the process of oxidative phosphorylation. Mitochondria are not autonomous bodies because they require their own genes and nuclear genes to function normally. The role of mitochondria in cytoplasmic inheritance was first understood when Ephrussi and his colleagues (1955) studied petite mutants (petite meaning small) in yeast (Saccharomyces cerevisiae). When a cell divides, approximately equal numbers of mitochondria pass into daughter cells. They can originate from pre-existing mitochondria, and can probably divide transversely. Several mutant strains of yeast are referred to as petite mutants because they form smaller colonies on agar as compared to normal yeast. The growth rate of petites is slow due to absence of respiratory enzymes, cytochromes a, b and c, and due to deficiency in some dehydrogenases present in normal mitochondria. The petite strains are of three types. In the first type called segregational petites when an individual is crossed with a normal strain, a 1 : 1 ratio of normal: petite results after segregation. This suggests Mendelian inheritance and the petite strain has originated due to a mutation in nuclear genes. The second and third types of petite strains arise when normal yeast cells are treated with the acridine dye euflavin or with ethidium bromide dyes which are known to intercalate in double stranded DNA. The petites so formed do not segregate regularly when crossed with normal yeast, and fall under two types—neutral and suppressive. When a neutral petite is crossed with a normal type, all the progeny in the next and successive generations are normal and the petite trait never reappears. If a haploid cell of the neutral petite strain is mated to a haploid vegetative cell of a normal strain, a diploid zygote is formed. The diploid cells produced by this zygote reproduce asexually, sometimes may even divide meiotically to form spores, but the petite trait is never visible in the progeny. The genetic explanation of the trait is that there is a cytoplasmic factor (p+) which is present in the normal strains of yeast but absent in petites (p–).  EXTRANUCLEAR TRANSMISSION OF TRAITS 93 The suppressive petites when crossed to normal cells of yeast show the petite trait in the progeny but in non-Mendelian ratios. Suppressive petite mutants are found to have mutant DNA in their mitochondria. The mutant mitochondria replicate and transmit the mutant phenotype to the progeny cells. Mitochondrial mutants are also known in Neurospora, Paramecium and Trypanosoma. THE GENETICS OF MITOCHONDRIA AND CHLOROPLASTS The presence of organelle-specific DNA in mitochondria and chloroplasts was identified with certainty over three decades ago. The occurrence of a specific chloroplast DNA was first demonstrated by Chun et al. in 1963, and of mitochondrial DNA a year later by Luck and Reich (1964). It is now fairly well established that organelle DNA is distinct from nuclear DNA in several respects. That it can carry out synthesis of RNA and protein on its own synthesising ‘machinery’ and replicate itself. There are mitochondrial and chloroplast ribosomes. The genomes of these organelles contain genes for the rRNA species present in both large and small subunits of their respective ribosomes. Mitochondrial DNA (mt DNA): Mitochondria contain an open or closed circular DNA molecule. DNA from different species shows a wide range in average base composition ranging between less than 20% to more than 50% of G + C content. It has been found that the amount of DNA present in mitochondria is sufficient to code for proteins. Grossman et al. (1971) have determined that a minimum base composition of 35% G + C is needed to code for an average protein. Thus regions of mt DNA containing very small amounts of G and C nucleotides (in cases where the average base composition is less than 30%) may not be able to code for a protein. Recent evidence indicates that there is one copy of the gene for each of the rRNA species associated with the large and small subunits of mitochondrial ribosomes. The positions of these two genes have been mapped in some lower eukaryotes. In the mold Neurospora crassa the two genes for rRNA are adjacent to each other, whereas in yeast (Saccharomyces cerevisiae) they are separated. With the use of radioactive isotopes it has been shown that in N. crassa and S. cerevisiae some of the mitochondrial inner membrane proteins are synthesised by the mitochondrial protein synthesising apparatus. The existence of mutations have been recorded in the mitochondrial genome. A study of the oligomycin resistance mutations have led to the identification of 3 distinct linkage groups in mtDNA. Recently Tzagaloff et al. (1975) have been able to identify a number of mutants with defects in cytochrome oxidase and coenzyme Q-cytochrome c reductase. mtDNA has also been studied in a limited number of higher plants. A closed circular DNA with molecular weight ranging between 60 – 80 × 106 daltons has been seen in spinach, lettuce and pea. However, higher plant mtDNA has not shown the presence of rRNA and tRNA genes. No protein synthesis is recorded for plant mtDNA. Chloroplast DNA (ctDNA): Circular DNA molecules have been observed in ctDNA of both higher plants and algae. It contains genes for both large and small rRNA species present in chloroplast ribosomes. Hybridisation studies have indicated one gene per chromosome for each rRNA in tobacco, whereas in bean, lettuce, pea, spinach and maize there are two genes per chromosome. Similar studies have demonstrated the presence of genes for about 25 tRNA species each of size about 25,000 daltons. Radioactive isotope studies in vivo have clearly shown that protein synthesis occurs in chloroplasts. One such protein is the large subunit of ribulose diphosphate carboxylase (termed  94 GENETICS fraction I protein) which is a major constituent of chloroplast as well as other membrane proteins. The large subunit of fraction I protein and some other membrane proteins are also synthesised by isolated chloroplasts. The small subunit of fraction I protein is coded for by nuclear DNA. Replication of Mitochondrial and Chloroplast DNA Both mtDNA and ctDNA replicate within their respective organelles. In vitro synthesis of DNA also occurs in the isolated organelles if the four deoxyribonucleotide triphosphates are supplied and is blocked by inhibitors of DNA synthesis such as ethidium bromide, acriflavin and actinomycin- D. Whereas nuclear DNA synthesis is limited to the S phase of the cell cycle, there is evidence for continuous synthesis of organelle DNA throughout the cell cycle. In vivo density label studies have shown that mt DNA in Hela cells and ct DNA in Chlamydomonas and Euglena replicates semiconservatively. As in bacteria, replication of organelle DNA appears to involve the membrane (called mesosome in bacteria). EXTRANUCLEAR GENES IN CHLAMYDOMONAS The unicellular green alga Chlamydomonas is haploid with a single nucleus, a chloroplast and several mitochondria (Fig. 9.3). It can reproduce asexually as well as sexually by fusion of gametes of opposite mating types (mt+ and mt–). The mating type is controlled by a single nuclear gene. There is a fusion of the two chloroplasts in the diploid zygote. The zygote divides meiotically to produce mt+ and mt– cells in the ratio 2:2. Flagellum Mitochondria Lamellae Nucleus Chloroplast Pyrenoid Fig. 9.3 The green alga Chlamydomonas. In 1954 Ruth Sager isolated a mutant resistant to streptomycin (sm-2) which showed a different pattern of segregation after meiosis. When a streptomycin resistant strain is crossed with a wild type strain, the resulting progeny depends upon which mating type contributed the  EXTRANUCLEAR TRANSMISSION OF TRAITS 95 gene for streptomycin-resistance. Thus when the mt+ parent carries the sm-2 allele, the progeny is almost all streptomycin-resistant instead of segregating in a 2 : 2 ratio. In the reciprocal cross when mt– parent has sm-2 allele, none of the progeny show streptomycin-resistance. Such uniparental transmission of a trait is typical of extrachromosomal inheritance. A noteworthy feature of the above crosses is the production of a very small percentage of ‘exceptional’ zygotes which receive the sm-2 allele from the mt– parent. These zygotes indicate biparental transmission. There is still another class of ‘exceptional’ zygotes, though rare, which transmit extranuclear genes only from the mt+ parent. A class of Chlamydomonas extranuclear mutants known as ‘minutes’ also show biparental inheritance. The ‘minutes’ are induced by treatment with acriflavin and ethidium bromide. The photosynthesis deficient mutants which depend upon acetate for growing in light, and the temperature sensitive mutants in Chlamydomonas also exhibit uniparental inheritance. All the four haploid products of meiosis are inherited from the mt+ parent. Exceptional zygotes showing biparental transmission occur also in these crosses at a frequency of less than 1 per cent. There is evidence that the inheritance of the extranuclear traits in Chlamydomonas is due to chloroplast genes. According to Sager uniparental transmission of chloroplast genes could be due to degradation of chloroplast DNA originating from the mt– parent in the zygote. When degradation does not occur exceptional zygotes showing biparental inheritance are formed. Such zygotes produce haploid progeny heterozygous for chloroplast genes. Subsequent mitotic divisions in the haploid progeny will show segregation of alleles, the members showing one or the other parental phenotype. When two or more chloroplast genes are considered in a cross, recombinants appear, their frequencies suggesting linkage between chloroplast genes. Sager and her colleagues have extended these studies and mapped positions of chloroplast genes. QUESTIONS 1. What criteria can be used to distinguish between cytoplasmic inheritance and nuclear inheritance? 2. In maize some male sterile plants were artificially pollinated with pollen from healthy male fertile plants. The seeds were used for raising progeny plants which again turned out to be male sterile. Explain. 3. Distinguish between (a) maternal influence and cytoplasmic inheritance; (b) chloroplast DNA and mitochondrial DNA. 4. (a) The killer trait in Paramecium is due to kappa particles in the cytoplasm. What results would be obtained if prolonged matings are allowed between (i) two killer strains; (ii) two sensitive strains; (iii) one killer and one sensitive strain. (b) Sometimes Paramecia of killer strain undergo fast multiplication and produce sensitives in their progeny. Explain. 5. Describe one example of a mutation in mitochondria that produces a visible phenotype in a fungus. 6. (a) Extrachromosomal traits are mostly transmitted through the maternal parent. Give reasons. (b) Is there an extranuclear trait showing (i) uniparental transmission through the male gamete? (ii) biparental transmission?