Core-12 PDF - Mendelian Laws & Genetics Experiments

# Expt. No. 1 ## AIM OF THE EXEPERIMENT: - Study of Mendelian laws. ## PROCEDURE: - To work out the practical aspects of Mendelian laws, the following procedure is followed: ### 1. Determine the type of inheritance: - Whether it is a monohybrid or dihybrid cross - Whether it exhibits comp...

# Expt. No. 1 ## AIM OF THE EXEPERIMENT: - Study of Mendelian laws. ## PROCEDURE: - To work out the practical aspects of Mendelian laws, the following procedure is followed: ### 1. Determine the type of inheritance: - Whether it is a monohybrid or dihybrid cross - Whether it exhibits complete dominance or incomplete dominance. ### 2. Determine the genotypes of individuals involved in the cross: - The genotype of the recessive individual is represented by small letters. For example, dwarfness in pea plants is represented by **tt**. Both alleles for the recessive character will be in homozygous state in the recessive individual. - Individuals with dominant characters may be homozygous and heterozygous. For example, tallness in pea plants may be **TT** or **Tt**. ### 3. Determine the types of gametes produced by either parent: - Homozygous individuals will produce only one type of gametes. - Individuals heterozygous for one pair of contrasting characters produce two types of gametes. - Dihybrid and polyhybrid individuals then produce four, and even eight types of gametes depending upon the type of cross. ### 4. Draw the Punnett square of checkerboard: - Place the gametes on the checkerboard and fill up the chambers of checkerboard to show the union of different types of gametes. # Expt. No. ## 5. Work out the result in the form of genotype. ## 6. Calculate the phenotype as a result of genotype. ## 7. Calculate the ratio between dominant and recessive characters ## PROBLEMS: ### 1. Gene T is dominant over gene t. What will be the phenotypic ratio in the offspring obtained from the following mating: - Tt x tt - TT x tt - Tt x Tt - Tt x TT **Solution: ** **1) For Tt x tt parents:** **Step 1:** Determine the type of inheritance. This is a monohybrid cross and parents are a simple case of complete dominance. Gene T is dominant on t. **Step 2:** Determine the genotype. One parent has genotype **Tt** and the other **tt**. **Step 3:** Types of gametes * Parent Tt produce two types of gametes with gene **T** and gene **t**. * Parent tt produce only one type of gamete with gene **t**. **Step 4:** Arrangement of gametes in checkered board.  * This is an example of a test cross. **2) For TT x tt parents:** **Step 1:** Determine the type of inheritance. This is a monohybrid cross and presents a simple case of complete dominance. Gene T is dominant on t. **Step 2:** Determine the genotype. One parent has genotype **TT** and the other **tt**. **Step 3:** Types of gametes. As parent **TT** produce only one type of gametes with gene **T** and parent **tt** produce only one type of gametes with gene **t**. **Step 4:** Arrangement of gametes in checkered board.  **3) For Tt x Tt parents:** **Step 1:** Determine the type of inheritance. This is a monohybrid cross and presente a simple case of complete dominance. Gene T is dominant on t. **Step 2:** Determine the genotype: One parent has genotype **Tt** and the other is also **Tt**. **Step 3:** Types of gametes: Both parents **Tt** produce 2 types of gametes with gene **T** and gene **t**. **Step 4:** Arrangement of gametes in checkered board:  **4) For Tt x TT parents:** **Step 1:** Determine the type of inheritance. This is a monohybrid cross and presents a simple case of complete dominance. Gene T is dominant on t. **Step 2:** Determine the genotype. One parent has genotype **Tt** and the other has **TT**. **Step 3:** Types of gametes: * Parent **Tt** produces 2 types of gametes with gene **T** and gene **t**. * Parent **TT** produces only one type of gametes with gene **T**. **Step 4:** Arrangement of gametes in checkered board:  * This is an example of a test cross. # Expt. No. ## 2. Red fruit (R) is dominant over yellow (r) and tallness (T) is dominant over short (t) in tomato plants. What phenotype and genotypic ratio would result if one of the parent plants is red homozygous and tall homozygous and other is red heterozygous and tall heterozygous? **Step 1:** Nature of Cross: dihybrid cross * Gene R for red color - Dominant * Gene r for yellow color - Recessive * Gene T for tall character - Dominant * Gene t for dwarf character - Recessive **Step 2:** Genotype: * Red homozygous and tall homozygous - **RRTT** * Red heterozygous and tall heterozygous - **RrTt** **Step 3:** Gametes: * Gametes from **RRTT** are only of one type **RT**. * Gametes from **RrTt** are of 4 types **RT, Rt, rT, rt**.  * **Homozygous for both characters:** **RRTT**, **RRtt** * **Homozygous for Red color:** **RRTT, RrTT, RRtt, RrTt** * **Heterozygous for both characters:** **RrTt** # Expt. No. ## 3. Mendel crossed pea plants producing round seeds with those producing wrinkled seeds from a total of 7324 F2 seeds, 5474 were round and 1850 were wrinkled. Using the symbols **W** and **w** for genes: * (a) Symbolize the original cross. * (b) The gametes. * (c) F1 progeny * (d) Represent a cross between two F1 plants. * (e) Symbolize the gametes and (f) summarize the expected F2 results under the headings: phenotypes, genotypes, genotypic frequency and phenotypic ratios. **Solution:** The cross between plants producing seeds with those producing wrinkled seeds can be represented as follows: **P Generation:**  **F1 Generation:**  **F2 Generation:**  * **3 Round: 1 wrinkled** * **5,474 round: 1,850 wrinkled** **Ratio: 3:1** # Expt. No. ## 4. In rabbits, black skin (B) is dominant over brown skin (b) and short hair (s) is dominant over long hair (s). If homozygous black-short haired male is crossed with a homozygous brown long haired female what will be the genotypes and phenotypes of F1 and F2 offspring? **Step-1:** Nature of cross: dihybrid cross. * Gene **B** for black skin - dominant * Gene **b** for brown skin - recessive * Gene **S** for short hair - dominant * Gene **s** for long hair - recessive **Step-2:** Genotypes: * Homozygous for black skin and short hair - **BBSS** * Homozygous for brown skin and long hair - **bbss** **Step-3:** * Sperm from father or male parent (homozygous dominant) - **BS** * Ova from mother or female parent (homozygous recessive) - **bs** **Step-4:**  * **F1:** All **BbSs** Black and short **Step-5:**  **F2:** This is a typical dihybrid cross. It’s a little easier to grasp with a Punnett Square and a diagram. * **9 Black & short** * **3 Black & long** * **3 Brown & short** * **1 Brown & long** **Ratio: 9:3:3:1** # Expt. No. ## 5. Let the allele for tallness be represented by ‘T’ and the allele for dwarfness by ‘t’ what will be the gametes produced by the parents and the height of the offspring (tall & dwarf) from each of the following crosses: * **a) Tt x tt** * **b) TT x Tt** * **c) Tt x Tt** * **a) Tt x tt (Test cross)** * **b) TT x Tt (Back cross)** * **c) Tt x Tt (Monohybrid cross)** # Expt. No. ## 6. In the table below, results are given for the separate mating of tomato plant phenotypes. What are the most probable genotypes for the parents in each instance? | Phenotypes of Plant | Number of Progens | | - | - | | Purple, cut x green, cut | 321 | 101 | 310 | 107 | | Purple, cut x purple, potato | 219 | 207 | 64 | 71 | | Purple, cut x green, potato | 404 | 0 | 387 | 0 | | Purple, cut x green, cut | 722 | 231 | 0 | 0 | | Purple, potato x green, cut | 70 | 91 | 86 | 77 | **Solution:** * **(a) Purple, cut x green, cut:** - Green is recessive, so it will have the genotype **pp**. - Some of the offspring are green and potato, so both parents have to be heterozygous. The genotypes will be **PpCc** for Purple, cut and **ppCc** for Green, cut. * **(b) Purple, cut x purple, potato:** - Some double recessive offspring are formed, which indicate that the parents are either homozygous recessive or heterozygous, but heterozygous for potato character, but homozygous and recessive for skin. - Therefore, the parents must have following genotypes: **Pp Cc** for Purple, cut and **PpCc** for Purple, potato. * **(c) Purple, cut x green, cut:** - Purple, cut is homozygous dominant for coat color and has the genotype **CC**. - Green, cut is homozygous recessive for both coat color and potato and has the genotype **ppcc**. * **(d) Purple, cut x green, potato:** - Purple, cut is homozygous for coat color and has the genotype **CC**. - Green, potato is homozygous recessive for both coat color and potato, so it will have the genotype **ppcc**. * **(e) Purple, potato x green, cut:** - Purple, potato is homozygous recessive for coat color and has the genotype **cc**. - Green, cut is homozygous recessive for potato and will have the genotype **cc**. # Expt. No. ## 7. Assuming that Aa and Bb allele pairs assort independently of each other, what would you expect to be the relative frequencies of the four gametes types in F2 generation? Make the checkerboard to explain the result. **Solution:** * This is a dihybrid cross. * Four types of gametes will be formed as follows: - AB - Ab - aB - ab  # Expt. No. ## 8. Pea plants with yellow, wrinkled seeds when crossed with plants with yellow, wrinkled seeds produce 145 yellow, wrinkled and 35 green, wrinkled - Give the genotype of parents. **Solution:** **Step 1:** Nature of cross: In pea plants: * Yellow color is dominant. * Green color is recessive. * Wrinkled character is recessive. **Step 2:** Genotype of parent: The parents are yellow and wrinkled, therefore, their genotype maybe **YYrr** and **Yyrr**. **Step 3:** Gametes: Yyrr parents will produce two types of gametes: **Yr** and **yr**. **Step 4:** Checkerboard:  **Step 5:** Conclusion: * The offspring produced from the cross are yellow, wrinkled and green, wrinkled in the ratio of 3:1 (145:35). * Yellow and green offspring are produced in 3:1 ratio. This means parents are heterozygous for yellow color, having genes **Yy**. * Therefore, both the parents have a genotype **Yyrr**. # Expt. No. ## 9. The following are some of Mendel's results with the hypothesis to which these were fitted. Test each for goodness of fit and indicate whether each is significantly different from the hypothesis. | Cross | Results | Hypothesis | | - | - | - | | Round x wrinkled seeds (F2) | 5,474:1,850 | 3:1 | | Violet x white flowers (F1) | 705:224 | 3:1 | | Round yellow x wrinkled green (F1) | 31:26:27:26 | 9:3:3:1 | | Round yellow (F1) x wrinkled green | 24:25:22:27 | 1:1:1:1 | **Solution:** * **a) Round seeds x wrinkled seeds** * **Step 1:** Nature of cross: The cross between round and wrinkled seed character, is an example of monohybrid cross, exhibiting phenomenon of complete dominance and segregation. Here, sound seed character is dominant over wrinkled seed characters. * **Step 2:** Genotype: * Parents bearing round seeds will have a genotype **RR**. * Parents bearing wrinkled seeds will have a genotype **rr**. **Step 3:** Gametes: * The gametes from parents bearing round seeds will have **R** gene. * The gametes from parents bearing wrinkled seeds will have **r** gene. **Step 4:** F1 result: * The F1 hybrids formed will have a genotype **Rr** and will produce gametes of two types: with R gene and r gene in equal proportion. * When F1 plants are interbred their own types of gametes have equal chances of uniting together either way, ie. R may unite with R or r. **From checkerboard:**  * The above results coincide with the experimental results, where the ratio between round and wrinkled seeds in F2 generation is approximately 3:1 (5474: 1850 or 2.96:1). # Expt. No. ## 10. In cattle, harmless condition (P) is dominant over horned (p). A certain bull is bred to 3 cows. With cow A, which is horned, a hornless calf is produced. With cow B, which is horned, a horned calf is produced. With cow C, which is hornless, a horned calf is produced. What are the genotypes of the 4 parents? **Solution:** * **Cow A (horned) is bred to a bull and the offspring is hornless.** * Cow A is horned, so it must be **pp**. * The offspring is hornless, so it can be either **PP** or **Pp**. * Since cow A is recessive, this means that the hornless bull is either homozygous **PP** or heterozygous **Pp**. * **Cow B (horned) is crossed to the same bull and the horned offspring is produced.** * The appearance of horned offspring shows that the bull is heterozygous **Pp**. * **Cow C (hornless) is crossed to the same bull and a horned offspring is produced.** * The cow C is hornless, so it can be either homozygous **PP** or heterozygous **Pp**. * The offspring is horned, so it must be **pp**. * Since the bull is heterozygous **Pp**, the cow C must also be heterozygous **Pp**. # Expt. No. 2 ## AIM OF THE EXPERIMENT: - To study gene interaction. ## THEORY: ### 1. INTERALLELIC OR INTRA-GENIC INTERACTION * This includes interaction between the genes of an allelic pair. * There are two main types: * Incomplete or partial dominance (1:2:1 Ratio): - After Mendel, several cases were recorded where F1 hybrid were not related to either of the parents, but the F1 hybrid in heterozygous condition show an intermediate phenotype. - It means that two genes of allelomorphic pairs are not related as dominance and recessive. - This was discovered by Carl Correns and was described as incomplete or partial dominance. - This is also called blending inheritance. - In such cases, F1 heterozygotes develop phenotype intermediate between the two homozygous phenotypes. - In F2 generation, the phenotypic ratio matches the genotypic ratio of 1:2:1 - **Example 1:** Inheritance of flower color in Mirabilis jalapa. - In the four-o’clock plant, Mirabilis jalapa, when plants with red flowers (RR) are crossed with plants having white flowers (rr) the F1 hybrid plants (Rr) bear pink flowers. - When these F1 plants with pink flowers are self-pollinated or are crossed among themselves, they produced plants with red, white and pink flowers in the ratio of 1:2:1 (F2 generation). * Codominance (1:2:1 Ratio): - In codominance, both the genes of an allelomorphic pair express themselves fully in F1 heterozygotes for combinant genes, exhibiting both the characters side by side. - These follow the law of segregation and F2 progeny exhibits both the characters side by side. - These follow the law of segregation and F2 progeny exhibits 1:2:1 ratio both genotypically as well as phenotypically. - **Example 1:** Codominance of coat color in cattle: - Gene **R1** stands for red coat color. - Gene **R2** stands for white coat color. - When red cattle (R1R1) are crossed with white cattle (R2R2), the F1 hybrids have roan colored skin. - The roan colored is actually expressed by a mixture of red and white hairs, which develop side by side in the heterozygous F1 hybrid. - In F2 generation, red, roan and white appear in the ratio of 1:2:1. - The phenotypic ratio coincides with the genotypic ratio R1R1, R2R2, (1:2:1) - **Example 2:** Codominance in Andalusian fowl. - In Andalusian fowl, a cross between pure black and pure white splashed varieties results in slate blue hybrids in F1. - These hybrids of F1 generation when crossed among themselves yield black, slate blue and white fowl in 1:2:1 ratio showing codominance of black and white. ### II. INTER-GENIC OR NON-ALLELIC INTERACTION: * This involves interaction between genes of different allelic pairs. * **1. Collaborator Genes: Novel phenotypes (9:3:3:1)** - In collaboration, two gene pairs, which are present on separate loci, but influence the same trait interact to produce completely new trait or phenotype that neither of the genes by itself can produce. - **Example 1:** Comb shape in chicken. - The inheritance of comb shape in chicken is the result of interaction between two pairs of genes. - The interaction between these two allelic pairs produces a new comb shape: - Genotype **R** - gives rise to : rose comb - Genotype **p** - gives rise to : pea comb - Genotype **rr** - gives rise to : single comb - Genotype **Rp** - gives rise to : walnut comb - Wyandotte variety of domestic chicken possesses rose comb, whereas Brahmas have pea comb. - Each of these types can be bred true. - Bateson crossed a comb shaped not expressed in either parent. - When these F1 chicken mated among themselves, the F2 chickens exhibited the familiar dihybrid ratio 9:3:3:1. - Walnut comb represented the novel phenotype or new phenotype. - These four phenotypes: walnut comb: rose comb: pea comb: single comb were in the ratio of 9:3:3:1. - Out of these four phenotypes, were different from those expressed in parents. ### 2. Epistasis: Inhibiting Genes. * Epistasis is the interaction between non-allelic genes (present at separate loci) in which one gene pair masks, inhibits or suppresses the expression of other gene pairs. * The gene that suppresses the expression of other genes is known as inhibiting or epistatic factor and the gene which is prevented from exhibiting itself is known as hypostatic. * For example: gene A masks the expression of gene B. The epistatic effect maybe in one direction or it may be occto in both directions between two genes pairs, ie. A -> B. * The epistatic crosses produce quite a number of modifications of 9:3:3:1 ratio. Although epistasis, two pairs of genes are involved, intragenic suppression. * Epistasis can be of two types: dominant epistasis and recessive epistasis. * **1. Dominant Epistasis:** - (12:3:1 or 13:3 Ratio) - In dominant epistasis, out of two pairs of genes, the single dominant allele (ie. gene A) of one gene masks the activity of other allelic pair (Bb). - Since the dominant epistatic gene A exerts its epistatic influence by suppressing the expression of gene B or b, it is known as dominant epistasis. * **2. Recessive Epistasis:** - (9:3:4 ratio) - Epistasis due to recessive gene is known as recessive epistasis. - Out of the two pairs of genes, the recessive epistatic gene locus, the dominant gene expresses itself only when the epistatic locus (also has the dominant gene. If the epistatic locus has recessive gene, the dominant gene fails to express. It means both acAA and ccaa individuals will have same phenotype, so that the overall phenotype becomes 9:3:4. ### 3. Complementary Genes or Duplicate Recessive Epistasis (9:7 ratio) * Epistasis due to recessive gene is known as recessive epistasis, ie, out of the two pairs of genes, the recessive epistatic gene masks the activity of the dominant gene of the other gene locus. * The dominant A expresses itself only when the epistatic locus C also has the dominant gene. * If the epistatic locus has recessive gene c, gene A fails to express. It means both CCAA and ccaa individuals will have same phenotype, so that the overall phenotype becomes 9:3:4. * The complementary genes are two pairs of nonallelic dominant genes (ie. present on separate gene loci), which interact to produce only one phenotypic trait, but either of them if present alone does not produce the phenotypic trait in the absence of another. * It means that, for the development of the dominant character in question, both the complementary genes must be represented at least by their own dominant allele, say A-B-. * The absence of even one of the two genes produces recessive phenotype (aab- or -A-bb). ### 4. Duplicate Genes or Duplicate Dominant Epistasis (15:1 Ratio) * When 2 pairs of genes determine the same, or nearly the same phenotype, so that either of the two genes is able to produce the character in question, these are said to be duplicate factors or duplicate genes. * The recessive phenotype develops only when gene pairs are double recessive.  ### III. PLEIOTROPIC EFFECT OF GENES: * The action of genes at a cellular level is unitary ie. one gene one action. * But, some genes produce a broad spectrum or phenotypic changes, so that a gene has multiple actions. * This is called pleiotropy. * Lethal genes usually have pleiotropic effects. ### Lethal Genes: * Certain genes are known to control the manifestation of some phenotypic trait as well as affect the viability of the organism. * Some genes have no effect on the viability alone. * These genes are known as lethal or semilethals depending upon their influence. * Complete lethal genes in homozygous condition will kill all or nearly all homozygous individuals, are able to survive. * The lethal genes are always recessive for their lethality and express the lethal effect only in homozygous condition. * **A. Dominant Lethal:** * The dominant lethal genes are lethal in homozygous condition and produce some detective or abnormal phenotype in heterozygous condition. * Their most serious effect in heterozygous condition may also cause death. * Following are the example of dominant lethal genes. * **B. Recessive Lethal:** - The recessive lethals produce lethal effects only in homozygous condition. - Their heterozygotes are normal. Hence, recessive lethals seem unnoticed in the population but are established in the population. - These are detected only when two heterozygous persons get married. - **Example:** Tay Sach’s Lethal * The recessive lethal gene for Tay Sach’s disease causes death of young children only in homozygotes, which are unable to produce enzymes needed for normal fat metabolism. - The accumulation of fat in nerve sheaths hampers transmission of nerve impulse, leading to poor muscular control and mental deficiency. ### IV. POLYGENIC INHERITANCE: * Quantitative inheritance (Polygenic Inheritance) - It is a type of inheritance controlled by one or more genes in which the dominant alleles have cumulative effect. - With each dominant allele expressing a part or unit of the trait, the full trait being shown only when all the dominant alleles are present. - The genes involved in quantitative inheritance are called polygenes. * Quantitative inheritance is therefore also called polygenic inheritance. * A few instances of quantitative inheritance occur in human: - kernel color in wheat, - cob length in maize - skin color in human beings, - human intelligence - milk and meat yield in animals - height in human beings - seed yield of crop plants, - including size, shape, and number of seeds or fruits per cent plant. * A polygene is defined as a gene whose dominant allele controls only a part or partial quantitative expression of a trait. * It is also termed as a gene in which a dominant allele individually produces a slight effect on the phenotype, but in the presence of similar other dominant alleles control the quantitative expression of a trait due to cumulative effect. * Hence, polygenes are also called cumulative genes. * In monogenic, as qualitative inheritance, the phenotypes are two (3:1) in case of a single gene pair and 4 (9:3:3:1) in case of 9 pairs of genes. * In polygenic or quantitative inheritance, the number of phenotypes is 3 (1:2:1) in case of one polygene pair, 5 (1:4:6:4:1) in case of two polygene pairs, and 7 (1:6:15:20:15:6:1) when 3 polygene pairs are involved. * Thus, we see that the number of intermediate types increases with the increase in the number of polygenes, but the number of parental types remain the same (2 in the above cases). ### V. Multiple Alleles * Mendel and his followers used the term ‘allele’ or ‘allelomorph’ to denote the alternative form of the normal gene. * It means that genes for all tall and dwarf characters of a pea plant are alleles. * The former is a normal allele or wild type and the later mutant allele. * A gene can mutate several times, producing several alternative expressions. * When three or more alleles are found for any particular gene, these are called multiple alleles. * They occupy the same locus in homologous chromosomes. * Many examples have been found in which more than two alternative alleles or multiple alleles for a character are present in a population. * Multiple alleles have been studied in rabbits, guinea pigs, mice, Drosophila, and man. ### VI. ABO Blood Groups: * Landsteiner (1900) discovered two kinds of antigens or agglutinogens. * These are named antigen A and antigen B. * There are present on the surface of red blood corpuscles. * Based on the presence and absence of these antigens human population is divided into the following four blood groups: * Persons with blood group A have antigen: A. * Persons with blood group B have antigen: B. * Persons with blood group AB have antigen: AB. * Persons with blood group O don’t have any antigen. * Normally, if a person has an antigen on his RBCs, his plasma has natural antibodies against the other antigen. * It means, person having antigen A in RBC, have antibody B in the plasma. * Person of blood group B have antigen A in their red blood corpuscles and antibody a (anti-a) in the plasma. * Persons of blood group AB have antigen both A and B in their red blood corpuscles, and no antibodies in the plasma. * Persons of group O have no antigen in red blood corpuscles, but have both A and B antibodies present in the plasma. * The following table represents the antigen and antibodies present in a person of different blood groups: | Blood groups | Antigen present in the RBC | Antibody Present in plasma | | - | - | - | | A | A | anti-B or b | | B | B | anti-A or a | | AB | A & B | No | | O | No | Anti-A & anti-B or ab | * **Genetics Basis Of ABO Blood Groups: ** * The synthesis of these antigens is controlled by three alleles: IA, IB, and i. Blood group O phenotype is wild type and recessive. Thus: * **Genotype of persons with blood group O = i i.** * **Genotype of persons with blood group A = IAIA or IAi.** * **Genotype of persons with blood group B = IBIB or IBi.** * **Genotype of persons with blood group AB = IAIB.** * The gene IA and IB are dominant to gene i, but are codominant to each other, and both are expressed when present together. # Expt. No. 3 ## AIM OF THE EXPERIMENT: - To study human karyotype (normal and abnormal) ## THEORY: * Of the 23 pairs of human chromosomes, twenty-two pairs are autosomes and one pair sex-chromosomes. * The autosomes of man and woman have similar appearance but sex chromosomes are different. * In a female, the two X chromosomes are identical and are represented as XX. * In a male, these are dissimilar and are represented as XY. * The Y chromosome is similar to X, and it is male-determining. ## Karyotyping: * Blood leukocytes are separated from the blood and are stimulated to divide by mitosis in vitro by adding phytohaemagglutinin. * Colchicine is added to arrest cell division at metaphase stage. * These cells are treated with hypotonic saline solution. * This makes the cells to swell and provides clarity to the chromosomes and helps in counting the chromosomes. * The chromosomes are stained with Giemsa technique to demonstrate banding pattern. * A suitable spaced of metaphase chromosomes is photographed. * The individual chromosomes are cut out from the photograph and are arranged in an orderly fashion in homologous pair. * This arrangement is called karyotype. * The following parameters are used for karyotypic arrangement: * Shape of chromosome. * Length of chromosome. * Proportion of arms, ie. the ratio between long and short arm of chromosomes. This ratio is 1:1 in a typical metacentric chromosome. * Centromeric index. - Determined by the ratio between the length of short arm and the total chromosome length. - Centromeric index = Short arm length / Total chromosome length. - The centromeric index for a metacentric chromosome is 0.5. - Depending upon the position of centromere and relative length of two arms, human chromosomes are of three types - metacentric, submetacentric, and acrocentric. - The photograph of chromosomes are artificially arranged in the order of descending length in seven groups from A to G by a convention of experts held at Denver, Colorado in 1960, as in Table (4). * **Karyotype and Idiogram:** - Karyotype is a systematized array of chromosomes of a single cell prepared either by drawing or by drawing or by photography, with the extension in meaning that the chromosomes of a single cell can typify the chromosomes of an individual or even a species. - The term idiogram is the diagrammatic representation of a karyotype, which may be based on measurements of chromosomes in several or many cells. - This sort of arrangement of chromosomes represents relative morphology of chromosomes, the karyotype. - This helps in proper identification and numbering of chromosomes. | SL No. | Group Size | Position of Centromere | Idiogram Number | Total No. of Chromosome in Diploid Cells | | - | - | - | - | - | | 1. | A | Large | Metacentric/ Submetacentric | 1, 2, 3 | 3 Pairs | | 2. | B | Large | Submetacentric | 4, 5 | 2 Pairs | | 3. | C | Medium | Submetacentric | 6, 7, 8, 9, 10, 11 | 6 Pairs + 1 in male (7 pairs + 1 pair in female) | | 4. | D | Medium | Acrocentric | 13, 14, 15 | 3 Pairs | | 5. | E | Small | Metacentric / Submetacentric | 16, 17, 18 | 3 Pairs | | 6. | F | Smallest | Metacentric | 19, 20 | 2 Pairs | | 7. | G | Small | Acrocentric | 21, 22, Y | 2 Pairs + 1 in male (3 pairs in female) | # Expt. No. ## 11. TH MEANS THEY POSSESS XXY-LEX-CHROMOSOMES INSTEAD OF NORMAL XY. * **Klinefelter’s syndrome** arises by the nondisjunction of XX-chromosomes. * When an abnormal egg-XX chromosome is fertilized with a sperm with Y-chromosome, the zygote possesses XXY sex chromosomes. * This develops into an abnormal male. * One out every 500 male births possess Klinefelter’s syndrome. ## 2) TURNER’S SYNDROME (45+X0) * The persons with Turner’s syndrome are phenotypic female, but their interphase nuclei are poorly developed. * They are sterile females with almost webbed necks and underdeveloped breasts. * They exhibit severe and broad chest, low set ears and broad chest. * Their intelligence is also below average. * In such women, instead of normal ovaries, only ridges of whitish tissue are present, which are known as streak gonads. * Many scientists use the term ‘gonadial dysgenesis’ to replace Turner’s syndrome. * They possess no Barr body; it means they possess only one X chromosome (XO). * Males with XXY (diplo-X), XXXY (triplo-X), XXXXY (tetra-X) or XXXXXY (penta-x) constitution were observed. * All these extra X arise as a result of non-disjunction of sex chromosomes. ## 3. Triplo-X Female: * Such females are known as super females. * These are usually retarded and infertile. * However, some triplo-X females are found to be fertile. * The children born to such females are found to be normal. * Presumably only X-bearing eggs are functional