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Q1. In garden pea (Pisum sativum), a plant with red flowers was crossed with a plant with white flowers. Work out the possible genotypes and phenotypes of F₁ and F₂ generations. State any one of Mendel's law which could be derived from this cross.

Solution

                                      Mendel's law of dominance can be deduced from the above cross. The hybrid is heterozygous containing both the alternative alleles (R and r). However, only one trait, i.e. red colour appeared and the white colour trait was suppressed in the F₁ generation. So, we can conclude that the red colour trait is dominant over the white colour trait. This explains Mendel’s law of dominance. 

Q2. In one family, each of the four children has a different blood group. The mother is group A and the father is group B. Explain this pattern of inheritance with the help of a cross along with their genotypes.

Solution

A cross is carried out between the heterozygous father (for blood group B) and the heterozygous mother (for blood group A) to get four children with different blood groups.         All the four blood groups are controlled by three allelic genes I^A, I^B, i which exhibit an instance of multiple allelism.  Both I^A and I^B are dominant over i. However, when together, both are dominant and show the phenomena of co-dominance, forming the blood group AB. Six genotypes are possible with the combination of these three alleles.  

Q3. The male fruit fly and female fowl are heterogametic, while the female fruit fly and the male fowl are homogametic. Why are they called so?

Solution

The male fruit fly has XY sex chromosomes and produces two types of gametes. Hence, it is called heterogametic. A female fowl has ZW sex chromosomes, thereby producing two types of gametes. Hence, it is called heterogametic. The female fruit fly has two X chromosomes, i.e. XX and produces similar gametes. Hence, it is called homogametic. Also, the male fowl has two Z chromosomes and produces similar gametes. Therefore, it is called homogametic.

Q4. Why did Mendel choose garden pea for his experiment? How did he make sure that the plants were true breeding?

Solution

Mendel choose garden pea for his experiment because of the following reasons: 1. It is an annual plant with a short lifecycle. 2. It has perfect bisexual flowers containing both male and female parts. 3. The flowers are self-pollinating, and so, it is easy to get a pure line for several generations. 4. It is easy to cross-pollinate the plant because pollen from one plant can be introduced to the stigma of another plant by removing of anthers. 5. Pea plant produces a large number of seeds in one generation. 6. The plants can be easily raised, maintained and handled. 7. Many easily detectable contrasting characters were available.

Q5. What is aneuploidy?

Solution

The phenomenon of failure of segregation of two homologous chromosomes of a particular pair during cell division as a result of gain or loss of chromosome(s) is called aneuploidy.

Q6. Mention the type of allele which expresses itself only in a homozygous state in an organism.

Solution

A recessive allele expresses itself only in the homozygous state in an organism.

Q7. Work out F₁ and F₂ generation to explain the inheritance of flower colour in Pisum sativum. Give the phenotypic and genotypic ratios (start with pure breeding).

Solution

This is a case of Mendel's monohybrid cross.                                          

Q8. Name one autosomal dominant and one autosomal recessive Mendelian disorder in humans.

Solution

Huntington's disease is an autosomal dominant disorder and haemophilia is an autosomal recessive disorder.

Q9. Mention the phenomenon of sex determination in the following cases: (a)  Male Drosophila - XY (b)  Female fowl - ZW (c) Male grasshopper - XO

Solution

(a) Male heterogamety (b) Female heterogamety (c) Male heterogamety

Q10. When a tall pea plant was self-pollinated, one-fourth of the progeny was dwarf. Give the genotype of the parent and dwarf progeny.

Solution

The genotype of the parent is Tt and the genotype of the dwarf progeny is tt.

Q11. Define mutation. Name one physical and one chemical mutagen.

Solution

Mutation is a rare, random, discontinuous, inheritable variation in the amount or the structure of the genetic material in the genotype of an organism. Physical mutagen: UV rays Chemical mutagen: 5-Bromouracil

Q12. Explain how does trisomy of the 21^st chromosome occurs in humans. List any four characteristic features in an individual suffering from it.

Solution

Down’s syndrome is also known as Trisomy 21. It is caused by the presence of an extra chromosome no. 21. Both the chromosomes of the pair 21 pass into a single egg due to non-disjunction during oogenesis in the mother’s ovary. Thus, the egg has 24 chromosomes instead of 23, and the offspring has 47 chromosomes (45+XY in male and 45+XX in female) instead of 46 chromosomes (44+XX or XY). Characteristic features of Down’s syndrome: (i) Short statured with small round head (ii) Partially open mouth with protruding furrowed tongue (iii) Broad palm with characteristic palm crease (iv) Slow mental development

Q13. During his studies on genes in Drosophila which were sex-linked, T. H. Morgan found that the population phenotypic ratios deviated from  the expected 9:3:3:1 ratio. Explain the conclusion he arrived at.

Solution

(i) T. H. Morgan observed that  when  the  two genes  in a dihybrid cross are  located  on  the  same chromosome, the  proportion of parental gene  combinations in  the  progeny was much  higher than  the non-parental or recombination of genes. (ii) Morgan and his colleagues found that when genes were grouped on the same chromosome, some genes are tightly linked and show less recombination. (iii) When the genes are loosely linked, they show higher recombination.

Q14. (i)  Why are grasshopper and Drosophila said to show male heterogamety? Explain. (ii) Explain female heterogamety with the help of an example.

Solution

(i) Drosophila exhibits XY type of sex determination. Males produce two types of sperms, one having X chromosome and the other having Y chromosome, whereas females have only X chromosome. Grasshoppers exhibit XO type of sex determination. Males produce two types of gametes, one with X chromosome and the other with no chromosome. Thus, both show male heterogamety. (ii) Female heterogamety can be seen in female birds. In these, the females have one Z and one W chromosome, whereas males have a pair of Z chromosomes besides the autosomes.

Q15. For flower colour in pea, the allele for purple flower (P) is dominant over the allele for white flower (p). A purple-flowered plant therefore could be of genotype PP or Pp. What genetic cross would you make to determine the genotype of a purple-flowered plant? Explain how your cross gives you the correct genotype of the purple-flowered plant.

Solution

The genotype of a purple-flowered plant can be determined by conducting a test cross.                      If the F₁ generation produces all purple flowers, the parent would be homozygous dominant, i.e. PP.                         If the F₁ generation produces purple and white flowers in a 1:1 ratio, the parent would be heterozygous, i.e. Pp.  

Q16. Name the phenomenon which leads to situations like 'XO' abnormality in humans. How do humans with 'XO' abnormality suffer? Explain.  

Solution

Absence of one X chromosome leads to ‘XO’ abnormality. The individuals are sterile females with rudimentary ovaries. They have shield-shaped thorax, webbed neck, show poor development of breasts, short stature, small uterus and puffy fingers.

Q17. A haemophilic son was born to normal parents. Give the genotypes of the parents.

Solution

Father: 44 + XY  Mother: 44 + XX^h (X^h = X chromosome with gene for haemophilia) 

Q18. Write the genotype of (i) an individual who is a carrier of the sickle cell anaemia gene but apparently unaffected, and (ii) an individual affected with the disease.

Solution

(i) An individual who is a carrier of the sickle cell anaemia gene but apparently unaffected: Hb^AHb^S (ii) An individual affected with the disease: Hb^SHb^S

Q19. (a) You are given tall pea plants with yellow seeds whose genotypes are unknown. How would you find the genotype of these plants? Explain with the help of a cross. (b) Identify a, b and  c in the table  given  below:   Sr. No. Pattern of inheritance Monohybrid F₁ phenotypic expression 1. Co-dominance a 2. b The progeny resembled only one of the parents 3. Incomplete dominance c    

Solution

(a) Test cross will be performed to know the genotype of the given tall pea plants with yellow seeds.       If all the plants of the F₁ generation are tall with yellow seeds, then the parent is homozygous dominant (Case i). If the plants in the F₁ generation are in the ratio 1:1:1:1, then the parent plant is heterozygous dominant. (b) a: Both the forms  of a trait are equally  expressed in the F1 generation. b: Dominance c: Phenotypic expression of the F₁ generation is somewhat intermediate between the two parental forms of a trait.

Q20. A human being suffering from Down's syndrome shows trisomy of 21^st chromosome. Mention the cause of this chromosomal abnormality.

Solution

In Down’ syndrome, due to non-disjunction, the 21^st pair of chromosome fails to separate during oogenesis. Therefore, the egg possesses 24 chromosomes instead of 23. When the egg fuses with a sperm, the zygote has three copies of chromosome 21 resulting in trisomy.

Q21. Why is it that the father never passes on the gene for haemophilia to his sons? Explain.    

Solution

Haemophilia is a sex-linked recessive disease. The defective gene is present on the X chromosome only and not on the Y chromosome. As the father always contributes a Y chromosome and never passes an X chromosome to his son, the gene for haemophilia can never be passed from a father to his son.

Q22. Explain the sex determination mechanism in humans. How is it different in birds?

Solution

The XY type of sex determination is seen in humans. Both males and females have equal number of chromosomes. Males have one X chromosome and a smaller Y chromosome, while females have a pair of X chromosomes. Hence, males have autosomes and XY sex chromosomes, while females have autosomes and XX sex chromosomes. This shows male heterogamety. Female heterogamety is observed in birds. They exhibit ZW type of sex determination. Both males and females have equal number of chromosomes. Female birds have one Z and one W chromosome, whereas males have a pair of Z chromosomes.

Q23. What is gene locus?

Solution

Gene locus is the particular location or position where a gene is located on a chromosome.

Q24. (i) State the principle of independent assortment. (ii) How would the following affect the phenomenon of independent assortment? (a) Crossing over (b) Linkage

Solution

(i) The law of independent assortment states that the alleles of different characters located in different pairs of homologous chromosomes are independent of one another in their segregation during gamete formation and in coming together into the offspring by fertilisation, both processes occurring randomly. (ii) (a) Crossing over: Crossing over influences linked genes. As a result, 50% recombination is obtained in the test cross progeny. (b) Linkage: It influences recombination which is less than 50%.

Q25. Explain the chromosomal theory of inheritance.

Solution

The chromosomal theory of inheritance was proposed independently by Walter Sutton and Theodore Boveri in 1902. According to this theory, 1. As the sperm and the egg cells serve as a connecting link from one generation to the other, all the hereditary characters must be carried in them. 2. The hereditary factors are present in the nucleus of cells. 3. Like Mendelian alleles, chromosomes are also found in pairs. 4. The sperm and egg having haploid sets of chromosomes fuse to re-establish the diploid state. 5. The genes are carried on the chromosomes. 6. Homologous chromosomes synapse during meiosis and get separated to pass into different cells. This forms the basis of segregation and independent assortment.

Q26. The map distance in certain organisms between gene A and B is 4 units, B and C is 2 units and gene C and D is 8 units. Which one of these gene pairs will show more recombination frequency? Give reasons in support of your answer.  

Solution

      The recombination frequency is directly proportional to the distance between the genes. In the above example, the distance between genes C and D is more, i.e. 8 units. So, the recombination frequency will be more between them.

Q27. Why in a test cross did Mendel cross a tall pea pant with a dwarf pea plant only?

Solution

In a test cross, Mendel crossed a tall pea plant with only a dwarf pea plant to determine the genotype of the F₂ generation.

Q28. What is non-disjunction?

Solution

The inability of homologous chromosomes to separate out during meiosis in an individual is called non-disjunction. Due to this, some gametes receive both the chromosomes, while the rest do not receive any chromosomes. 

Q29. Name a disorder, give the karyotype and write the symptoms where a human male suffers as a result of an additional X-chromosome.

Solution

In Klinefelter's syndrome, a human male suffers as a result of an additional X-chromosome. His karyotype is 44+XXY.  Symptoms of Klinefelter's syndrome are (i) The sex of the individual is male, but the person possesses feminine characters. (ii)  Gynaecomastia, i.e. development of breasts. (iii) Poor beard growth and often sterile. (iv)  Feminine pitched voice.

Q30. A garden pea plant (A) produced inflated yellow pod, and another plant (B) of the same species produced constricted green pods. Identify the dominant traits.

Solution

Inflated green pod is the dominant trait and constricted yellow pod is the recessive trait.

Q31. Define heterozygous and homozygous.

Solution

When dissimilar or different pairs of alleles are present for a character, it is called heterozygous condition, e.g. Tt. When similar pairs of alleles are present for a character, it is called homozygous, e.g. TT.

Q32. A haemophilic man marries a normal homozygous woman. What is the probability that their daughter will be haemophilic?  

Solution

A haemophilic man (X^hY) marries a normal homozygous woman (XX). So, their daughter would have the genotype (X^hX). As haemophilia is a recessive trait, the daughter would always be a carrier of the trait and can never be haemophilic.

Q33. What is crossing over?

Solution

Crossing over is the mutual exchange of the corresponding segments of the adjacent paternal and maternal chromatids of the synapsed homologous chromosomes in the pachytene stage of meiosis-I, producing new combinations of genes. 

Q34. A non-haemophilic couple was informed by their doctor that there is a possibility of a haemophilic child to be born to them. Explain the basis on which the doctor conveyed this information. Give the genotypes and the phenotypes of all possible children who could be born to them?

Solution

The doctor must have conveyed this information on the basis of pedigree analysis. Pedigree analysis is a strong tool used to trace the inheritance of a specific trait, abnormality or a disease. Because  both  the parents are non-haemophilic, their  genotypes will be                                                

Q35. (i) How does a chromosomal disorder differ from a Mendelian disorder? (ii)  Name any two chromosomal aberration-associated disorders. (iii)  List the characteristics of disorders mentioned above which help in their diagnosis.

Solution

(i) Difference between Mendelian disorder and chromosomal disorder   Mendelian disorder Chromosomal disorder 1. This disorder is mainly due to alteration or mutation in a single gene. 1. This disorder is caused due to absence or excess or abnormal arrangement of one or more chromosomes. 2. It follows Mendel's  principles  of inheritance. 2. It does not follow Mendel's principles of inheritance. 3. It may be recessive or dominant in nature. 3. It is always dominant in nature. 4. Examples: Haemophilia,  sickle-cell anaemia 4. Example: Turner's syndrome   (ii) Two chromosomal aberration-associated disorders are Down's syndrome and Klinefelter's syndrome. (iii) (a) Down's syndrome: The individuals have overall masculine development, but they express feminine characteristics such as development of breasts,  i.e. gynaecomastia. They are sterile. (b) Klinefelter's syndrome: The females are sterile as ovaries are rudimentary. Other secondary sexual characters are also lacking.

Q36. Mendel crossed plants which bred true for yellow seeds with plants which bred true for green seeds. All seeds in the F₁ generation were yellow. Work out the inheritance involved in this cross by using symbols for the trait. Which trait was dominant?

Solution

                    We can conclude that the yellow seed colour is dominant over green because it is expressed in the F₁ generation.

Q37. Inheritance pattern of ABO blood group in humans shows dominance, co-dominance and multiple allelism. Explain each concept with the help of blood group genotypes.

Solution

Dominance: The alleles I^A and I^B are dominant over allele i as I^A and I^B form antigens A and B, respectively, but i does not form any antigen. Co-dominance: Both the alleles I^A and I^B are co-dominant as both of them are able to express themselves in the presence of each other in blood group AB (I^AI^B) by forming antigens A and B. Multiple allelism: It is the phenomenon of occurrence of a gene in more than two allelic forms on the same locus.  The ABO blood group in humans is determined by three different allelic forms I^A, I^B and i. The above three explanations prove that the inheritance of ABO blood group in humans shows dominance, co-dominance and multiple allelism.   Genotype Surface antigen Blood group I^Ai A A I^AI^A A A I^Bi B B I^BI^B B B I^AI^B AB AB ii - O    

Q38. (a) Sickle-cell anaemia in humans is a result of point mutation. Explain. (b) Write the genotypes of both the parents who have produced a sickle-celled anaemic offspring.

Solution

(a) In sickle-cell anaemia, due to point mutation, there is a substitution of a single nitrogen base at the sixth codon of the globin chain of haemoglobin.             (b) The genotypes of both the parents would be Hb^AHb^S and Hb^AHb^S.  

Q39. What are linked genes? How can a pair of linked genes be identified?

Solution

The genes which tend to transmit together as a unit as they are closely located on the same chromosome are called linked genes. The linked genes modify the Mendelian dihybrid ratio from 9:3:3:1 to 3:1 and modify the dihybrid test ratio from 1:1:1:1:1 to 1:1.

Q40. (a) Explain a monohybrid cross taking seed coat colour as a trait in Pisum sativum. Work out the cross up to the F2 generation. (b) State the laws of inheritance which can be derived from such a cross. (c) How is the phenotypic ratio of the F₂ generation different in a dihybrid cross?

Solution

        (b) Law of dominance: When two alternative forms of a trait or character (genes or alleles) are present in an organism, only one factor expresses itself in the F₁ progeny and is called dominant and the other which remains masked is called recessive. Law of segregation: The factors or alleles of a pair segregate from each other during gamete formation such that a gamete receives only one of the two factors. They do not show any blending. (c) Phenotypic ratio of the F2 generation in a monohybrid cross is 3:1, whereas in a dihybrid cross, it is 9:3:3:1.