Questions - Answers
Very short answer type questions
1. What is the cross between the F1 progeny and the homozygous recessive parent called? How is it useful?
A: Test cross.
It is used to assess the genotype of parent organisms and test the purity of parents.
2. Do you think Mendel’s laws of inheritance would have been different if the characters that he chose were located on the same chromosome?
A: Yes.
3. Who proposed the Chromosome Theory of Inheritance?
A: Walter Sutton and Theodore Boveri.
4. Define true breeding. Mention its significance.
A: True breeding is continuous self-pollination to arrive at a constant genotype. It is useful to show the stable trait inheritance and expression for several generations.
5. Explain the terms phenotype and genotype.
A: Phenotype: External characters of an organism.
e.g.: Tall, Dwarf.
Genotype: Genetic compostion of a character.
e.g.: Tall = (TT) or (Tt), Dwarf = tt.
The phenotype tall has two genotypes TT and Tt, while the phenotype dwarf has only one genotype tt.
6. What is point mutation? Give an example.
A: The mutations that arise due to change in a single base pair of DNA are called ‘Point Mutations’.
e.g.: Sickle cell anemia
7. A person has to perform crosses for the purpose of studying inheritance of a few traits/characters. What should be the criteria for selecting the organisms?
A: The following criteria are necessary for selecting the organism:
* It should be a short duration plant (annual).
* The characters should be well defined.
* It should be easy to grown and cross it easily.
* Presence of bisexual flowers with both male and female parts.
* It should be suitable for both self-pollination and cross pollination.
* It should have a short life cycle and produce many offsprings.
8. In order to obtain the F1 generation, Mendel pollinated a purebreeding tall plant with a pure breeding dwarf plant. But to get the F2 generation, he simply self - pollinated the tall F1 plants. Why?
A: It is necessary to cross two F1 plants to produce F2. Self pollinating tall F1 plants is equal to crossing two F1 plants hence Mendel simply self - pollinated tall F1 plants.
9. How alleles of a particular gene are different from each other? Explain its significance.
A: The alleles of a gene exist as two different entities. One of them may produce an enzyme responsible to act on a substrate through which a phenotype is expressed. The allele that produces this enzyme and is unmodified is the dominant allele. It represents the original phenotype. The modified allele may produce less efficient enzyme or no enzyme which causes a different phenotypic expression. Such modified allele is a recessive allele.
10. In a monohybrid cross between red and white flowered plants, a geneticist got only red flowered plants. On self - pollinating these F1 plants, he got both red and white flowered plants in 3 : 1 ratio. Explain the basis of using RR and rr symbols to represent the genotype of plants of parental generation.
A: When Red flowers are obtained in F1 it indicates that Red colour is dominant hence using RR for red genotype is correct, while White colour is not expressed in F1 hence it is recessive and represented as rr. As the ratio is 3 : 1 the parents can be assumed to be true breeding types and hence homozygous TT and tt can be used as genotypes for parental generation.
11. What is the genetic nature of wrinkled phenotype of pea seeds?
A: Wrinkled phenotype of pea seeds are genetically rr.(homozygotic as r is recessive gene it is expressed only in homozygous condition).
Short answer type questions
1. In a Mendelian monohybrid cross, the F2 generation shows identical genotypic and phenotypic ratios. What does it tell us about the nature of alleles involved? Justify your answer.
A: * Identical genotypic and phenotypic ratios are found in incomplete dominance or co dominance.
* The alleles involved in such a cross may show incomplete dominance that means the F1 is heterozygotic with intermediate character between two alleles. e.g.: Colour of flowers is pink in F1 individuals when White flowered and Red flowered plants are crossed (Snap dragon). When F1 is self pollinated we obtain 1 : 2 : 1 ratio of red, pink and white flowered plants whose genotypes are RR, Rr and rr. Hence both phenotypic and genotypic ratios are 1 : 2 : 1.
* In other case if the alleles show co-dominance the F1 individual resembles both parents that means both the alleles are expressed in F1 to some extent. e.g.: Spotted and Dotted seed character in Lentil. When a Spotted (seed) plant is crossed with Dotted (seed) plant the F1 produces seed with both spots and dots. The selfing of F1 produces F2 with 1 : 2 : 1 ratio of spotted, spotted and dotted and spotted progeny with genotypes (CsCs) (CsCd) and (CdCd). Here also both phenotypic and genotypic ratios are 1 : 2 : 1.
2. Mention the advantages of selecting pea plant for experiment by Mendel.
Advantages of selecting pea plant by Mendel for his hybridization experiment
A: * Pea plant is an annual with well-defined characters.
* It can be grown and crossed easily.
* It has bisexual flowers with androecium and gynoecium.
* It is suitable for self-fertilization and cross fertilization.
* It has a short life cycle and produces large number of offsprings.
* It shows number of contrasting characters which can be studied easily.
3. Differentiate between the following.
a) Dominant and Recessive
b) Homozygous and Heterozygous
c) Monohybrid and Dihybrid
A: a. Dominant and Recessive:
In a pair alleles only one appears in F1 generation which is called dominant allele or factor.
e.g.: T for tallness appears in F1 so it is dominant over ‘t’ for dwarfness which does not appear in F1 but reappears in F2.
In a pair of dissimilar alleles the factor that does not appear in F1 but reappears in F2 is called recessive allele or factor. ‘t’ for dwarfness is recessive.
b. Homozygous and Heterozygous:
If a allelic pair of genes are identical (similar) this condition is said to be homozygous condition.
e.g.: TT (Tall), tt (Dwarf). If a allelic pair of genes is non-identical or dissimilar this condition is called heterozygous condition.
e.g.: Tt (Tall).
c. Monohybrid and Dihybrid:
A cross between two parents differing in one pair of contrasting characters is called a monohybrid cross. For example A cross between a tall pea plant (TT)and a dwarf pea plant (tt). A monohybrid cross results in a F1 offspring called a monohybrid.
e.g.: (Tt).
A cross between two plants which differ in two pairs of contrasting characters is called a dihybrid cross.
e.g.: A cross between yellow and round seeded pea plant (YYRR) with a green and wrinkled seeded pea plant (yyrr). The F1 produced by by a dihybrid cross is called a dihybrid (YyRr).
4. Explain the Law of Dominance using a monohybrid cross.
A: Monohybrid cross is across between two true breeding lines which differ in a pair of contrasting characters. Mendel crossed a homozygous tall plant (TT) with a homozygous dwarf plant (tt) and produced heterozygous tall plants (Tt) which are called monohybrids. The details of the monohybrid cross are given hereunder:
The phenotypic ratio of Mendel’s monohybrid cross is 3 : 1 (Tall : Dwarf) while the genotypic ratio is 1 : 2 : 1 (homozygous tall : heterozygous tall : homozygous dwarf). The phenotypic probability of tall plants in F2 is 0.75 and dwarf plants if 0.25. The genotypic probalility of tall plants and dwarf plants is 0.25 homozygous tall, 0.50 heterozygous tall and 0.25 homozygous dwarf plants.
Law of Dominance
* “When a cross is made between two parents differing in a pair of contrasting characters, one of the characters appears in F1 generation which is called dominant character; the other character which does not express in F1 but appears in F2 is called recessive character.”
* The law of dominance explains about the expression of one parental character in F1.
* It also explains 3 : 1 ratio in F2.
5. Define and design a test cross.
A: A test cross is a cross between F1 individual with its recessive parent. It is done to identify the purity of parents. (or) A cross between F1 with its recessive parent to know the genotype composition is called a test cross.
e.g.: Tt × tt.
If a dihybrid test cross is made between a F1 dihybrid (Yellow and Round YyRr) and its recessive parent (yyrr), the following ratio is obtained.
Therefore the test cross ratio of a dihybrid cross would be 1 : 1 : 1 : 1.
6. When a cross is made between tall plant with yellow seeds (TtYy) and a tall plant with green seeds (Ttyy) what proportion of phenotype in the offspring is expected to be
a) Tall and green or
b) Dwarf and green
A: a. Tall and green 3/8
b. Dwarf and green 1/8
Explanation:
From the above data any question may be answered
Tall yellow = 3/8
Tall green = 3/8
Dwarf yellow = 1/8
Dwarf green = 1/8
7. Explain the following terms with examples.
a) Co-dominance
b) In complete dominance
A: a. Co-dominance:
This is a type of inheritance where the F1 individual resembles both parents.
e.g.: 1. blood group patterns in human beings (A, AB, B and O)
2. Seed coat pattern and size in Lentil (Lens culinaris) plants.
Seed coat pattern in Lentil (Lens culinaris)
Phenotypic and genotypic ratio = 1 : 2 : 1 (Spotted : Spotted and Dotted : Dotted)
In the above example when a pure bred spotted lentil is crossed with pure bred dotted lentil the F1 heterozygotes show both spotted and dotted seed coat pattern.
This shows that neither spotted nor dotted allele is dominant or recessive to each other.
As the alleles express equally in heterozygote they are termed ‘codominant’.
Self-pollination of spotted and dotted F1 progeny produces F2 progeny in the ratio 1 : 2 : 1 (spotted : spotted and dotted : dotted).
b. Incomplete dominance
Sometimes F1 phenotypes do not resemble either of the parental characters but show an intermediate character which is known as incomplete dominance.
e.g.: Flower colour in Snapdragon (Antirrhinum species)
Both phenotypic ratio and genotypic ratio of incomplete dominance are 1 : 2 : 1 (Red flowers : Pink flowers : White flowers).
8. Write a brief note on chromosomal mutations and gene mutations.
A: The alteration of genes which results in changes in the genotype or phenotype of an organism is called a ‘Mutation’.
Mutation leads to variation in DNA.
Mutation was first observed by Hugo de Vries in Oenothera lamarckiana (Evening primrose).
Mutations may take place within the genes or chromosomes.
Mutations may occur as deletions or insertions in a segment of DNA which are called gene mutations.
The mutations that arise due to change in a single base pair of DNA are called ‘Point Mutations’. e.g.: Sickle cell anemia
Mutations that occur due to deletions and insertions of base pairs of DNA are called frame-shift mutations.
The abonormalities or abberations in chromosome structure or number are called chromosomal mutations. e.g.: Mutations in cancer cells.
The chemical and physical factors that induce mutations are called ‘Mutagens’.
e.g.: UV radiation.
9. Define Law of Segregation and Law of Independent Assortment.
Law of Segregation:
A: It states that “when two alleles of a gene are present together in a heterozygous state, they do not fuse or blend in any way, but remain distinct and segregate during meiosis or in the formation of gametes so that each gamete will carry only one of them”.
Law of Independent Assortment:
It states that “when two pairs of traits are combined in a hybrid, segregation of one pair of characters is independent of the other pair of characters”.
10. With the help of an example differentiate between incomplete dominance and co-dominance.
A: In Incomplete dominance F1 phenotypes do not resemble either of the parental characters but show an intermediate character which is known as incomplete dominance. Incomplete dominance is a condition when one allele of a gene is not completely dominant over the other allele and results in the heterozygotes having phenotype different from the dominant and recessive homozygotes. e.g.: Flower colour in Snapdragon (Antirrhinum species)
Both phenotypic ratio and genotypic ratio of incomplete dominance are 1 : 2 : 1 (Red flowers : Pink flowers : White flowers)
Where as in Co-dominance the F1 individual resembles both parents. It is a phenomenon where heterozygotes have features of both the homozygotes, that means an allele is neither dominant nor recessive to the other.
e.g.: 1. Blood group patterns in human beings (A, AB, B and O)
2. Seed coat pattern and size in Lentil (Lens culinaris) plants.
Seed coat pattern in Lentil (Lens culinaris)
Phenotypic and genotypic ratio = 1 : 2 : 1 (Spotted : Spotted and Dotted : Dotted)
Long answer type questions
1. In a plant, tallness is dominant over dwarfness and red flower is dominant over white. Starting with the parents work out a dihybrid cross. What is standard dihybrid ratio? Do you think the values would deviate if the two genes in question are interacting with each other?
A:
F2 = F1 × F1 = Tall Red (Tt Rr) × Tall Red (Tt Rr)
Gametes TR, Tr, tR, tr TR, Tr, tR, tr
Probability of F2 generation
Through the above Punnet square it is evident that the phenotypic ratio of F2 generation is 9 : 3 : 3 : 1 and the genotypic ratio is 4 : 2 : 2 : 1 : 2 : 1 : 2 : 1 : 1.
If the genes in the above question interact (show linkage) then the above ratio changes. We may expect more parental combinations than recombinations.
Exercises
1. What will be the phenotypic ratio in the offsprings obtained from the following crosses.
a. Aa × aa
b. AA × aa
c. Aa × Aa
d. Aa × AA
A: a) 1 : 1
b) 100%Aa
c) 3 : 1
d) AA : Aa 1 : 1
2. In garden pea, the gene T for tall is dominant over its allele for dwarf. Give the genotypes of the parents in the following crosses.
a) Tall × dwarf producing all tall plants
b) Tall × tall producing 3 tall and 1 dwarf plants
c) Tall × dwarf producing half tall and half dwarf number of plants.
A: a. TT × tt
b. Tt × Tt
c. Tt × tt
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 R and r for genes, predict the:
a) The parental (p) genotypes
b) The gametes
c) F1 progeny
d) The cross between F1 hybrids
e) Genotypes, phenotypes, genotypic frequency, phenotypic ratio of F2 progeny.
A: a. RR × rr
b. R and r
c. Rr
d. Rr × Rr
e. RR (Round), Rr (Round), rr (wrinkled), genotypic frequency 1 : 2 : 1, Phenotypic ratio = 3 : 1.
4. The following data was obtained from an experiment on peas. The grey coloured seed is dominant over white coloured seed. Use the letter G for grey and g for white traits. Predict genotypes of the parents in each of the following crosses.
Parent Progeny
Grey White
a) Grey × White 164 156
b) Grey × grey 59 19
c) White × white 0 100
d) Grey × grey 180 0
A: a. Test cross between Gg × gg (1 : 1)
b. F1 × F1 Gg × Gg
c. gg × gg
d. GG × Gg or GG × GG
5. In tomatoes red fruit colour (R) is dominant to yellow (r). Suppose a tomato plant homozygous for red is crossed with one homozygous for yellow. Determine the appearance of the following:
a) The F1
b) F2
c) The offspring of a cross of the F1 back to the red parent.
d) The offspring of a cross of the F1 back to the yellow parent.
A: a. Red
b. Red : Yellow (3 : 1)
c. All red
d. Red : yellow (1 : 1)
6. In pea axillary position of flowers (T) is dominant over its terminal position (t). Coloured flowers (C) are dominant to white flowers (c). A true breeding plant with coloured flowers in axils is crossed to one with white terminal flowers. Give the phenotypes, genotypes and expected ratios of F1, F2 back cross and test cross progenies. What genotypic ratio is expected in the F2 progeny?
A: Phenotype of F1 = coloured axillary (all)
Genotype of F1 = Cc Tt (all)
Phenotypes of F2 = Coloured axillary, coloured terminal, white axillary
And white terminal ( 9 : 2 : 2 : 1)
Genotypes of F2 = CCTT, CCTt, CcTT, CcTt, CCtt, Cctt, ccTT, ccTt,
cctt . (1 : 2 : 2 : 4 : 1 : 2 : 1 : 2 : 1)
Phenotypes of back cross = All coloured axillary (all)
Genotypes of back cross = CCTT, CCTt, CcTT, CcTt (1 : 1 : 1 : 1)
Phenotypes of test cross = Coloured axillary, coloured terminal, white axillary and white terminal (1 : 1 : 1 : 1)
Genotypes of test cross = CcTt, Cctt, ccTt, cctt (1 : 1 : 1 : 1)
Genotypic ratio of F2 progeny = 1 : 2 : 2 : 4 : 1 : 2 : 1 : 2 : 1
7. In summer squash, a plant with white flowers and disc-shaped fruits is crossed to a plant with yellow flowers and sphere-shaped fruits. The F1 hybrids had white flowers and disc shaped fruits. Which phenotypes are dominant? Give the phenotypes of the parents and the hybrids. If these hybrids were selfed and 256 progeny were obtained, what would be the frequencies of the various phenotypes?
A: White flowers and disc-shaped fruits are dominant.
Parent phenotypes = White disc and yellow sphere
Phenotypes of hybrids = White disc
Frequencies of F2 would be 144(white disc) : 48(white sphere) : 48 (yellow disc) : 16 (yellow sphere).
8. Give the ratios of the following.
a) Monohybrid test cross
b) Dihybrid test cross
c) F2 phenotypic ratio of monohybrid cross
d) F2 phenotypic ratio of dihybrid cross
e) F2 genotypic ratio of monohybrid cross
f) F2 genotypic ratio of dihybrid cross
A: a. 1 : 1
b. 1 : 1 : 1 : 1
c. 3 : 1
d. 9 : 3 : 3 : 1
e. 1 : 2 : 1
f. 1 : 2 : 2 : 4 : 1 : 2 : 1 : 2 : 1
9. A diploid organism is heterozygous for 4 loci. How many types of gametes can it produce?
A: 16
10. What is crossing over? In which stage of cell division crossing over occurs? What is its significance?
A: The exchange of chromatid segments between non-sisters chromatids of homologous chromosomes is called crossing over. It occurs during pachytene stage of prophase - I during meiosis - I. Crossing over leads to new combinations of genes on the chromosomes which is called recombination. Recombinations leads to variations.
11. “Genes contain the information that is required to express a particular trait”. Explain.
A: Gene is a functional unit of inheritance which comprises of a particular sequence of DNA nucleotides that help in transcription and lead to protein synthesis. The traits are expressed based on the structural or functional proteins. Hence the information on the gene is required to express a particular trait via protein synthesis.
12. For the expression of traits genes provide only the potentiality and the environment provides the opportunity. Comment on the veracity of the statement.
A: In the studies of gene control it is observed that many factors like substrate influence the function of the genes (e.g.: Lac operon). Hence genes provide potentiality for the expression of genes and environment provides opportunity.
13. Two heterozygous parents are crossed. If the two loci are linked what would be the distribution of phenotypic features in F1 generation for a dihybrid cross?
A: The phenotypic distribution would be in such a way that parental combination are more and frequency of recombinations is less depending on the strength of linkage.
