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Karnataka 2nd PUC Biology Question Bank Chapter 5 Heredity and Variation
2nd PUC Biology Heredity and Variation One Mark Questions and Answers
Question 1.
Give the meaning of the term allele.
Answer:
The alternate forms of a gene that occupies the same position on the homologous chromosomes are called alleles.
Question 2.
What is genetics?
Answer:
Branch of biology which deals with the study of heredity and variations is called genetics.
Question 3.
Define Phenotype.
Answer:
The external appearance of an individual which has resulted due to the interaction between the genotype and the environment.
Question 4.
Define genotype.
Answer:
The total genetic constitution both expressed and suppressed of an organism is known as genotype.
Question 5.
Give the reason for Down’s syndrome.
Answer:
Trisomy of 21st chromosome.
Question 6.
Define test cross. What is its significance?
Answer:
It is a cross made between the organism of unknown genotype with double recessive individual to determine the nature of unknown genotype (ie., homozygous/heterozygous).
Question 7.
If a diploid organism is heterozygous for 4 loci, how many types of gametes can be produced?
Answer:
A diploid organism heterozygous for 4 loci, will produce 4 × 4 or 16 types of gametes.
Question 8.
Why is the blood group ‘O’ called as universal donor?
Answer:
‘O’ blood group does not have any antigen and it is accepted by all the other blood group individuals. So it is regarded as universal donor,
Question 9.
Write the phenotypic ratio of Monohybrid cross.
Answer:
3 : 1.
Question 10.
Mention the phenotypic ratio of a dihybrid cross.
Answer:
9 : 3 : 3 : 1.
Question 11.
What are multiple alleles?
Answer:
The phenomenon of occurrence of 3 or more alleles which have arisen as a result of mutation of normal genes and which occupy the same locus on homologous chromosomes are known as multiple alleles.
Question 12.
What genetic principle could be derived from a monohybrid cross?
Answer:
Law of dominance.
Question 13.
Which one change is the cause of sickle-cell anaemia?
Answer:
It is caused due to a point mutation at 6th position in B-chain of hemoglobin in which glutamic acid is replaced by valine.
Question 14.
Name any one plant and its feature that shows the phenomena of incomplete dominance?
Answer:
Mirabilis jalapa which, shows incomplete dominance in the colour of the flower.
2nd PUC Biology Heredity and Variation Two Marks Questions and Answers
Question 1.
What are multiple alleles? Give an example.
Answer:
When a gene expresses itself in more than 2 allelic forms, then the alleles are called multiple alleles.
e.g: The gene ‘I’ controls the human blood group. It exist in 3 different alleleic forms i.e., IA, IB, and i, and is responsible for the four human blood groups viz, A, B, O and AB.
Question 2.
Give the chromosomal constitution and the resulting sex in each of the following syndrome.
(a) Turner’s syndrome
(b) Klinefelter’s syndrome
Answer:
Turner’s syndrome – 22 pairs of autosomes and one X chromosome.
(44 + XO = 45 chromosome)
The resulting individual is a female.
Klinefelter ’s syndrome – 22 pairs of autosomes and XXY = [44 + XXY = 47 chromosomes]
The resulting individual is male with more female character.
Question 3.
Differentiate between the following:
(a) Dominance and Recessive genes
(b) Homozygous and Heterozygous
(c) Monohybrid and Dihybrid.
Answer:
(a) Dominant gene – A gene which has the capacity to express in the hybrid condition.
Recessive gene – A gene which fails to express in the hybrid due to the presence of a dominant gene
(b) Homozygous – It is a situation of an individual in which a particular character is controlled by a pair of similar genes.
Heterozygous – It is a situation of an individual in which a particular character is controlled by a pair of dissimilar genes.
(c) Monohybrid : An hybridisation in which inheritance of only one character is followed from parents to offsprings, is known as a monohybrid cross.
Dihybrid : An hybridisation in which inheritance of two characters are followed from. one generation to another is called as dihybrid cross.
Question 4.
Explain the following terms with examples:
(a) Co-dominance
(b) Incomplete dominance.
Answer:
(a) Co-dominance: It is a phenomenon where both the alleles in a heterozygous condition express themselves equally.
(b) It is a process where the dominant gene is incompletely dominant over the recessive gene and produces a phenotype which is intermediate to the parental type.
Question 5.
What is DNA polymorphism? Mention its significance.
Question 6.
Write any four characteristics of Down’s, syndrome.
Answer:
Characteristics of Down’s syndrome include the following.
- Short stature, large head, small nose with flat nasal bridge.
- Open mouth with a protruding tongue.
- Eyes that slant upward with epicanthus fold.
- Stubby parted fingers and toes.
- Mental retardation.
Question 7.
Mention the possible blood groups of the progency whose mother is heterozygous ‘ for Group A and father is heterozygous for Group B.
Answer:
A, B, AB and O.
Question 8.
List the antigens and antibodies of ‘A’ blood group and ‘O’ blood group?
Answer:
Question 9.
Write any four characters of Turners syndrome?
Answer:
- Broad chest and short body stature.
- Low hairline at the back of the neck.
- Shortened fourth and fifth fingers.
- Heart abnormalities.
- Hand and feet may be swollen at birth.
Question 10.
Write the chromosomal complement and two symptoms of Klinefelter’s syndrome. Chromosomal complement 44AA + XXY
Answer:
Symptoms:
- Gynoecomastia – Feminine characters like absence of facial hairs and presence of enlarged breast.
- Underdeveloped testis.
- Absence of spermatogenesis.
- Sterility.
Question 11.
Differentiate between incomplete dominance and co-dominance.
Answer:
Incomplete dominance | Codominance |
It is a phenomenon in which neither of the parental characters were completely expressed, instead a new intermediate character is expressed in the F1 generation. | It is a phenomenon where both the alleles in a heterozygous condition express themselves equally. |
Question 12.
Define linkage. Who discovered linkage in Drosophila?
Answer:
Morgan and others 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 combinations (also called recombination) of genes.
Question 13.
Define Co-dominance? Give an example.
Answer:
It is a phenomenon where both the alleles in a heterozygous condition express themselves equally.
- The blood group in humans is controlled by a single gene (IA) with three alleles, IA, IB and i
- The gene IA and IB are dominant over the recessive allele i, whereas the allele IA and IB are co-dominant i.e. they are expressed together without being influenced by each other.
- Genetically, a person having blood group ‘A’ may be homozygous(IA IA) or heterozygous (IAi)
- A person having blood group ‘B’ will be homozygous (IBIB) or heterozygous (IBi)
- A person having blood group ‘AB’ will be (IA, IB), heterozygous co-dominant.
- A person having blood groups ‘O’ will be (ii), homozygous recessive
Question 14.
Define the terms autosomes and allosomes.
Answer:
Autosomes | Allosomes |
are known as somatic chromosomes. | are known as sex chromosomes. |
contain genes for somatic characters. | are responsible for sex determination. |
2nd PUC Biology Heredity and Variation Three Marks Questions and Answers
Question 1.
What is incomplete dominance? Describe with one example.
Answer:
It is a process where the dominant gene is incompletely dominant over the recessive gene and produces a phenotype which is intermediate to the parental type.
In Mirabilis or 4° clock plant, incomplete dominance has been reported with respect to colour of the flower. In these plants some produce red flowers and some others produce white flowers. When a plant producing red flowers is crossed with a plant producing white flowers, in the F1 generation all plants produced pink flowers.
It is a heterozygous condition and the pink character is an intermediate or a blend between red and white This colour is produced because the red is being only partially dominant over white.
When F1 pink flower plants are self crossed to raise F2 progeny 3 types of plants are produced i.e, Red, Pink and White, in the ratio of 1 : 2 : 1 respectively.
By using appropriate symbols, the incomplete dominance can be represented as shown below. This cross is illustrated as indicated below by using appropriate letters.
Thus in this case, parental characters are expressed in homozygous conditions, and intermediate character in heterozygous condition.
Question 2.
Define the disorder of phenyketonuria.
Answer:
- It is caused by a recessive mutant allele on chromosome 12.
- The affected individuals lack an enzyme that catalyses the conversion of the amino acid phenylalanine into tyrosine.
- Consequently phenylalanine is metabolised into phenly pyruvate and other derivatives.
- Accumulation of these chemicals in the brain results in mental retardation.
- These are also excreted in the urine as they are not absorbed by the kidney.
Question 3.
Mention the advantages of selecting the pea plant for experiment, by Mendel.
Answer:
Mendel selected pea plant for his experiments because of the following reasons:
1. A large number of true breeding varieties with observable alternative forms for a trait were available.
2. Peas are normally self-pollinated. However, pea flowers can be readily cross pollinated (hybridised) if self pollination is prevented. This is achieved by removing the anthers (emasculation) before the pollen grains mature, and dusting the stigma of these flowers (female parent) with pollen from the desired plant (male parents).
3. Pea is an annual plant which gives results within a year.
4. A large number of seeds are produced by a pea plant, that helps in drawing correct conclusions.
5. Pea plants are easy to cultivate and do not require aftercare except at the time of pollination.
Question 4.
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?
Answer:
(only two types of gametes will be produced from each parent as the two loci are linked.)
Question 5.
How is the sex determined in human beings?
Answer:
In human beings, XY type of sex-determination is seen.
- The males have an X-chromosome and another small, but characteristically-shaped Y- chromosome, i.e., males have XY chromosomes along with other autosomes.
- The females have two X-chromosomes along with the other autosomes.
- It is a case of male heterogamety, where the males produce two types of sperms, 50% of sperms having one X-chromosome and the other 50% with one Y-chromosome.
- The females arc homogametic and produce all ova with one X-chromosome.
- Sex of an individual is decided at the time of fertilization by the type of sperm fertilizing the ovum.
It is as given below :
Question 6.
A child has blood group O. If the father has blood group A and mother blood group B, work out the genotypes of the parents and the possible genotypes of the other offsprings.
Answer:
A person having blood group O has genptype (ii). He/she receives one allele from each of the parents for this trait. Therefore, his/her father with blood group A must have genotype (IAi) and mother with blood group B must have genotype (IB i).
The other possible genotype of their offsprings will be as follows :
Question 7.
By using Punnet square, schematically represent the dihybrid cross experiment conducted by Mendel using seed color and seed shape of pea as characters.
Answer:
In a dihybrid cross, Mendel selected two pairs of contrasting characters. He selected the shape and colour of the seed as two characters, for his cross. He recognised round and wrinkled shape of the seeds as one pair of contrasting characters and yellow and green colour as another pair of contrasting characters.
He crossed a plant having Yellow and Round seeds with a plant having Green and Wrinkled seeds and got F1 generation. In the F1, he got all the plants producing Round and Yellow seeds. Thus he found that, yellow and round traits were dominant over green and wrinkled, respectively.
Later, he self-crossed the F1 hybrids and raised F2 generation. In this generation, he got four types of plants producing Yellow-Round, seed Yellow-Wrinkled seed, Green Round seed, and Green- Wrinkled seed. Thus he got two new characters-Yellow-Wrinkled seed and Green Round seed, in addition to the parental characters.
2nd PUC Biology Heredity and Variation Five Marks Questions and Answers
Question 1.
Explain the Law of Dominance using a monohybrid cross.
Answer:
MONOHYBRID CROSS : Mendel selected two plants. One with tall and the other with dwarf to get F1 generation. To his surprise, he got all tall plants in the F1 generation.
Parents : Tall plant × Dwarf plant
F1 : → All tall plants
Thus the tall character appeared in the F1 generation but dwarf character did not appear. He self crossed the F1 plants and raised the F2 generation. Surprisingly he got both tall and dwarf plants in the ratio of 3:1 respectively. This ratio is called Phenotypic ratio. He repeated the experiment by taking other contrasting characters and got the same results.
Based on these results, he concluded that the tall character is dominant and the dwarf character is recessive. Then he used appropriate symbols to designate the dominant and recessive factors. He used capital letters for dominant factors and small letters of the same type to denote the recessive alleles. The monohybrid cross can be represented as below.
Number of plants of alleles involved -1 pair
Base Number – 4
Phenotypic ratio – 3 : 1 (3 tall: I dwarf
Genotypic ratio – 1 : 2 : 1
Phenotypic classes – Two (Tall and dwarf only)
Genotypic classes – Three
[1 homozygous tall, 2.heterozygous tall and 3. homozygous dwarf]
Question 2.
Using a Punnett Square, workout the distribution of phenotypic features in the first filial generation after a cross between a homozygous female and a heterozygous male for a single locus.
Answer:
Question 3.
When a cross in made between a tall plant with yellow seeds (TtYy) and another tall plant with green seeds (Ttyy) , what proportions of phenotype in the offspring could be expected to be (a) tall and green (b) dwarf and green.
Answer:
Question 4.
Write about chromosomal complement, cause and symptoms of Down’s syndrome.
Answer:
Down’s syndrome is called as mongoloid idiocy or 21st trisomy. It is due to autosomal hyperploids i.e.., trisomy in 21st pair where the chromosomal compliment is 45AA + xy for males and 45AA + xx for female.
Symptoms include:
- Flattened nose.
- Protruding tongue.
- Upward slanting eyes.
- Epicanthal fold (inner comer of eyes may have a rounded fold of skin).
- Hands are short and broad with short fingers.
- Hypotonia (Low muscular strength in infants).
- Short stature.
- Small or malformed ears.
- Mental dullness.
Question 5.
Explain Mendel’s dihybrid cross.
Answer:
It is a cross between two plants differing in 2 pairs of contrasting characters.
Mendel selected shape of the seed and colour of the cotyledons as the two pairs of contrasting characters.
Round seed coat is dominant over wrinkled seed coat and yellow colour of cotyledon is dominant over the green colour.
Mendel crossed a pure breeding garden pea plant producing round and yellow seeds with a pure breeding garden pea plant producing wrinkled and green seeds.
Round seed coat and yellow cotyledons = 9
Wrinkled seed coat and yellow cotyledons = 3
Round seed coat and green cotyledons = 3
Wrinkled seed coat and green cotyledons = 1
Dihybrid phenotypic ratio = 9: 3 : 3 :1.
Genotypic ratio: 1 : 2 : 2 : 4 : 1 : 2 : 1 : 2 : 1.
Thus in the F2 generation, he got Yellow Round, Yellow wrinkled, Green round and Green wrinkled in the ration of 9 : 3 : 3 : 1, respectively.
Question 6.
Explain sickle cell Anaemia.
Answer:
This is an inborn error where the normal biconcave erythrocytes of an individuals are transformed into crescent or sickle shape. This is due to the production of defective S-haemoglobin. The production of normal Hb is controlled by the gene Hb whereas the other allele Hb controls the production of s-haemoglobin or sickle cell haemoglobin.
Therefore, this disease is not due to recessive gene but due to partially dominant pairs of alleles. When Hb occurs in homozygous condition (HbA HbA) normal hemoglobin is produced in the body and the person will be healthy. But the presence of Hbs in homozygous (Hbs Hbs) condition makes the person to suffer from chronic haemolytic anaemia.
It is due to the production of S- hemoglobin, which contains the amino acid valine instead of glutamic acid. This defective S-hemoglobin cannot transport oxygen. As a result, the RBC are deformed into sickle shape. Such RBCs are not able to move through the blood capillaries and therefore cause internal bleeding and pain. Such patients become anaemic and die.
The genetic code and the codon for glutamic acid is as follows:
Results: Glutamic acid is in the 6th position.
But under abnormal condition, instead of glutamic acid at the 6th position, valine, whose code and codon is as follows:
Result: Valine in the 6th position.
In heterozygous condition, both HbA and Hbs genes are present: Such individuals produce both normal and S-haemoglobin This indicates that the genes HbA and Hbs are co-dominant such individuals suffer from periodic discomfort and anaemia at higher altitudes.
Question 7.
What are chromosomal disorders? Describe the following chromosomal disorders in human beings
a) Klinefelter’s syndrome
b) Turner’s syndrome.
Answer:
a) Klinefelter Syndrome: This syndrome was reported by Harry Klinefelter. It is caused due to the presence of additional X in a genotype of XY. (Trisomy) The extrachromosomal number may go upto two or three in some instances XXXY and XXXXY.
Klinefelter syndrome is one of the most common causes of hypogonadism, affecting one in 500 births. In two thirds of cases, the chromosomal karyotype is 47, XXY, in males. Expression of the syndrome varies with t number of X-chromosomes and the degree of endocrinologic dysfunction. The patient’s history and physical examination provide enough information diagnose the condition in puberty, but most patients do not seek medical attention until adulthood, when problems such as impotency or infertility become an issue.
Symptoms:
- Small penis
- Diminished pubic, axillary, and facial hair
- Enlarged breast tissue (called gynecomastia)
- Learning disabilities
- Simian crease (a single crease in the palm)
- Abnormal body proportions (long legs, short trunk)
- Small firm testicles
- Sexual dysfunction
- Tall stature
- Personality impairment.
Note : The severity of symptoms may vary.
Tests may include:
- Karyotyping showing 47 chromosomes with XXY
- Semen exam showing low sperm count
- Decreased serum testosterone level
- Increased serum luteinizing hormone and increased serum follicle stimulating hormone.
b) Turner’s syndrome: Harry Turner reported this syndrome. It is an allosomal numerical abnormality caused due to a loss of X chromosome in a female genotype of XX.
Turner syndrome is a genetic disorder affecting only females, in which the patient has one X – chromosome in some cells; or has two X – chromosomes but one is damaged. Turner syndrome affects approximately 1 Out of every 2,500 female births worldwide. It embraces a broad spectrum of features, from major heart defects to minor cosmetic issues.
Some individuals with Turner syndrome may have only a few features, while others may have many. Almost all people with Turner syndrome have short stature and loss of ovarian function, but the severity of these problems varies considerably amongst individuals.
Signs of Turner syndrome include:
- Broad chest.
- Short stature.
- Shortened with a webbed appearance.
- Low hairline at the back of the neck.
- Low – set ears.
- Delayed growth of the skeleton.
- Shortened fourth and fifth fingers.
- Heart abnormalities.
- 1-lands and feet may be swollen or puffy at birth.
- Often have soft nails that turn upward at the ends when they are older.
- Many defects are due to obstruction of the lymphatic system during foetal development.
- A characteristic cosmetic feature is the presence of multiple pigmented navy blue colored spots on the skin.
- Women with Turner syndrome are usually infertile due to ovarian failure.
Diagnosis:
- Blood test
- Karyotype