Very Short Answer Type Questions

Question. Give an example of a human disorder that is caused due to a single gene mutation.
Answer : Sickle cell anaemia/Thalassemia/Phenylketonuria.

Question. What is a Mutagen? Name a physical factor that can be a Mutagen. 
Answer : All the physical and chemical factors that induce mutation, UV radiation/X rays. 
Detailed Answer :
All the physical and chemical factors that induce mutation are called mutagens.
The physical factors which can be a mutagen are the ionizing radiations like X-rays, gamma rays & the non-ionizing radiations like ultraviolet rays.

Question. Give an example of a sex-linked recessive disorder in humans.
Answer : Colour blindness. 

Question. Give an example of an organism that exhibits haplodiploid sex-determination system.
Answer : Honey bees.

Question. Indiscriminate diagnostic practices using X-rays etc., should be avoided. Give one reason.
Answer : Indiscriminate diagnostic practices using X-rays, gamma rays etc. are ionizing radiations which usually produce breaks in the chromosomes and chromatids and abnormal mitosis in the irradiated cells. They cause abnormal functioning of the cells, mutations resulting in the development of various types of cancers specially blood cancer or leukemia etc. 

Question. Write the chromosomal defect in individuals affected with Klinefelter’s syndrome.
Answer : Male – Additional copy of X chromosome / XXY. 

Question. How many chromosomes do drones of honeybee possess ? Name the type of cell division involved in the production of sperms by them.
Answer : 16, Mitosis.  

Question. Why do normal red blood cells become elongated sickle shaped structures in a person suffering from sickle cell anaemia?
Answer : The mutant haemoglobin molecule (substitution of Glutamic acid by valine) undergoes polymerization, under low oxygen tension causing the change.
Detailed Answer :
The defect is caused by the substitution of Glutamic acid (Glu) by Valine (Val) at the sixth position of the beta globin chain of the haemoglobin molecule.
The substitution of amino acid in the globin protein results due to the single base substitution at the sixth codon of the beta globin gene from GAG to GUG.
The mutant haemoglobin molecule undergoes polymerisation under low oxygen tension causing the change in the shape of the RBC from biconcave disc to elongated sickle like structure. 

Question. The son of a haemophilic man does not get this genetic disorder. Mention the reason.
Answer : The son gets Y chromosome from the father and X chromosome from the mother. Therefore, as the gene for haemophilia is located on X chromosome, a son cannot get the disease from his father. 

Question. What is the cause of Down’s syndrome in humans ?
Answer : This syndrome develops due to trisomy of chromosome 21. The non-disjunction of the 21 chromosome during meiosis causes the trisomy of 21st chromosome and results in Down’s syndrome. 

Question. Mention the combination (s) of sex chromosomes in a male and a female bird.
Answer : Male – ZZ, Female – ZW
Detailed Answer :
The type of sex chromosomes in a female bird is ZW and in case of male bird is ZZ.

Question. Name the disorder caused due to the absence of one of the X-chromosomes in a human female.
Answer : Turner’s syndrome
Detailed Answer :
A syndrome that occur due to monosomy is called turner’s syndrome. It occurs due to union of an allosome free egg (22 + 0) and a normal X sperm or a normal egg and an allosome free sperm (22 + 0).
The individual has 2n = 45 chromosomes (44 + XO) instead of 46.

Question. Observe the pedigree chart and answer the following questions :

(i) Identify whether the trait is sex-linked or autosomal.
(ii) Give an example of a disease in human beings which shows such a pattern of inheritance.
Answer : (i) Sex-linked.
(ii) Haemophilia/colour blindness. 

Question. Write the genotype of (i) an individual who is carrier of sickle cell anaemia gene but apparently unaffected and (ii) an individual affected with the disease.
Answer : (i) Hb^A, Hb^S
(ii) Hb^S Hb^S 

Short Answer Type Questions

Question. Differentiate between ‘ZZ’ and ‘XY’ type of sexdetermination mechanisms.
Answer : ZZ type of sex determination mechanism is found in birds, reptiles and fishes. In this type, the females have heteromorphic sex chromosomes (ZW), while males have homomorphic sex-chromosomes (ZZ). Females are heterogametic i.e. produce two dissimilar types of eggs while males produce only one type of sperms. The egg determines the sex of the individual.
XY type of sex determination mechanism is found in human beings. In this type, the male individuals have heteromorphic sex chromosomes (XY) and are therefore heterogametic i.e. producing two types of sperms are with X and the other carrying Y chromosome. The females have homomorphic sex chromosome (XX) and homogametic i.e. produce only one type of eggs. The sex of the offspring is determined by type of sperm taking part in fertilization. 

Question. Is haemophilia in humans a sex linked or autosomal disorder ? Work out a cross in support of your answer.
Answer : Haemophilia in humans is a sex linked disorder.

Question. Why is the possibility of human female suffering from haemophilia rare ? Explain.
Answer :

Rare because mother should be atleast carrier and father haemophilic (non viable at later stage). 
Detailed Answer :
Haemophilic females can be produced only in the homozygous condition i.e. when the genes for haemophilia are present on both of the X chromosomes (X^hX^h). Marriage between a haemophilic man (X^hY) and carrier woman (XhX) will produce haemophilic female (X^hX^h) but however this condition is lethal and therefore haemophilic female dies in the foetus state. Thus, the possibility of human female suffering from haemophilia is rare.

Question. This is the pedigree of a family tracing the movement of the gene for haemophilia. Explain the pattern of inheritance of the disease in the family. 

Answer : In the given pedigree of a family, a normal female is crossed with a haemophilic male. The offspring generated are carrier females as haemophilia is an X-linked disease and female has two X chromosomes. These carrier females when crossed with normal males produced carrier females and haemophilic males.

Question. (i) Why are grasshopper and Drosophila said to show male heterogamety ? Explain.
(ii) Explain female heterogamety with the help of an example. 
Answer : (i) In grasshopper, males have one X only (XO type), in Drosophila males have one X and one Y (XY type) – Males in both cases produce 2 different kinds of gametes so heterogametic.
(ii) In birds female has ZW, produce two kinds of gametes and so heterogametic.
OR
(i) Male heterogamety, Grasshopper.
(ii) Female heterogamety, Birds.
Detailed Answer :
(i) In grasshoppers, there is XX—XO type of sex determination. Males have only one sex chromosome and therefore produce two types of male gametes : one with X chromosome (A + X) and the other without sex chromosome (A + O), thus shows heterogamety.
In Drosophila, there are two sex chromosomes (XY) in males which are heteromorphic and therefore two types of male gametes : one with X chromosome (A + X) and another with Y chromosome (A + Y).
(ii) Production of dissimilar types of gametes by females is called as female heterogametic. It is usually found when the sex chromosomes are heteromorphic. In female bird, the sex chromosomes are ZW and produce two types gametes (A + Z) and (A + W), thus showing heterogamety.

Question. During a cytological study conducted on the chromosomes of the insects, it was observed that only 50% of the sperms had a specific structure after spermatogenesis. Name the struture and write its significance in sex determination of insects. 
Answer : X body / X factor / X chromosome.
In insects the sex chromosome consists of XX female; XO-Males 1 + 1
Detailed Answer :
The specific structure is X chromosome or X factor or X body.
In insects, all eggs bear additional X chromosome.
Male have only one X chromosome besides autosomes where as females have a pair of X chromosomes e.g., grasshopper.

Question. A couple with normal vision bear a colour blind child. Workout a cross to show how it is possible and mention the sex of the affected child.
Answer :

Affected child is male. 
Detailed Answer :

Question. One of the twins born to parents having normal colour vision was colour blind whereas the other twin had normal vision. Work out the cross. Give two reasons how it is possible.
Answer :

Genes that lead to colour blindness located on X-chromosome. 
Gene is recessive and is suppressed in heterozygous mother (female) but expressed in male in single dose. 

Question. Explain with the help of an example each– Male and female heterogamety mechanisms of sex determination. 
Answer : Male produces 2 different types of gametes
XO – e.g. grasshopper.
XY – e.g. human, it is the type of sperm fertilising the egg that determine the sex of the offspring.
Female produces 2 different types of gametes.
ZW – eg. : Birds, it is the type of egg getting fertilised with the sperm that determine the sex of the chick.

Question. Identify ‘a’, ‘b’, ‘c’, ‘d’, ‘e’ & ‘f’ in the table given below :

Answer : (a) Short statured / small round head / furrowed tongue / partially open mouth / palm is broad / physical development retarded / psychomotor development retarded / mental development retarded. 
(b) both / male and female 
(c) Klinefelter’s syndrome 
(d) male 
(e) sterile ovaries / rudimentary ovaries, lack of secondary sexual characters. 
(f) female. 

Question. Given below is the representation of amino acid composition of the relevant translated portion of b-chain of haemoglobin, related to the shape of human red blood cells.

(i) Is this representation indicating a normal human or a sufferer from certain genetic disease ? Give reason in support of your answer.
(ii) What difference would be noticed in the phenotype of the normal and the sufferer related to this gene ?
(iii) Who are likely to suffer more from the defect related to the gene represented – the males, the females or both males and females equally ? And why ? 
Answer : (i) This representation (HbA peptide) indicates a normal human, because the glutamic acid in the sixth position is not substituted by Valine.
(ii) The sufferer’s RBCs become elongated and sickle shaped as compared to the normal biconcave RBCs. 
(ii) Both males and females are likely to suffer from the disease equally, as this is not a sex linked disease. It is an autosomal linked recessive trait. 
Detailed Answer :
(i) This micrograph representation (HbA peptide) is of amino acid composition of a part of b-chain of haemoglobin molecule of a normal human because the sixth codon of b-globin mRNA is GAG and therefore there is glutamic acid at the sixth position of b-globin chain.
(ii) The sufferer’s RBCs get elongated and becomes sickle shaped because this gene, if simulated cause the haemoglobin molecule to undergo polymerization due to oxygen tension resulting in the change of shape of RBCs from concave disc to elongated sickle like structure. Such RBCs can not pass through narrow blood capillaries, therefore they tend to slow down the blood flow, clot, degenerate and thus causing sickle cell anaemia.
(iii) The males and females both are likely to suffer from this disorder or defect because this is not a sex linked disorder. It is an autosomal linked recessive disorder due to single base substitution resulting in the change at the sixth codon of beta globin chain from GAG to GUG resulting in the substitution of glutamic acid to valine.

Question. (i) Name the genetic disorder in a human female having 44 + XO karyotype. Mention the diagnostic features of the disorder.
(ii) Explain the cause of such chromosomal disorder.
Answer : (i) (a) Turner’s syndrome.
(b) 44 with XO chromosomes — such females are sterile as ovaries are rudimentary. Other features include lack of other secondary sexual characters, short stature and under developed feminine characters.
(ii) Such a disorder is caused due to the absence of one of X chromosomes. 

Question. A cross between a normal couple resulted in a son who was haemophilic and a normal daughter. In course of time, when the daughter was married to a normal man, to their surprise, the grandson was also haemophilic.
(i) Represent this cross in the form of a pedigree chart. Give the genotypes of the daughter and her husband.
(ii) Write the conclusion you draw from the inheritance pattern of this disease. 
Answer :

Construction of pedigree chart 
(ii) Sex – linked recessive inheritance pattern 
Detailed Answer :
(i) Genotype of daughter is XX^h
Genotype of her husband is XY
(ii) Conclusion drawn from the inheritance pattern :
(a) XX^h females are carriers of haemophilia. They are not suffering from haemophilia but can pass on the gene X^h to offspring.
(b) X^hY males suffer from haemophilia. They seldom reach reproductive age.
(c) XX females are normal.
(d) XY males are normal too.
(e) X^hXh females die in embryonic stage.
(f) Inheritance of haemophilia follows a criss-cross pattern.

Question. Haemophilia is a sex linked recessive disorder of humans. The pedigree chart given below shows the inheritance of Haemophilia in one family.
Study the pattern of inheritance and answer the questions given.
(a) Give all the possible genotypes of the members 4, 5 and 6 in the pedigree chart.
(b) A blood test shows that the individual 14 is a carrier of haemophilia. The member numbered 15 has recently married the member numbered 14.
What is the probability that their first child will be a haemophilic male? Show with the help of Punnett square. 
Answer : (a) Genotypes of member 4 – XX or XX^h
Genotypes of member 5 – X^hY
Genotypes of member 6 – XY

(b) The probability of first child to be a haemophilic male is 25%.
Detailed Answer:
(a) Same as in marking scheme.
(b) Probability of first child to be hemophilic male is 25 %.

Question. Why is that the father never passes on the genes for haemophilia to his son ?
Answer : Haemophilia is a sex linked trait. The gene for it is located on X chromosome only. Since, father contributes only Y chromosome to the son, he never passes haemophilic gene to his son.