Sickel cell anaemia is
(1) an autosomal linked dominant trait
(2) caused by substitution of valine by glutamic acid in the β-globin chain of haemoglobin.
(3) caused by a change in base pair of DNA
(4) characterized by elongated sickle like RBCs with a nucleus.
Point mutation involves
(1) insertion
(2) change in single base pair
(3) duplication
(4) deletion
Study the pedigree chart given below
What does it show?
1. | Inheritance of a sex-linked inborn error of metabolism like phenylketonuria |
2. | Inheritance of a condition like phenylketonuria as an autosomal recessive trait |
3. | The pedigree chart is wrong as this is not possible |
4. | Inheritance of a recessive sex-linked disease like haemophilia |
Select the incorrect statement from the following
(1) linkage is an exception to the principle of independent assortment in heredity
(2) galactosemia is an inborn error of metabolism
(3) small population size results in random genetic drift in a population
(4) baldness is a sex limited trait
Which one of the following condition in human is correctly matched with its chromosomal abnormality/linkage?
(1) Klinefelter's syndrome—44 autosomes + XXY
(2) Colourblindness —Y-linked
(3) Erythroblastosis foetalis— X-linked
(4) Down syndrome—44 autosomes + XO
A human male produces sperms with the genotypes AB, Ab, aB and ab pertaining to two diallelic characters in equal proportions. What is the corresponding genotype of this person ?
(1) AaBb
(2) AaBB
(3) AABb
(4) AABB
In the hexaploid wheat, the haploid (n) and basic (x) numbers of chromosomes are:
1. n = 7 and x = 21
2. n = 21 and x = 21
3. n = 21 and x = 14
4. n = 21 and x = 7
Inheritance of skin colour in humans is an example of :
(1) chromosomal aberration
(2) point mutation
(3) polygenic inheritance
(4) codominance
Two genes R and Y are located very close on the chromosomal linkage map of maize plant. When RRYY and rryy genotypes are hybridised, then F2 segregation will show :
(1) higher number of the recombinant types
(2) segregation in the expected 9 : 3 : 3 : 1 ratio
(3) segregation in 3 : 1
(4) higher number of the parental types
A common test to find the genotype or hybrid is by :
(1) crossing of one F2 progeny with male parent
(2) crossing of one F2 progeny with female parent
(3) studying the sexual behaviour of F1 progenies
(4) crossing of one F1 progeny with recessive parent