In case of starch synthesis in pea seeds [controlled by B gene], BB homozygotes produce large and round starch grains; bb homozygotes produce smaller and wrinkled starch grains and Heterozygotes produce round seeds of intermediate size. What can be inferred from this observation?
(1) The gene B mutates at a faster rate.
(2) The gene B is easily influenced by the environment
(3) Dominance is not an autonomous feature of a gene or the product that it has information for
(4) Some genes do not follow the law of segregation.
When Morgan hybridised yellow-bodied, white-eyed females to brown-bodied, red-eyed males and intercrossed their F1 progeny, the F2 ratio deviated very significantly from the 9:3:3:1 ratio. this can be attributed to the fact that:
(1) The genes are located on X and Y chromosomes
(2) Fruit fly has abnormal chromosomes
(3) The genes are located on the X chromosome
(4) The genes exhibit incomplete dominance
What is incorrect for Hemophilia?
1. In this disease, a single protein that is a part of the cascade of proteins involved in the clotting of blood is affected.
2. In an affected indlvidual a simple cut will result in non-stop bleeding.
3. The heterozygous female (carrier) for haemophilia may transmit the disease to sons.
4. The possibility of a female becoming a haemophilic is extremely rare because mother of such a female has to be hemophilic and the father should be a carrier.
Sickle cell anaemia results from.
1. A chromosomal aberration
2. Non disjunction of autosome
3. A point mutation
4. Blood transfusion reaction
Which of the following is not a feature of Down's Syndrome?
1. It is caused by a non-disjunction in an autosome
2. The affected individual has trisomy of chromosome 21
3. The aflected individual has a characteristic simian palmar crease
4. The mental development of affected individual is normal
Match each item in Column I with one item in Column II and chose your answer from the codes given below:
Column I Disorder |
Column II Feature |
I. Phenylketonuria II. Sickle Cell anemia III. Down's Syndrome IV. Turner's Syndrome |
(i) Rudimentary ovaries (ii) Gynecomastia (iii) Trisomy 21 (iv) Lack of enzyme PAH (v) Lack of tyrosinase (vi) Mutation GAG to GUG (vii) Mutation GUG to GAG |
1. I - v; II - vii; III - iii; IV - i
2. I - iv; II - vi; III - iii; IV - i
3. I - v; II - vi; III - iii; IV - ii
4. I - v; II - vi; III - iii; IV - i
In a monohybrid cross F1 progeny resemble neither of the parents. What would be true in this case?
1. The parental traits would not appear in any of the F2 -progenies
2. The F2 phenotypic ratio will be different from the F2 genotypic ratio
3. It could be a case of incomplete dominance
4. The F2 phenotypic ratio will be similar to any Mendelian monohybrid cross
The two alleles of a gene pair are located on:
1. Homologous sites on homologous chromosomes
2. Heterologous sites on homologous chromosomes
3. Homologous sites on heterologous chromosomes
4. Heterologous sites on heretologous chromosomes
The trait shown in the given pedigree chart is most likely a/an:
1. Autosomal recessive trait
2. Autosomal dominant trait
3. Sex linked recessive trait
4. Sex linked dominant trait
Aneuploidy results from :
1. Point mutations
2. Gross structural changes in chromosomes
3. Failure of cytokinesis after telophase stage of cell division
4. Failure of segregation of chromatids during cell division