# MASTER THE GENETIC CODE (CLASS XII)

|Mar 15th, 2012 12:21pm

Genetics is one of the most difficult but interesting subjects in biology. A good understanding of genetics requires knowledge about the structure and function of cell, cell division and reproduction. Here are a few tips on how to master this subject.

• The unit, Genetics and Evolution amounts to 18 marks in the biology paper.
• This unit consists of three chapters namely; Principles of Inheritance and Variation (Chapter 5), Molecular Basis of Inheritance (Chapter 6) and Evolution (Chapter 7).
• Figure out the critical aspects of genetics and try to perfect them.
• An analysis of previous years question papers shows that many questions are asked as it is from the NCERT textbooks.
• The table given below shows the exact question numbers of the questions in the question papers and the page numbers in the textbook where the answers to these questions are available.
 2009 Set 1 2009 Set 2 2009 Set 3 2010 Set 3 2011 Set 1 2011 Set 3 Marks Chapter 5 Chapter 6 Chapter 7 Q. No. Pg. No Q.No Pg. No Q.No Pg.No 1 2 3 24 88 22 106-107 25 133 5 29 100 1 1 130 2 3 6(OR)6 88 77 5 1 3 111 2 130 2 3 5 8 (OR) 76 8 105 1 4 112 6 127 2 17 76 3 19 84 20 24 131 5 30 89 30 (OR) 136 1 7 112 2 7 99 3 20 22 75,76, 77, 89 21 97 5 29 (OR) 29 134,135, 141 1 7 112 1 2 9 106 3 21 106 5
 Cross Phenotypic Ratio Genotypic Ratio Monohybrid Cross 3:1 1:2:1 Dihybrid Cross 9:3:3:1 4:2:2:2:2:1:1:1:1 Incomplete Dominance 1:2:1 1:2:1
• Constructing a Punnett square is an essential part of solving problems based on Mendelian genetics.
• In order to understand Punnett square, let us consider the given example.
 Let us now understand the cross involved, XHXh  x  XhY Gametes: XH, Xh  and Xh, YPunnett Square: The gametes of the parents are taken on the horizontal and the vertical columns, here is the Punnett square for the given cross. Therefore, the probability that the male child will be affected is 25% while 50% of their total progeny would be affected.In case of a dihybrid cross: The Punnett square for the F2 generation in a dihybrid cross consists of 16 boxes. Each square gets one parent at the top and the left. Choose the parental genotype in such a way that the cross obtained possesses a set pattern. For example, let us take a cross between seed colour and seed shape. Yellow colour and round shape is dominant over green colour and wrinkled shape. Therefore, the cross would be Here you can note that there is a set pattern; The original homozygous parental genotype occupies the first and the last boxes. The hybrids, heterozygous for both the traits are formed along the diagonal. The rest of the boxes are occupied by the hybrids that are either heterozygous for seed shape or seed colour.

Questions on Pedigree Analysis:

• Pedigree analysis questions are usually asked on the inheritance of diseases such as sickle cell anaemia, haemophilia and colour blindness.
• The inheritance pattern of such diseases involves the understanding of these traits as to whether they are sex-linked or autosomal.
• The inheritance of sex-linked disorders differ in males and females while in case of autosomal disorders, there is no distinction in males and females.
• The inheritance of autosomal disorders depends on whether the trait is dominant or recessive. Here are some examples of sex-linked and autosomal disorders.
• The given table enlists some common genetic diseases.
• Here is an example of a question on pedigree analysis.
 2009 Set 1(Delhi Region): Question 24 Haemophilia is a sex-linked recessive disorder of humans. The pedigree chart given below shows the inheritance of haemophilia in one’s family. Study the pattern of inheritance and answer the question 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? Answer: (a) 4 − Carrier female with genotype X*X 5 − Affected male with genotype X*Y 6 − Normal male with genotype XY Here, X* represents the trait for haemophilia. (bThere is a 25% probability that their first child will be a haemophilic male.
• For questions on processes, try to give your answers in points or flow charts and then elaborate on them if the question asks for it.
• Now let us take a look at the chapter-wise important topics in this unit.
 Principles of Inheritance Molecular Basis of Inheritance Evolution and Variation Chapter / Marks 1 2 3 and 5 Mendelian ratios Mendelian crosses Laws of Mendel Codominance Test cross Incomplete dominance Linkage Sex Determination, Male and female Heterogamety Mendelian and chromosomal disorders Genetic code – start and stop codons Base complementarity Rule Griffith’s experiment, Hershey and Chase Experiment Messelson and Stahl experiment Splicing Splicing Replication histone proteins and DNA packaging Post-transcriptional Modifications (capping, Tailing) Transcription, translation Mutation Features of the genetic code Structure of tRNA lac operon Goals and features of HGP DNA Fingerprinting Origin of life Convergent and Divergent Evolution Adaptive radiation Homologous and Analogous organs Natural selection and Saltation Natural selection and Saltation Human evolution Hardy-Weinberg Principle Human evolution

Follow these simple tips to achieve great marks in the exam.

All the Best!

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• 1. aishwarya  |  March 14th, 2014 at 9:33 pm

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