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How To Make Gametes For Genetic Crosses
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Dr Praveen Kumar Agrawal is teaching live on Unacademy Plus

Dr Praveen Kumar Agrawal
Ex - Faculty, Allen Kota, 22 Yrs Experience. Author - 17 books Known for best explanation Youtube : Concept Gurukul

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amazing explanation sir G
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easy to understand..... thank you so much
sir confusing to h genetics pls slowly bolne ki koshish kijiye..
  1. COURSE Principles of inheritance and variations Lesson 6 How to make gametes for genetic crosses Dr. Praveen Kumar Agrawal M.Sc., Ph.D., CSIR NET (JRF). SRF, GATE 21Years of Pre-medical teaching experience


  2. How to Make Gametes? Method of Gamete Formation Rule 1 . Quality i.e., type of gamete is important. Quantity is not important at all. Rule 2 The number of alleles in gamete is half of the parental alleles. Gametes must contain alleles for each character, shown in parental genotype. Example .If parental genotype is AABB. Here the parent has 4 alleles, so a gamete will have 2 alleles, but the gametes in this case can never be AA and BB. The correct gamete should be AB. Rule 3 When all alleles in the genotype of the parent are in homozygous condition, only one type of gametes are produced. In such cases, take one allele for each character. Example: Parental genotype AA, gamete will be A. Parental genotype aa, gamete will be a. . Parental genotype AABB, gamete will be AB. .Parental genotype AAbbDD, gamete will be AbD.


  3. Rule 4 . Heterozygosity determines the types of gametes to be formed, i.e., types of gametes will be as much as the number of heterozygous characters. Note The types of gametes will be 2n, where, n is the number of heterozygous pairs (allele pairs). Example: 1. If parental genotype is AABB In this case, none of the character is in heterozygous condition. The value of n is 0. So, by formula 2n, the types of gametes will be 20 1 2. If parental genotype is AABb In this case, only one allele pair (i.e., Bb) is in heterozygous condition. So, according to formula, types of gametes should be 212. 3. If parental genotype is AaBbCc. In this case, all three allele pairs are in heterozygous condition. So, according to formula, types of alleles should be 23 8. So, eight types of gametes would be produced. Rule 5 .For preparing the types of alleles, use Bifurcation method Write both capital and small alphabet of one allele pair at a short distance. Now write both capital and small alphabet of other heterozygous allele pair (in bifurcating manner) against each alphabet of previous pair. Repeat this process for remaining allele pairs. Now, make the gametes


  4. Example: 1. Parental genotype is MmNn. Since, there is no homozygous pair, select both pairs. Write as follows MN Mn mN mn The correct gametes are MN, Mn, mN and mn (4 types). 2. Parental genotype is PpRrSsNN. In this case leave the NN as it is in homozygous condition Now, write as below: PRS PRs PrS Prs pRS pRs prS prs Now add N to each gamete (For homozygous pair). So, the true gametes of this case will be NPRS, NPrs, NPrS, NPrs, NpRS, NpRs, NprS and Nprs Total number is eight types


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