Part B (2.5 pts) Assuming that there is NO meiotic recombination between homolog

Question

Part B (2.5 pts) Assuming that there is NO meiotic recombination between homologous chromosomes during meiosis I during the formation of gametes, what is the number of different gametes possible in humans due to independent assortment of the chromosomes (assume heterozygosity between maternal and paternal homologues)? e.g. how many different chromosome combinations are possible? Now, using this information, how many different zygotic combinations are possible upon fertilization between a human egg and sperm? The answer of this question basically answers what is the possibility of two independent fertilization events producing identical twins. SHOW YOUR WORK AND EXPLAIN YOUR ANSWER

Explanation / Answer

The cell contains 22 auto chromosomes of female and 22 auto chromosomes of the male. Individually those 22 chromosomes have possible chances of combinations are 22! or factorial of 22. These combinations will occur during the cell division.

Therefore, the possible chromosomal patterns are-

22! = 22x 21x20x19x18x17x16x15x14x13x12x11x10x9x8x7x6x5x4x3x2x1 = 11.24 x 1021 chances

The numbers of combinations for the zygotic condition:-

Now in the availability of the two types of chromosomes male and female, the possible combination then will be =22! x 22!

= 1.26 x 1042

Therefore, the probability of producing identical twins is 2/ 1.26 x 1042

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---