Question

A male rabbit of genotype AABBDD is crossed with a female rabbit of genotype aabbdd to produce ${ F }_{ 1 }$ hybrid offspring. How many genetically different gametes can be produced by this ${ F }_{ 1 }$ hybrid
(a) 4
(b) 8
(c) 16
(d) 32

Answer 1

Hint: We all know that a diploid individual possesses a pair of alleles for any particular trait, and each of those parents passes one of these randomly to its offspring. And for the calculation of gametes simply place a 2 above each heterozygous gene pair and a one above each homozygous gene pair and then multiply the numbers together to get the total number of different possible gametes.

Complete step by step answer:
Mendel discovered a theory and stated a law called the Law of Segregation which said that a diploid organism passes a randomly selected allele for a trait to its offspring where from each parent the offspring receives one allele. He observed those true-breeding pea plants with contrasting traits which gave rise to ${ F }_{ 1 }$ generations that all expressed the dominant trait and ${ F }_{ 2 }$ generations that expressed the dominant and recessive traits in a manner i.e 3:1 ratio. According to this law of segregation, each individual has different traits and for that, a diploid has a pair of alleles for each character and each parent passes an allele at random to their offspring which results in a diploid organism. The phenotype of the offspring is determined by the allele that contains the dominant trait. In a few words, the law states that copies of genes separate or segregate so that each gamete receives only one allele.
When we say genotype, it refers to the genetic makeup, composition, or structure of a specific organism and each gamete carries half a genotype since each gamete (whether this is a male gamete or a female gamete) is a haploid(i.e a single set of chromosomes) . Both parents produce $25\%$ each of gametes. The different types of gametes possible can be calculated by ${ 2 }^{ n }$ (where n is the number of heterozygous genes). The genotype of the ${ F }_{ 1 }$ hybrid is AaBbDdEe. The number of heterozygous genes is 4. So, the number of different types of gametes possible are: ${ 2 }^{ 4 }{ = }{ 16 }$.
So. the correct answer is, ’16’.

Additional information:
1) During meiosis, each gamete acquires one of the two alleles as chromosomes separate into different gametes.
2) The heterozygotes possess one dominant and one recessive allele, which receives each allele from either parent and will look identical to homozygous dominant individuals.
3) After a fine observation on the pea plants with two different traits produced offspring that all expressed the dominant trait, Mendel proposed the Law of Segregation which follows the 3:1 ratio.

Note: The first division of meiosis is the physical basis of Mendel’s law of segregation in which the homologous chromosomes with their different versions of each gene are segregated into daughter nuclei and the behavior of those homologous chromosomes during meiosis the segregation of the alleles can account at each genetic locus to different gametes. The two different alleles for a particular gene also segregate during meiosis as chromosomes separate into different gametes so that each gamete acquires one of the two alleles, but during the lifetime of Mendel, the scientific community never understood the role of the meiotic segregation of chromosomes in sexual reproduction.