BIOL 1090 Lecture Notes - Lecture 5: Punnett Square, Dihybrid Cross, Mendelian Inheritance
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We just look at one gene at a time if 1 parent was homozygous, he/she would generate only 1 kind of gamete. If the other parent was heterozygous, then the equation would be 1x2=2 possible genotypes. This assort(cid:373)e(cid:374)t for me(cid:374)del"s peas results i(cid:374) this pu(cid:374)(cid:374)ett square. Predicting the outcomes of a di-hybrid cross with a punnett square: for 1 gene with 2 alleles there are: 21 = 2 possible haploid genotypes (gametes) for 2 genes with 2 alleles there are: 22 = 4 possible haploid genotypes (gametes) Every time figure out the # of loci (e. g. genes) and for each of these loci the number of alleles. Question 2: you cross 2 pea plants, both of which have purple flowers. You are surprised to see that in the first generation, there are plants with both purple flowers and white flowers.