BIOL 1090 Lecture Notes - Lecture 5: Punnett Square, Dihybrid Cross, Mendelian Inheritance

22 views5 pages

Document Summary

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.

Get access

Grade+
$40 USD/m
Billed monthly
Grade+
Homework Help
Study Guides
Textbook Solutions
Class Notes
Textbook Notes
Booster Class
10 Verified Answers
Class+
$30 USD/m
Billed monthly
Class+
Homework Help
Study Guides
Textbook Solutions
Class Notes
Textbook Notes
Booster Class
7 Verified Answers

Related Documents

Related Questions