General Biology 1 Fall 2024 Introduction to Genetics PDF

General Biology 1 Fall 2024 Introduction to Genetics Part -2 Prof : Dr. Vincent Gagnon Book Raven, Biology, 13th edition  Introduction to Genetics Study guide: Topic...

General Biology 1 Fall 2024 Introduction to Genetics Part -2 Prof : Dr. Vincent Gagnon Book Raven, Biology, 13th edition  Introduction to Genetics Study guide: Topic References (13th Ed) Section 12.1 to 12.4, Identify Mendel's Principles. Be prepared to support the use of garden peas pages 231-239 as an experimental tool. Study all figures in this chapter Demonstrate how Mendel determined each principle using the mono- and Section 12, Figures dihybrid crosses. Be able to identify the parental and first or second filial 12.5 and 12. 7 pages generation in any genetic cross. 235 & 237. Relate Mendel's model to the scientific method and identify his hypotheses Section 12, pages 231- for the monohybrid and dihybrid experiment. 236. Define test cross and understand its importance in determining genotypes Section 12, pages 236- of dominant phenotypes. Use a Punnett square to determine the outcome 237, Figure 12.8 of a cross and calculate phenotypic ratios.  Introduction to Genetics Monohybrid crosses Cross to study only 2 variations of a single trait. Example: flower colour True breed parental generations Dihybrid crosses Examination of 2 separate traits in a single cross (genes located on different chromosomes). Produced true-breeding lines for 2 traits. Example: - Seed colour (Yellow or Green) - Seed shape (Round or wrinkle)  Introduction to Genetics Dihybrid crosses Examination of 2 separate traits in a single cross. True breeding parent of each traits Dominant Recessive RR YY rr yy The F1 generation of a dihybrid cross shows only the dominant phenotypes for each trait. Punnett squares ry ry RY RrYy RrYy RY RrYy RrYy  Introduction to Genetics Dihybrid crosses F1 self-fertilizes Self pollination of F1 hybrid Rr Yy Rr Yy The F2 generation shows all four possible phenotypes in a set ratio 9:3:3:1 The phenotype ratio of F2 show the independent assortment of the gene of each trait (they are on different chromosomes)  Introduction to Genetics Principle of Independent Assortment In a dihybrid cross, the alleles of each gene assort independently. The segregation of different allele pairs is independent. Independent alignment of different homologous chromosome pairs during metaphase I leads to the independent segregation of the different allele pairs. P p In other word the genes that we are testing are on different chromosomes.  Introduction to Genetics How to calculate the probability of multiple characters at once? Imagine that you want to know the probability of having a yellow wrinkle seed with a green pod colour. That is a lot of criteria to calculate using a Punnett square. So to find the solution you can use math using the law of probability!  Introduction to Genetics How to calculate the probability! Law 1: If it is certain that an event will occur, its probability is one (P = 1) or 100%. If it is certain that it will not happen, its probability is zero (P = 0). If it is possible for it to happen, its probability is necessarily between zero and one.  Introduction to Genetics How to calculate the probability! Law 1: If it is certain that an event will occur, its probability is one (P = 1) or 100%. If it is certain that it will not happen, its probability is zero (P = 0). If it is possible for it to happen, its probability is necessarily between zero and one. Law 2: If we define q as the probability of an event occurring, then (1 - q) is the probability (p) that this event does not occur (Colour of seed, q = yellow or p = not yellow).  Introduction to Genetics How to calculate the probability! Law 1: If it is certain that an event will occur, its probability is one (P = 1) or 100%. If it is certain that it will not happen, its probability is zero (P = 0). If it is possible for it to happen, its probability is necessarily between zero and one. Law 2: If we define q as the probability of an event occurring, then (1 - q) is the probability (p) that this event does not occur (Colour of seed, q = yellow or p = not yellow). Law 3: The addition of the probabilities of all possible outcomes of a dependent event is necessarily equal to 1, i.e. p + q = 1 (probability of yellow + probability of green = 1).  Introduction to Genetics How to calculate the probability! Law 1: If it is certain that an event will occur, its probability is one (P = 1) or 100%. If it is certain that it will not happen, its probability is zero (P = 0). If it is possible for it to happen, its probability is necessarily between zero and one. Law 2: If we define q as the probability of an event occurring, then (1 - q) is the probability (p) that this event does not occur (Colour of seed, q = yellow or p = not yellow). Law 3: The addition of the probabilities of all possible outcomes of a dependent event is necessarily equal to 1, i.e. p + q = 1 (probability of yellow + probability of green = 1). Law 4: The probability that mutually independent events (seed shape and flower colour) all occurrence is equal to the multiplication of their respective probabilities (P = P1 × P2 × P3...).  Introduction to Genetics Example of probability calculation Question: What is the probability to have face when flipping a coins?  Introduction to Genetics Example of probability calculation Question: What is the probability to have face when flipping a coins? Answer: Probability of ½, since the coins as only two sides, so you have 1 chance out of 2 to have face  Introduction to Genetics Example of probability calculation Question: Consider a six-sided die, what is the probability for any roll of the die?  Introduction to Genetics Example of probability calculation Question: Consider a six-sided die, what is the probability for any roll of the die? Answer: Probability of 1/6, since the die as 6 sides, so you have 1 chance out of 6 to have any number  Introduction to Genetics Example of probability calculation Question: Consider a six-sided die, what is the probability for any roll of the die? Answer: Probability of 1/6, since the die as 6 sides, so you have 1 chance out of 6 to have any number Question: If you have a die what is the probability to role a 2 or 6?  Introduction to Genetics Example of probability calculation Question: Consider a six-sided die, what is the probability for any roll of the die? Answer: Probability of 1/6, since the die as 6 sides, so you have 1 chance out of 6 to have any number Question: If you have a die what is the probability to role a 2 or 6? Answer: Addition of 1/6 and 1/6 = 2/6 or 1/3 Law 3: The addition of the probabilities of all possible outcomes of a dependent event is necessarily equal to 1, i.e. p + q = 1  Introduction to Genetics Example of probability calculation Question: If you have 2 dices what is the probability to role a 2 and 6?  Introduction to Genetics Example of probability calculation Question: If you have 2 dices what is the probability to role a 2 and 6? Answer: Multiplication of 1/6 and 1/6 Law 4: The probability that mutually independent events all occurrence is equal to the multiplication of their respective probabilities (P = P1 × P2 × P3...).  Introduction to Genetics Example of probability calculation Question: What is the probability of having the genotype Pp when crossing two heterozygote Pp together? + Pp Pp  Introduction to Genetics Example of probability calculation Question: What is the probability of having the genotype Pp when crossing two heterozygote Pp together? + Punnett squares Pp Pp P p ¼ P PP Pp ¼ p Pp pp Genotype  Introduction to Genetics Example of probability calculation Question: What is the probability of having the genotype Pp when crossing two heterozygote Pp together? + Punnett squares Pp Pp Answer: probability of obtaining Pp (¼), PLUS probability of obtaining pP (¼). P p ¼ P PP Pp ¼ + ¼ = ½. Law 3: ¼ p Pp pp The addition of the probabilities of all possible outcomes of a dependent event is necessarily equal to 1, i.e. p + q = 1 Genotype  Introduction to Genetics Example of probability calculation Question: What is the probability of having the phenotype purple when crossing two heterozygote Pp together? + Punnett squares Pp Pp What genotype give you purple: PP, Pp ? P p P PP Pp p Pp pp Genotype  Introduction to Genetics Example of probability calculation Question: What is the probability of having the phenotype purple when crossing two heterozygote Pp together? + Punnett squares Pp Pp What genotype give you purple: PP, Pp ? P p ¼ ¼ P PP Pp ¼ p Pp pp Genotype  Introduction to Genetics Example of probability calculation Question: What is the probability of having the phenotype purple when crossing two heterozygote Pp together? + Punnett squares Pp Pp What genotype give you purple: PP, Pp ? P p Answer: probability of obtaining PP (¼), PLUS ¼ ¼ probability of obtaining Pp (¼) PLUS P PP Pp probability of obtaining pP (¼) ¼ p Pp pp ¼ + ¼ + ¼ = 3/4. Genotype  Introduction to Genetics Example of probability calculation Question: Following self-pollination of a pea plant that is heterozygous for seed color and texture, knowing that yellow and round are dominant alleles, what is the probability that an individual of the offspring will have round and green seeds. Y R Remember: In Mendelian genetic all trait are independent and located on different chromosomes.  Introduction to Genetics Example of probability calculation Question: Following self-pollination of a pea plant that is heterozygous for seed color and texture, knowing that yellow and round are dominant alleles, what is the probability that an individual of the offspring will have round and green seeds. First what is the genotype of the parents:  Introduction to Genetics Example of probability calculation Question: Following self-pollination of a pea plant that is heterozygous for seed color and texture, knowing that yellow and round are dominant alleles, what is the probability that an individual of the offspring will have round and green seeds. First what is the genotype of the parents: YyRr  Introduction to Genetics Example of probability calculation Question: Following self-pollination of a pea plant that is heterozygous for seed color and texture, knowing that yellow and round are dominant alleles, what is the probability that an individual of the offspring will have round and green seeds. Second what are the possible genotype of a smooth and green seeds: (yyRr) + = (yyrR) (yyRR) (YyRr) (YyRr)  Introduction to Genetics Example of probability calculation Question: Following self-pollination of a pea plant that is heterozygous for seed color and texture, knowing that yellow and round are dominant alleles, what is the probability that an individual of the offspring will have round and green seeds. What are the possible of getting a round phenotype? RR or Rr or rR  Introduction to Genetics Example of probability calculation Question: Following self-pollination of a pea plant that is heterozygous for seed color and texture, knowing that yellow and round are dominant alleles, what is the probability that an individual of the offspring will have round and green seeds. Punnett squares What are the possible of getting a round phenotype? RR or Rr or rR R r ¼ ¼ R RR Rr ¾ round ¼ phenotype r rR rr Genotype  Introduction to Genetics Example of probability calculation Question: Following self-pollination of a pea plant that is heterozygous for seed color and texture, knowing that yellow and round are dominant alleles, what is the probability that an individual of the offspring will have round and green seeds. What are the possible of getting a green phenotype? yy  Introduction to Genetics Example of probability calculation Question: Following self-pollination of a pea plant that is heterozygous for seed color and texture, knowing that yellow and round are dominant alleles, what is the probability that an individual of the offspring will have round and green seeds. Punnett squares What are the possible of getting a green phenotype? yy Y y Y YY Yy ¼ green ¼ phenotype y yY yy Genotype  Introduction to Genetics Example of probability calculation Question: Following self-pollination of a pea plant that is heterozygous for seed color and texture, knowing that yellow and round are dominant alleles, what is the probability that an individual of the offspring will have round and green seeds. (yyRr) + = (yyrR) (yyRR) (YyRr) (YyRr) Answer: probability of obtaining round phenotype (3/4), probability of obtaining green phenotype (¼) Since these results are independent of each other (colour and shape are on different 3/4 x 1/4 = 3/16. chromosomes), we must multiply them to obtain the probability that the seeds are both smooth and green.