BIOL 310 1st Edition Lecture 3 Outline of Previous Lecture- MendelismI. Mendel’s ExplanationsII. The Law of SegregationIII. Mendel’s Next QuestionsIV. Mendel’s Conclusions A. the Law of Independent AssortmentB. The Central DogmasOutline of Current Lecture:I. A re-examination of Mendel’s Observations and ConclusionsII. Alternative method for Dihybrid CrossesIII. Mixed CrossesIV. 4 trait mixed crossi. A re-examination of Mendel’s Observations and ConclusionsWhat type of gametes do Rr Yy Individuals make? And in what ratios?- Traditionally, to answer this question you would make a HUGE punnent square that looksa little like this… RY Ry rY ryRY RRYY RRYy RrYY RrYyRy RRYy RRyy RryY RryyrY rRYY rRYy rrYY rrYyry rRyY rRyy rryY rryyo NOTE: the mothers alleles are written first in the boxes, this does not matter for small crosses but some characteristics are genetically imprinted to a mom or dads alleleThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.- Now, while this method DOES work, it is not the most efficient way. Dr. E said that finishing the exam is virtually impossible using this method and can get quite lengthy when give 3+ traits: 4^3 =64 so a cross with 3 traits will have 64 squares.ii. Alternative Method for Dihybrid crosses- Here are some important short hand that Dr. E will be using during his lectures and on tests…o RR and Rr have the same phenotype and are denoted as R_o rr has a distinct phenotype and is denoted as such rr.- Cannot be r_ because the dash indicates that any allele will work here which is not the case. If you substitute R for _ it will change the phenotype to a dominant phenotype.- These short hands may be used for any alleles, Y, T, Z, X, S, etc.- To tackle the cross:o Work with one allele cross at a time.  Try this one…. Rr Yy Tt × Rr Yy Tt - Rr=round/wrinkled Yy=yellow/green Tt=tall/short (this could be self pollination or simply a trihybrid cross.)R_ = 3/4 Y_ = 3/4T_ = 3/4rr= 1/4 yy = 1/4 tt = 1/4- What proportion of these are Round, Yellow, and Tall?o R_= 3/4o Y_ = 3/4o T_ = 3/4o Round, Yellow and Tall = 27/64 because 3×3×3 = 27 and 4×4×4= 64- What about Round, Green, and Short?o R_ = 3/4o yy = 1/4o tt = 1/4o Round, green and short = 3/64 because 3×1×1 = 3R rR RR Rrr rR rrY yY YY Yyy yY yyT tT TT Ttt tT ttiii. Mixed Crosses- Now what happens when you take a mixed cross? it is the same steps just with different ratios!- Try Rr Yy Tt × rr Yy Tt **because dad can ONLY make rr in the first box you can either do a 4 box square or reduce it to 2R_ = 1/2Y_ = 3/4T_ = 3/4rr = 1/2 yy = 1/4 tt = 1/4- So, What proportion are heterozygous (het) in the R, Y, and T boxes?o Rr = 1/2o Yy = 1/2o Tt = 1/2o So, heterozygous = 1/8 - What propotion are homozygous for all traits?o rr & RR = 1/2o YY & yy = 1/2o TT & tt = 1/2o So, homozygous = 1/8 remember, you can have 2 kinds of homozygous, homozygous dominant and homozygous recessive where as you can only have 1 kind of heterozygousiv. 4 Trait Mixed Cross- Lets try Rr Yy Tt Ss PP × Rr yy Tt SS pp- **In Genetics, we assume that 50% of the offspring will be male and50% will be female.**Y yY YY Yyy yY yyT tT TT Ttt tT ttrR Rrr rrR r R RR Rrr rR rryY Yyy yyS S SSs sSRr = round/ wrinkledYy= yellow/ greenTt= tall/ shortSs= smooth/ indentedPp=purple/ yellowWith this in mind….. - What Proportion will be….o Round, Green, Tall, Indented, Purple 3/4 × 1/2 × 3/4 × 0 × 1 = 0 because there will be no indented peas, even though there are indented alleles, indented is recessive to smooth; same with purple and yellow, there will only be purple plants so you have 1 representing all offspring or 4/4o Female, Round, Green, Tall, Indented, Purple1/2 × 3/4 × 1/2 × 3/4 × 0 × 1 = 0o Proportion of female progeny that are Round, Green, Tall, Indented, Purple This one doesn’t even have to be worked because all of the offspring are considered the female progeny and the ratios were worked above in #1o Dominant phenotype at all loci: R_ =3/4 Y_ = 1/2 T_ = 3/4 S_ = 1 P_ = 1 so, 9/32 offspring are dominant phenotype at all loci because..- 3/4 × 1/2 × 3/4 × 1 × 1=9/32o Heterozygous at each locus: Rr = 1/2 Yy= 1/2T t T TT Ttt tT ttpP Pp Tt= 1/2 Ss= 1/2 Pp =1  So, 1/16 is heterozygous at each locuso Revessive phenotype at all loci: 0 because there is no recessive phenotypes in the purple P box therefore there cannot be a recessive at all