AP Biology Unit 5: Heredity Notes Pt. 1

Mitosis vs meiosis

    Type down some differences between the meiosis and mitosis, that you talked about in your group: Mitosis creates two genetically identical daughter cells that are identical to the parent cell and are diploid. They make somatic cells or any cells that are genetically identical to the original parent cell. They only go through mitosis once. Meiosis creates four haploid cells that have genetic variation because there are steps like crossing over and the independent assortment that happens in meiosis 1. They are haploid cells that are only produced by sex organs for sexual reproduction. The haploid cells that are produced are gametes. They go through two steps, meiosis one and meiosis two. Meiosis one causes genetic variation but meiosis two is just like mitosis, with the lack of genetic material being duplicated in interphase.

1- Pea plants heterozygous for flower position and stem length (AaTt) are allowed to self-pollinate, and 400 of the resulting seeds are planted. Draw a punnett square for this cross. How many offspring would be predicted to have terminal flowers and be dwarf?

Gametes Male AaTt - AT,At,aT,at

Gametes Female AaTt - AT,At,aT,at

The possible gametes for each parent are AT,At,aT,at. Since terminal flower and dwarf are both recessive traits, the genotype would have to be aatt. Using the punnett square, we are able to find that 1 out of 16 offspring would be predicted to have terminal flowers and be dwarf. This means that 25 out of 400 offspring would be predicted to have terminal flowers and be dwarf.  

2- If P1 is PpYyRr X Ppyyrr. What fraction of offspring from this cross are predicted to exhibit the recessive phenotypes for at least two of the three characters?

Gametes Male PpYyRr: PYR,PYr,PyR,Pyr,pYR,pYr,pyR,pyr

Gametes Female Ppyyrr: Pyr,pyr

The offspring that are predicted to exhibit the recessive phenotypes would need two recessive alleles. Since two of the three characters have to be shown in this question, the possible combinations are PPyyrr, Ppyyrr, ppYyrr, ppyyRr, ppyyrr, Ppyyrr. The first punnett square shows that there is 6/16 chance which is 0.375 or ⅜.

Another way to find the probability of pp, when Pp and Pp are crossed, it is a ¼ chance. When Yy and yy are crossed, there is a 50% chance that the recessive trait is expressed. Then when Rr and rr are crossed, there is also a  50% chance that the recessive trait is expressed. The chance all 3 are recessive is 0.0625. Since we are only looking for two out of the three recessive traits to be exhibited, you only have to multiply 2 so 0.5 times 0.5 which is 0.25 and 0.5 times 0.25 which is 0.125. When added together, the total is 0.375 or ⅜. 

3- Three characters (flower color, seed color, and pod shape) are considered in a cross between two pea plants (PpYyIi x ppYyii). What fraction of offspring are predicted to be homozygous recessive for at least two of the three characters?

The possibility of recessive pp is 50%. The possibility of recessive yy is 25%. The possibility of recessive ii is 50%. The probability of having recessive pp and yy is 0.125.  The probability of having recessive pp and ii is 0.25. The sum of these probabilities are 0.375. 

4- complete the cross for the snapdragon flower that your teacher has started. Write down the genotypic and phenotypic ratios for the F2.

Since the offspring of the first breeding creates all heterozygous pink plants, the pink snapdragon is self pollinated, and creates the F2 which has RR as 25%, rr as 25% and Rr as 50%. The genotype ratio is RR:rr:Rr=1:1:2. The phenotypic ratio of f2 is red:white:pink=1:1:2. Two offspring are pink, one is red and one is white since it shows incomplete dominance.