EXERCISE 8 DIHYBRID CROSS ar, you have worked on Punnett squares that show monoh

Question

EXERCISE 8 DIHYBRID CROSS ar, you have worked on Punnett squares that show monohybrid crosses-probabilities for an offspring from parents involving a single trait, such as tongue e his research, Mendel devised something rolling When conductin es in peadihybrid cross, which allowed him to consider two different traits. Mendel's work on dihybrid croa bea plants helped him realize that traits are inherited separately (the law of independent assortment) Prom ty ee ft lbut canaso fechaue rans Frion hybrid cross with Mendel 's pea plants Consider two traits in pea plants: tallness (T) and flower exor Cise 7 SCENARIO A dont- Code ontlodiesS allness (7) is dominant over shortness (t), and purple flowers (P) are dominant over white flowers (p). If a plant erozygous for tallness (Tt) and homozygous recessive for flower color (pp) is mated with a plant that is homo- cose is an example of a dihybrid cross with Mender Pe ifs Conasde to owet color (P). dominant for tallness (TT) and for flower color (PP], what are the probabilities for the offspring's genotypes? The first step is to determine the parents' genotypes. Here, one plant is Ttpp and the other is TTPP STEP 2 Next, we have to determine all the possible gametes each plant might contribute. To do this, we write down the T and P combinations that may be present in each plant's gametes. » In plant 1, the possible gametes are Tp, Tp, tp, and tp. » In plant 2, the possible gametes are TP, TP, TP, and, TP STEP 3 Last, we mate these possible gametes in a large Punnett square: In this example, the results are as Plant 1: Ttpp Tp TTPp follows: » There is a 50% chance of an off- tp TIPP TtPp TtPp TtPp tp spring that is TTPp There is a 50% chance of an off- spring that is TtPp. On the phenotypic level, there is a 100% likelihood that the offspring will be tall and have purple flower » TTPp TP TP TP TP » TIPP TTPp TTPp TIPP TTPp TTPp

Explanation / Answer

Suzy's genotype : FfWw

Jose's genotype : FFWW

Required Punett's square :

Fw FW fw fW

FW FFWw FFWW FfWw FfWW

FW FFWw FFWW FfWw FfWW

FW FFWw FFWW FfWw FfWW

FW FFWw FFWW FfWw FfWW

There is a 25% chance of a baby that is FFWw, 25% chance of a baby that is FFWW, 25% chance of a baby that is FfWw and 25% chance of a baby that is FfWW.

Answer :

1. Will have freckles(FF) but no widows peak (Ww) = 25 %

2. Not have freckles ( Ff ) but will have a widows peak (WW) = 25 %

3. Will have freckles (FF) and a widows peak (WW) = 25%

4. Will not have freckles (Ff) and widows peak (Ww) = 25%

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