This shows T dominant allele is responsible for tasting and t is responsible for not tasting.
So, we have to calculate the allelic frequency of T and t allele.
Total people = 25
People with tt phenotype = 7
So, tt genotype frequency = 7/25 = 0.28
For t allele frequency = ✓0.28 = 0.529
Individuals with TT that is homozygous taster = 5+1 = 6
Frequency of TT genotype = 6/25 = 0.24
Therefore, frequency of T allele that is taster allele is ✓0.24 = 0.48
Thus, frequency of taster allele is 0.48
Frequency of non taster allele is 0.52.
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Class Results Genotype Phenotype Strong Taster Weak Taster Non-taster TT 5 1 Tt 1 10 1...
Activity 1: What traits do you have? Trait Class Totals Freckles # freckles #no freckles (yes/no) No 교교 a Paper Taster? (es/no) No I4t It Sodium Benzoate Paper Taster? (yes/no) # taste 121 #nontaster Control Paper Taster? (ges/no) # taster #nontaster ト O 고구 #brown-50m。 Eye Color #blue-green forown Peak #widow #straight ig Hairline Widow's peak/Straight widowa Tongue Rolling ability Roller/non-roller #non-roller #Roller Non Kollen 2: Determine Phenotypic Proportions of Different Traits and e Corresponding Frequencies of the Determining Alleles...
Using the following information below, fill out Table 1 for you. PTC: taster (T-), non-taster (tt) Sodium benzoate: taster (B-), non-taster (bb) Hair type: Is your hair naturally wavy (CC), wavy (Cc), or straight (cc)? Hair color: Dark hair (DD or Dd) is dominant over light colored hair (dd) Widow's peak: the presence of a window's peak is dominant (WW or Ww) over absent (ww) Eyebrows: bushy (BB or Bb) dominant over fine (bb) Eye color: non-blue (EE or Ee)...
Question 5 (1 point) Describe this: Tt Question 5 options: A) homozygous phenotype B) heterozygous genotype C) homozygous genotype D) heterozygous phenotype Question 6 (1 point) Which of these is a prenatal genetic test? Question 6 options: A) in vitro fertilization B) nuclear transplantation C) chorionic villi sampling D) DNA profiling Question 7 (1 point) All the offspring of a white hen and a black rooster are gray. The simplest explanation for this type of inheritance is: Question 7 options:...
4. Genotype Results Were you successful in isolating genomic DNA and obtaining a PCR product? (Describe size and estimated amount.) If answering No to either, explain why! a. /) enimenon. TC alle )/.tet e t。 /1</i J What is your genotype as revealed by the Fnu4HI digest of your PCR product and if it matches your phenotype? b. c. Why might an individual's phenotype not match their genotype? DOPULATION GENETI ORY 7 AROSE GEL ELECTROPHORESis On the template below, indicate...
Use the survey information to help form a hypothesis of whether you are phenotypically a “taster” or “non-taster”. Look at your scores for orange juice, carrots, eggplant, green beans, sprouts, and potatoes crossing them off as you go down the list. – Do you dislike most of these? Perhaps you are a picky eater. The rest of the foods have been placed on the list based on reports that some tasters find the food bitter and/or due to the known...
hedrozygay heolvozyggay 1. Cross a homezygous tall (Tt) plant with a homoavgous dwarf plant (tt). Show the Punnett square in the space below. (5 pts) a. Which is the dominant allele? How do you know? (1 pt) b. What is the % of tall plants? What is the % of dwarf plants? (1 pt) 2. You are breeding Labrador retrievers. The female is Bbee and the male is heterozygous black BbEe. Show the results in a Punnet square. (10 pts)...
i wouldnt mind if someone could check my work on the first page. but whats confusing me is section B. it starts on the bottom of the first page and carries onto the second page. A. Estimation of allele frequency for a specific trait Fill in the tables below for taster and allele frequencies in your class section. Taster/Non-taster Frequency=Genotype frequency (1 pt) # of students #Tasters Frequency of # Non-tasters in Lab Section Tasters (p+2pg) 16 1 = 5125...
The occurrence of the NN blood group genotype in the US population is 1 in 400, consider NN as the homozygous recessive genotype in this population. You sample 1,000 individuals from a large population for the MN blood group, which can easily be measured since co-dominance is involved (i.e., you can detect the heterozygotes). They are typed accordingly: BLOOD TYPE GENOTYPE NUMBER OF INDIVIDUALS RESULTING FREQUENCY M MM 490 0.49 MN MN 420 0.42 N NN 90 0.09 Using the...
5. Examine the following fabricated data for a population of oak trees. Genotype Height Number of individuals with this genotype in generation 1 Number of individuals with this genotype in generation 2 A1A1 5 meters 100 200 A1A2 10 meters 100 A2A2 15 meters 100 200 TOTAL: 300 450 a. (2 pts) What is the allele frequency of A1 in the first generation? b. (2 pts) is this population in Hardy-Weinberg Equilibrium? Show your work.
Traits Symbol Dominant Phenotype od Dimples Your phenotype Your genotype $ 1. Facial dimples 2. Bent little finger 3. Eye Color Bent little finger Brown eyes 4. Free ear lobe E, e F.f Hih Free ear lobe 5. Mid-digital hair Presence of hair 6. Hand clasping Left on top 7. Widow's peak Widow's peak Ww Tt 8. Tongue Rolling Tongue roll се Cleft in chin 9. Chin cleft 10. Hitchhiker's Thumb 11. Handedness Hi, Straight (180" RT Right 12. Finger...