#3. In garden pea plants, the yellow seed colour (Y) is dominant over the green see colour (y). What are the genotypes of heterozygous yellow seeds and homozygous recessive seeds repectively? a). Yy&YY. b).yY&yy.c).Yy &yy d).Yy& yY. I think that the answer they are looking for is A but i am not sure.
#4.In garden pea plants, the purple flower color (P) is dominant over the white color flower (p). What are the phenotypes of homozygous dominant and heterozygous flowers, respectively? Select: a). White and purple. b). Purple & qhite c). White & white. d). Purple, purple. I think it is b but im not sure
#5. In fruit flies, grey body color (G) is dominant over black body clour (g). A cross between a heterozygous grey fruit fly and a black fruit fly can be represented by a). Gg x Gg b). Gg x GG c). Gg x gg. d). GG x gg. I think it is D but i am not sure.
#6. In fruit flies, long wings (L) are dominant over short wings(l). LL,Ll and ll represent a). Three genotypes. b). three dominants. c). three phenotypes. d). three true breeding crosses. I think it is b.
#7. when a trait is only expressed in the presence of two identical alleles, the genotype is: a). Homozygous dominant. b). Homozygous recessive. c). Heterozygous recessive. d). Hetrozygous dominant. I think it is a but again not sure.
#8. In sheep a hairy coat (H) is dominant over a wooly fleece(h). The phenotype of a heterozygous sheep and a homozygous recesive sheep are repspectively, a). Hairy and hairy. b). hairy and wooly. c). HH and hh. d). Hh and hh.
#9. In garden pea plants, the yellow seed colour (Y) is dominant over the green seed color(y). A true breeding yellow-seeded plant was crossed with a green-seeded plant. All of the F1 generation plants had yellow seeds. Which of the following states is true for the F1 generation? a). None of the F1 generation plants will have the same allele for green seeds. b). Half of the F1 generation plants will have the same allele seeds for yellow seeds. c). All of the F1 generation plants will have the allele for green seeds. d). None of the F1 genertation plants will have allele for yellow seeds.
#10. In mice, the waltzing allele (w) that causes the mouse to run in circles due to an inner ear defect is recessive to the non- waltzing allele(W). If a heterozygous mouse mated wit ha walting mouse. If a heterozygous mouse mated with a waltzing mouse, a). 100% of the offspring would have a copy of the recessive allele. b). 75% of the offspring would have the dominant trait. c).100% of the offspring would have the same copy of the dominant allele. d). 75% of the offspring would have the recessive trait.
#11. Use the following info for question 12 as well. In mice, albinism (g) IS recessive to grey coat colour (G). A grey mouse mated with an albino mouse. Over the course of several months, 38 pups were born. Out of 38 pups, 17 had a grey coat and 21 were albino. Which of the following rows correcrly identifies the parental and F1 generation genotypes? a). Parental corss GgxGG F1 Genotypes GGxGg. b). Parental cross GGxgg F1 generation Gg. c).Ggxgg F1 generation GG, Gg and gg. d).Gg x gg F1 generation F1Gg and gg.
#12. Theoretically what is the excepted result from the parental cross of a grey mouse and albino mouse, and why does teh excpeted result not match the result stated in the information given above? a). Theoretically, there should be a 1:1 ratio of grey mice to albino mice, or grey coat mice and 19 albino mice. With more litters, the actual results would match the expected results more closely. b). Theoretically, 100% of the mice should have grey coats so all 38 offspring should have grey coats. There was an experimental error that can account for the deviation.c). Theoretically, there should be a 3:1 ratio of grey mice to albino mice, or 29 grey coat mice and 9 albino mice. With more litters, the actual results would match the expected results more closely.d). Theoretically, there should be a 3:1 ratio of grey mice to albino mice, or 28.5 grey coat mice and 9.5 albino mice. There was an experimental error that can account for the deviation.
#13. Which of the following statements describes the law of segregation? a). A Punnett square can be used to predict the outcome of a parental cross. b). the dominant allele is represented with an uppercase letter and the recessive allele is represented with alowercase letter.c). Dominant alleles are always expressed in a homozygous or heterozygous individual.d). A Bb individual will produce B and b alleles, while a bb individual will only produce a b allele.
Hi there, i need help with these questions. Please help me with this when you see my post. Please and thank you :)
6 years ago
Answered By Gregory S
#3 Y=dominant trait (yellow) and y= recessive trait. (green) with that, a dominant trait just needs one to be expressed while recessive needs both. Heterozygous means both traits are there (Yy) while home homozygous means only one is present. (YY or yy) for a recessive trait to be expressed it must be yy. So the answer to 3 is Heterozygous Yy and homozygous recessive yy
#4. The trick I use to remember the difference between genotype and phenotype is genotype= Garden and phenotype = playground. Garden is inside the fence (genotype is inside the cell) and playgrounds are in the open (phenotype is what you see) so with this one you have heterozygous and homozygous dominant phenotype. (PP and Pp) so because you only need one dominant trait to see that trait. Both flowers are purple. None white.
That's all the time I have. But if you have this many questions it might be time to consider hiring a tutor.