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QUESTION 3 A test cross is set up to establish the genotype of a plant with rounded peas. It is crossed with a plant that is
U THUMNA Would not be degraded properly, so she would see high levels throughout the nucleus and cytoplasm. QUESTION 9 Which
LIVE. WELL... WEITE, Ulu 1995 MUMU LIL U..W p s//m.cha.plus/... Chem 130 e-book CHEM 130 HW P sych 242 htts/ Question Complet
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Answer #1

1) the seed shape is considered here

let the alleles be R- round, r-wrinkled

round seed shape is dominant to the wrinkled seeds if the round seeded plant was homozygous dominant RR

RR * rr

Rr

all the progenies would have round seeds.

if the round seeded plant is Rr

Rr * rr

R r
r Rr ( round) rr ( wrinkled)
r Rr ( round) rr ( wrinkled)

the expected ratio of round; wrinkled=1:1

so the answer is B) heterozygous for the pea shape.

9) pleiotropic gene is the gene which affects multiple phenotypes,

the epistatic gene affects the phenotype produced by the other locus.

so the answer is B) A gene involved in multiple phenotypes like pigmentation, hearing and intestinal traits.

10) colorblindness is a sex-linked recessive trait. so a female is affected when both the X chromosomes have mutated alleles and the normal man has no mutated allele.

let the alleles be XC- normal vision, Xc- colorblind

the genotype of the woman is XcXc and genotype of the man is XCY

XCY * XcXc

Xc Xc
XC XCXc ( carrier, phenotypically normal female) XCXc( carrier, phenotypically normal female)
Y XcY ( colorblind son) XcY (colorblind son)

here all male children are colourblind so the answer is c) all male children will be colourblind.

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