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BIOD27H3 Chapter Notes - Chapter 1: Codex Corbeiensis I, Fibronectin

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jademole404
9 Nov 2011
School
UTSC
Department
Biological Sciences
Course
BIOD27H3
Professor
Catherine Nash

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Related Questions

Cystic fibrosis is inherited as an autosomal recessive trait. Two phenotypically normal parents have a child with cystic fibrosis.
F f
F| FF Ff
f| Ff ff
If they have four additional children, what is the probability of 2 phenotypically normal children and 2 children with cystic fibrosis?

1. Genes F and Q are linked. In an individualof genotype Fq/fQ, what possible gamete genotypes can beproduced, assuming that meiosis occurs normally?

A. F, f, Q, q

B. Ff, Ff, Qq, Qq

C. FQ, fq, Fq, fQ

D. Ff, Qq, FQ, fq, Fq, fQ

E. Fq, fQ

2. In question #1, what possible gamete genotypes can beproduced if recombination never occurs?

A. F, f, Q, q

B. Ff, Ff, Qq, Qq

C. FQ, fq, Fq, fQ

D. Ff, Qq, FQ, fq, Fq, fQ

E. Fq, fQ

18. (Problem 28b) The human MN blood type is determined by two codominant alleles, LM and LN. The frequency of LM in Eskimos on a small Arctic island is 0.80. If the inbreeding coefficient for this population is 0.05, what are the expected frequencies of the M, MN, and N blood types on the island?
f(MM) = 0.64, f(MN) = 0.304, f(NN) = 0.04
f(MM) = 0.64, f(MN) = 0.32, f(NN) = 0.04
f(MM) = 0.648, f(MN) = 0.048, f(NN) = 0.304
f(MM) = 0.648, f(MN) = 0.304, f(NN) = 0.048

f(MM) = 0.048, f(MN) = 0.304, f(NN) = 0.648

20. (Problem 12, part 3) Briefly describe underdominance.
The heterozygote has lower fitness than either homozygote.
The heterozygote has greater fitness than either homozygote.
All alleles have equal fitness.
One allele has greater fitness than another.
The selection coefficients are all zero.