Page 68 - Genetics_From_Genes_to_Genomes_6th_FULL_Part1

60 Chapter 3 Extensions to Mendel’s Laws

Figure 3.15 Dominant alleles of two genes needed for purple color in sweet peas. (a) White and purple sweet pea flowers. (b) The
9:7 ratio of purple to white F 2 plants indicates that at least one dominant allele of each gene is necessary for the development of purple color.
© William Allen/National Geographic Creative
(a) Lathyrus odoratus (sweet peas) (b) A dihybrid cross showing reciprocal recessive epistasis

P AA bb aa BB

Gametes A b a B

F 1 (all identical) Aa Bb Aa Bb

F 2 A B A b a B a b

A B AA BB AA Bb Aa BB Aa Bb 9 A– B– (purple)

A b AA Bb AA bb AaBb Aa bb (3) A– bb
7 (3) aa B– (white)
a B Aa BB Aa Bb aa BB aa Bb (1) aa bb

a b Aa Bb Aa bb aa Bb aa bb

Figure 3.16 A biochemical explanation for reciprocal
genotype hh for the second gene, do not make substance H at recessive epistasis in the generation of sweet pea color.
all. Thus, even if these people make an enzyme that would Enzymes specified by the dominant alleles of two genes are both
add A or B to this polysaccharide base, they have nothing to necessary to produce pigment. The recessive alleles of both genes
add it onto; as a result, Bombay-phenotype individuals appear specify no enzymes. In aa homozygotes, no intermediate precursor
to be type O. For this reason, homozygosity for the recessive 2 is generated, so even if enzyme B is available, it cannot produce
purple pigment.
h allele of the H-substance gene masks the effects of the ABO AA, Aa BB, Bb
gene, making the hh genotype epistatic to any combination of
A
B
I , I , and i alleles (except for ii.). Enzyme A Enzyme B
B
A
B
A person who carries I , I , or both I and I but is also Colorless Colorless Purple
A
pigment
precursor 2
precursor 1
an hh homozygote for the H-substance gene may appear to
A
be type O, but he or she will be able to pass along an I or bb
B
I allele in sperm or egg. An offspring receiving, let’s say, AA, Aa
an I allele for the ABO gene and a recessive h allele for the No enzyme B
A
H-substance gene from the father plus an i allele and a Colorless Enzyme A Colorless No purple
dominant H allele from the mother would have blood type precursor 1 precursor 2 pigment
A
A (genotype I i Hh), even though neither of the parents is
phenotype A or AB (Fig. 3.14b). aa
White sweet pea flowers In the first decade of the twenti- No enzyme A BB, Bb
eth century, William Bateson conducted a cross between No
two lines of pure-breeding white-flowered sweet peas Colorless colorless Enzyme B No purple
(Fig. 3.15a). Quite unexpectedly, all of the F 1 progeny were precursor 1 precursor 2 pigment
purple (Fig. 3.15b). Self-pollination of these novel hybrids
produced a ratio of 9 purple: 7 white in the F 2 generation. aa bb
The explanation? Two genes work in tandem to produce
purple sweet pea flowers, and a dominant allele of each No enzyme A No enzyme B
gene must be present to produce that color. Colorless No No purple
A simple biochemical hypothesis for these results is precursor 1 colorless pigment
shown in Fig. 3.16. Because it takes two enzymes catalyzing precursor 2