Pea plants were particularly well suited for use in Mendel s breeding experiments for all of the following reasons except that peas show easily observed variations in a number of characters such as pea shape and flower color ID: 594129
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Slide1
Mendel and the Gene IdeaSlide2
Pea plants were particularly well suited for use in Mendel
’s breeding experiments for all of the following reasons except that
peas show easily observed variations in a number of characters, such as pea shape and flower color. it is possible to control matings between different pea plants. it is possible to obtain large numbers of progeny from any given cross. peas have an unusually long generation time. many of the observable characters that vary in pea plants are controlled by single genes.Slide3
Pea plants were particularly well suited for use in Mendel
’s breeding experiments for all of the following reasons except that
peas show easily observed variations in a number of characters, such as pea shape and flower color. it is possible to control matings between different pea plants. it is possible to obtain large numbers of progeny from any given cross. peas have an unusually long generation time.
many of the observable characters that vary in pea plants are controlled by single genes.Slide4
A pea plant is heterozygous at the independent loci for flower color (white versus purple) and seed color (yellow versus green). What types of gametes can it produce?
two gamete types: white/white and purple/purple
two gamete types: white/yellow and purple/greenfour gamete types: white/yellow, white/green, purple/yellow, and purple/greenfour gamete types: white/purple, yellow/green, white/white, and purple/purpleone gamete type: white/purple/yellow/greenSlide5
A pea plant is heterozygous at the independent loci for flower color (white versus purple) and seed color (yellow versus green). What types of gametes can it produce?
two gamete types: white/white and purple/purple
two gamete types: white/yellow and purple/greenfour gamete types: white/yellow, white/green, purple/yellow, and purple/greenfour gamete types: white/purple, yellow/green, white/white, and purple/purpleone gamete type: white/purple/yellow/greenSlide6
A cross between homozygous purple-flowered and homozygous white-flowered pea plants results in offspring with purple flowers. This demonstrates
the blending model of genetics.
true breeding. dominance. a dihybrid cross. the mistakes made by Mendel.Slide7
A cross between homozygous purple-flowered and homozygous white-flowered pea plants results in offspring with purple flowers. This demonstrates
the blending model of genetics.
true breeding. dominance. a dihybrid cross. the mistakes made by Mendel.Slide8
Imagine a genetic counselor working with a couple who have just had a child who is suffering from
Tay-Sachs disease. Neither parent has Tay
-Sachs, nor does anyone in their families. Which of the following statements should this counselor make to this couple?“Because no one in either of your families has Tay-Sachs, you are not likely to have another baby with Tay-Sachs. You can safely have another child.”
“
Because you have had one child with
Tay
-Sachs, you must each carry the allele. Any child you have has a 50% chance of having the disease.
”
“
Because you have had one child with
Tay
-Sachs, you must each
carry the allele. Any child you have has a 25% chance of having the disease.
”
“
Because you have had one child with
Tay
-Sachs, you must both carry the allele. However, since the chance of having an affected child is 25%, you may safely have thee more children without worrying about having another child with
Tay
-Sachs.
”“You must both be tested to see who is a carrier of the Tay-Sachs allele.”Slide9
Imagine a genetic counselor working with a couple who have just had a child who is suffering from
Tay-Sachs disease. Neither parent has Tay
-Sachs, nor does anyone in their families. Which of the following statements should this counselor make to this couple?“Because no one in either of your families has Tay-Sachs, you are not likely to have another baby with Tay-Sachs. You can safely have another child.”
“
Because you have had one child with
Tay
-Sachs, you must each carry the allele. Any child you have has a 50% chance of having the disease.
”
“
Because you have had one child with
Tay
-Sachs, you must each carry the allele. Any child you have has a 25% chance of having the disease.
”
“
Because you have had one child with
Tay
-Sachs, you must both carry the allele. However, since the chance of having an affected child is 25%, you may safely have thee more children without worrying about having another child with
Tay
-Sachs.
”
“You must both be tested to see who is a carrier of the Tay-Sachs allele.”Slide10
Imagine a locus with four different alleles for fur color in an animal. The alleles are named
Da,
Db, Dc, and Dd. If you crossed two heterozygotes, Da
D
b
and
D
c
D
d
, what genotype proportions would you expect in the offspring?
25%
D
a
D
c
, 25%
D
a
D
d, 25% DbDc, 25% DbDd
50%
D
a
D
b
, 50% DcDd 25% DaDa, 25% DbDb, 25% DcDc, 25% DdDdDcDd50% DaDc, 50% DbDd25% DaDb, 25% DcDd, 25% DcDc, 25% DdDdSlide11
Imagine a locus with four different alleles for fur color in an animal. The alleles are named
Da,
Db, Dc, and Dd. If you crossed two heterozygotes, DaD
b
and
D
c
D
d
, what genotype proportions would you expect in the offspring?
25%
D
a
D
c
, 25%
D
a
D
d
, 25% DbDc, 25% DbDd
50%
D
a
D
b
, 50% DcDd 25% DaDa, 25% DbDb, 25% DcDc, 25% DdDdDcDd50% DaDc, 50% DbDd25% DaDb, 25% DcDd, 25% DcDc, 25% DdDdSlide12
John, age 47, has just been diagnosed with Huntington’s disease, which is caused by a dominant allele. His daughter, age 25, now has a 2-year-old son. No one else in the family has the disease. What is the probability that the daughter will contract the disease?
0%
25%50%75%100%Slide13
0%
25%50%
75%100%
John, age 47, has just been diagnosed with Huntington’s disease, which is caused by a dominant allele. His daughter, age 25, now has a 2-year-old son. No one else in the family has the disease. What is the probability that the daughter will contract the disease?Slide14
An individual with the genotype
AaBbEeHH is crossed with an individual who is aaBbEehh. What is the likelihood of having offspring with the genotype
AabbEEHh?1/81/161/321/64That genotype would be impossible.Slide15
An individual with the genotype
AaBbEeHH is crossed with an individual who is aaBbEehh. What is the likelihood of having offspring with the genotype
AabbEEHh?1/81/161/321/64That genotype would be impossible.Slide16
ABO blood type in humans exhibits
codominance and multiple alleles. What is the likelihood of a type A father and a type A mother having a type O child?
It is impossible.25% if both parents are heterozygous50% if both parent are heterozygous25% if only the father is heterozygous25% if only the mother is heterozygousSlide17
ABO blood type in humans exhibits
codominance and multiple alleles. What is the likelihood of a type A father and a type A mother having a type O child?
It is impossible.25% if both parents are heterozygous50% if both parent are heterozygous25% if only the father is heterozygous25% if only the mother is heterozygousSlide18
If roan cattle (incomplete dominance) are allowed to breed, what ratio of phenotypes is expected in the offspring?
1:1 red:white
all roan1:2:1 red:roan:white3:1 red:white1:1:1 red:roan:whiteSlide19
If roan cattle (incomplete dominance) are allowed to breed, what ratio of phenotypes is expected in the offspring?
1:1 red:white
all roan1:2:1 red:roan:white3:1 red:white1:1:1 red:roan:whiteSlide20
Imagine that the last step in a biochemical pathway to the red skin pigment of an apple is catalyzed by enzyme X, which changes compound C to compound D. If an effective enzyme is present, compound D is formed and the apple skin is red. However, if the enzyme is not effective, only compound C is present and the skin is yellow. What can you accurately say about a heterozygote with one allele for an effective enzyme X and one allele for an ineffective enzyme X?
The phenotype will probably be yellow
but cannot be red.The phenotype will probably be red but cannot be yellow.The phenotype will be a yellowish red.The phenotype will be either yellow or red.The phenotype will be either yellowish
red or red.Slide21
Imagine that the last step in a biochemical pathway to the red skin pigment of an apple is catalyzed by enzyme X, which changes compound C to compound D. If an effective enzyme is present, compound D is formed and the apple skin is red. However, if the enzyme is not effective, only compound C is present and the skin is yellow.
What can you accurately say about a heterozygote with one allele for an effective enzyme X and one allele for an ineffective enzyme X?
The phenotype will probably be yellow but cannot be red.The phenotype will probably be red but cannot be yellow.The phenotype will be a yellowish red.The phenotype will be either yellow or red.
The phenotype will be either yellowish
red or red.Slide22
Examine this genetic pedigree. What mode of inheritance does this trait follow?
autosomal dominantautosomal recessive
sex-linked recessivemitochondrialSlide23
Examine this genetic pedigree. What mode of inheritance does this trait follow?
autosomal dominantautosomal recessive
sex-linked recessivemitochondrialSlide24
Examine this genetic pedigree. What mode of inheritance does this trait follow?
autosomal dominantautosomal recessive
sex-linked recessivemitochondrialSlide25
Examine this genetic pedigree. What mode of inheritance does this trait follow?
autosomal dominantautosomal recessive
sex-linked recessivemitochondrialSlide26
The following offspring were observed from many crossings of the same pea plants. What genotypes were the parents?
465 purple axial flowers
152 purple terminal flowers 140 white axial flowers 53 white terminal flowers
PpAa
x
PpAA
PpAa
x
ppAA
PPAA
x
ppaa
PpAa
x
PpAa
PPaa
x
ppAASlide27
The following offspring were observed from many crossings of the same pea plants
. What genotypes were the parents?
465 purple axial flowers 152 purple terminal flowers 140 white axial flowers 53 white terminal flowers
PpAa
x
PpAA
PpAa
x
ppAA
PPAA
x
ppaa
PpAa
x
PpAa
PPaa
x
ppAA