Lecture 10 Flashcards
Given the genotypic ratios that may appear among the progeny of simple crosses, give the genotypes of the parents that may give rise to each ratio:
a. 1:2:1
b. 1:1
c. Uniform progeny
a. 1:2:1
Aa x Aa
b. 1:1
Aa x aa
AA x Aa
c. Uniform
aa x aa
AA x aa
AA x AA
Given the phenotypic ratios that may appear among the progeny of simple crosses, give the genotypes of the parents that may give rise to each ratio:
a. 3:1
b. 1:1
c. Uniform progeny
a. 3:1
Aa x Aa
b. 1:1
Aa x aa
c. uniform aa x aa AA x AA AA x aa AA x Aa
What is the principle of independent assortment? How is it related to the principle of segregation?
According to the principle of independent assortment, genes for different characteristics that are at different loci segregate independently of one another. The principle of segregation indicates that the two alleles at a locus separate; the principle of independent assortment indicates that the separation of alleles at one locus is independent of the separation of alleles at other loci.
In which phases of mitosis and meiosis are the principles of segregation and independent assortment at work?
In anaphase I of meiosis, each pair of homologous chromosomes segregate independently of all other pairs of homologous chromosomes. The assortment is dependent on how the homologs line up during metaphase I. This assortment of homologs explains how genes located on different pairs of chromosomes will separate independently of one another during anaphase I. Anaphase II results in the separation of sister chromatids and subsequent production of gametes carrying single alleles for each gene locus as predicted by Mendel’s principle of segregation. Mendel’s principles of independent assortment and segregation do not apply to mitosis, which produces cells genetically identical to each other and to the parent cell.
How is the chi-square goodness-of-fit test used to analyze genetic crosses? What does the probability associated with a chi-square value indicate about the results of a cross?
The goodness-of-fit chi-square test is a statistical method used to evaluate the role of chance in causing deviations between the observed and the expected numbers of offspring produced in a genetic cross as predicted from the null hypothesis. The probability value obtained from the chi-square table refers to the probability that random chance produced the deviations of the observed numbers from the expected numbers.
In cucumbers, orange fruit color (R) is dominant over cream fruit color (r). A cucumber plant homozygous for orange fruit is crossed with a plant homozygous for cream fruit. The F1 are intercrossed to produce the F2.
a. Give the genotypes and phenotypes of the parents, the F1, and the F2.
b. Give the genotypes and phenotypes of the offspring of a backcross between the F1 and the orange fruited parent
c. Give the genotypes and phenotypes of a backcross between the F1 and the cream-fruited parent.
a. Solution:
The cross of a homozygous cucumber plant that produces orange fruit (RR) with a homozygous cucumber plant that produces cream fruit (rr) will result in an F1 generation heterozygous for the orange fruit phenotype.
Intercrossing the F1 will produce F2 that are expected to show a 3:1 orange-to- cream-fruit phenotypic ratio.
P0: RR x rr
F1.: Rr ( all orange fruit)
F2: 1RR:2Rr:1rr
(3:1 orange:cream)
b. Solution: The backcross of the F1 orange offspring (Rr) with homozygous orange parent (RR) will produce progeny that all have the orange fruit phenotype. However, one-half of the progeny will be expected to be homozygous for orange fruit and one-half of the progeny will be expected to be heterozygous for orange fruit.
Rr (F1) x RR (P0)
–> 1:1 RR:Rr (all orange)
c. Solution: The backcross of the F1 offspring (Rr) with the cream parent (rr) is also a testcross. The product of this testcross should produce progeny, one-half of which are heterozygous for orange fruit and one-half of which are homozygous for cream fruit.
Rr (F1) x rr(P0)
1:1 Rr:rr (1:1 orange:cream)
J. W. McKay crossed a stock melon plant that produced tan seeds with a plant that produced red seeds and obtained the following results (J. W. McKay. 1936. Journal of Heredity 27:110–112).
Cross: tan ♀ x red ♂
F1: 13 tan seeds
F2: 93 tan, 24 red seeds
a. Explain the inheritance of tan and red seeds in this plant.
b. Assign symbols for the alleles in this cross and give genotypes for all the individual plants.
a. The F1 generation contains all tan seed producing progeny and is the result of crossing a tan-seed-producing plant with a red-seed-producing plant. The F1 result suggests that the tan phenotype is dominant to red. In the F2 generation, the ratio of tan-seed-producing to red-seed-producing plants is about 3.9 to 1, which is similar but not identical to a 3 to 1 ratio expected for monohybrid cross involving dominant and recessive alleles. The F2 ratio suggests that the F1 parents are heterozygous dominant for tan color.
b. We will define the tan allele as “R” and the recessive red allele as “r.”
Tan-seed-producing ♀ parent: RR
Red-seed-producing ♂ parent: rr
F1 tan-seed-producing offspring: Rr
F2 tan-seed-producing offspring: RR or Rr
F2 red-seed-producing offspring: rr
In cats, blood-type A results from an allele (IA) that is dominant over an allele (iB) that produces blood-type B. There is no O blood type. The blood types of male and female cats that were mated and the blood types of their kittens follow. Give the most likely genotypes for the parents of each litter. Let “M” represent “male” and “F” represent “female”
a. Parents: M = A, F = B
Kittens: 4A, 3B
b. Parents: M = B, F = B
Kittens: 6B
c. Parents: M = B, F = A
Kittens: 8A
d. Parents: M = A, F = A
Kittens: 7A, 2B
e. Parents: M = A, F = A
Kittens: 10A
f. Parents: M = A, F = B
Kittens: 4A, 1B
a. male with blood type A × female with blood type B
Solution: Because the female parent has blood type B, she must have the genotype iBiB. The male parent could be either IAIA or IAiB. However, as some of the offspring are kittens with blood
type B, the male parent must have contributed an iB allele to these kittens. Therefore, the male must have the genotype of IAiB.
b. male with blood type B × female with blood type B
Solution: Because blood type B is caused by the recessive allele iB, both parents must be homozygous for the recessive allele or iBiB. Each contributes only the iB allele to the offspring.
c. male with blood type B × female with blood type A
Solution: Again, the male with type B blood must be iBiB. A female with type A blood could have either the IAIA or IAiB genotypes. Because all of her kittens have type A blood, this suggests that she is homozygous for the for IA allele (IAIA) and contributes only the IA allele to her offspring. It is possible that she is heterozygous for type A blood, but if so it is unlikely that chance alone would have produced eight kittens with blood type A.
d. male with blood type A × female with blood type A
Solution: Because kittens with blood type A and blood type B are found in the offspring, both parents must be heterozygous for blood type A, or IAiB. With both parents being heterozygous, the offspring would be expected to occur in a 3:1 ratio of blood type A to blood type B, which is close to the observed ratio.
e. male with blood type A × female with blood type A
Solution:Only kittens with blood type A are produced, which suggests that each parent is homozygous for blood type A (IAIA), or that one parent is homozygous for blood type A (IAIA), and the other parent is heterozygous for blood type A (IAiB). The data from the offspring will not allow us to determine the precise genotype of either parent.
f. male with blood type A × female with blood type B
Solution: On the basis of her phenotype, the female will be iBiB. In the offspring, one kitten with blood type B is produced. This kitten would require that both parents contribute an iB to produce its genotype. Therefore, the male parent’s genotype is IAiB. From this cross, the number of kittens with blood type B would be expected to be similar to the number of kittens with blood type A. However, due to the small number of offspring produced, random chance could have resulted in more kittens with blood type A than kittens with blood type B.
Joe has a white cat named Sam. When Joe crosses Sam with a black cat, he obtains one-half white kittens and one-half black kittens. When the black kittens are interbred, all the kittens that they produce are black. On the basis of these results, would you conclude that white or black coat color in cats is a recessive trait? Explain your reasoning.
The black coat color is likely recessive. When Sam was crossed with a black cat, one-half the offspring were white and one-half were black. This ratio potentially indicates that one of the parental cats is heterozygous dominant while the other parental cat is homozygous recessive—a testcross. The interbreeding of the black kittens produced only black kittens, indicating that the black kittens are likely to be homozygous, and thus the black coat color is the recessive trait. If the black allele was dominant, we would have expected the black kittens to be heterozygous, containing a black coat color allele and a white coat color allele. Under this condition, we would expect one-fourth of the progeny from the interbred black kittens to have white coats. Because this did not happen, we can conclude that the black coat color is recessive.
- Alkaptonuria is a metabolic disorder in which affected persons produce black urine. Alkaptonuria results from an allele (a) that is recessive to the allele for normal metabolism (A). Sally has normal metabolism, but her brother has alkaptonuria. Sally’s father has alkaptonuria, and her mother has normal metabolism.
a. Give the genotypes of Sally, her mother, her father, and her brother.
b. If Sally’s parents have another child, what is the probability that this child will have alkaptonuria?
c. If Sally marries a man with alkaptonuria, what is the probability that their first child will have alkaptonuria?
a. Give the genotypes of Sally, her mother, her father, and her brother.
Solution: Sally’s father, who has alkaptonuria, must be aa. Her brother, who also has alkaptonuria, must be aa as well. Because both parents must have contributed one a allele to her brother, Sally’s mother, who is phenotypically normal, must be heterozygous (Aa). Sally, who is normal, received the A allele from her mother but must have received an a allele from her father. The genotypes of the individuals are: Sally (Aa), Sally’s mother (Aa), Sally’s father (aa), and Sally’s brother (aa).
b. If Sally’s parents have another child, what is the probability that this child will have alkaptonuria?
Solution: Sally’s father (aa) × Sally’s mother (Aa) Sally’s mother has a one-half chance of contributing the a allele to her offspring. Sally’s father can contribute only the a allele. The probability of an offspring with genotype aa and alkaptonuria is therefore ½ × 1 = ½.
c. If Sally marries a man with alkaptonuria, what is the probability that their first child will have alkaptonuria?
Solution: Since Sally is heterozygous (Aa), she has a one-half chance of contributing the a allele. Her husband with alkaptonuria (aa) can only contribute the a allele. The probability of their first child (as well as for any additional child) having alkaptonuria (aa) is ½ × 1 = ½.
Hairlessness in American rat terriers is recessive to the presence of hair. Suppose that you have a rat terrier with hair. How can you determine whether this dog is homozygous or heterozygous for the hairy trait?
We will use h for the hairless allele and H for the dominant. Because H is dominant to h, a rat terrier with hair could be either homozygous (HH) or heterozygous (Hh). To determine which genotype is present in the rat terrier with hair, cross this dog with a hairless rat terrier (hh). If the terrier with hair is homozygous (HH), then no hairless offspring will be produced. However, if the terrier is heterozygous (Hh) then we would expect one-half of the offspring to be hairless.
What is the probability of rolling a six sided die and obtaining the following numbers?
a. 2
b. 1 or 2
c. an even number
d. any number but six
a. 2
Solution: 1/6
Because 2 is only found on one side of a six-sided die, then there is a 1/6 chance of rolling a 2.
b. 1 or 2
Solution: 2/6 = 1/3
The probability of rolling a 1 on a six-sided die is 1/6. Similarly, the probability of rolling a 2 on a six-sided die is 1/6. Because the question asks for the probability of rolling a 1 or a 2, and these are mutually exclusive events, we should use the additive rule of probability to determine the probability of rolling a 1 or a 2: (p of rolling a 1) + (p of rolling a 2) = p of rolling either a 1 or a 2 1/6 + 1/6 = 2/6 = 1/3 probability of rolling either a 1 or a 2
c. even number
Solution: 3/6 = 1/2
The probability of rolling an even number depends on the number of even numbers found on the die. A single die contains three even numbers (2, 4, 6). The probability of rolling any one of these three numbers on a six-sided die is 1/6. To determine the probability of rolling a 2, a 4, or a 6, we apply the additive rule: 1/6 + 1/6 + 1/6 = 3/6 = ½.
d. Any number but a six
Solution: 5/6
The number 6 is found only on one side of a six-sided die. The probability of rolling a 6 is therefore 1/6. The probability of rolling any number but 6 is (1 – 1/6) = 5/6.
What is the probability of rolling two six-sided dice and obtaining the following numbers?
a. 2 and 3
b. 6 and 6
c. At least one six
d. Two of the same number
e. An even number on each die
f. An even number on at least one die
a. 2 and 3
Solution: 2/36 = 1/18
There is a 1/6 chance of rolling a 2 on the first die and a 1/6 chance of rolling a 3 on the second die. The probability of this taking place is therefore 1/6 × 1/6 = 1/36. There is also a 1/6 chance of rolling a 3 on the first die and a 1/6 chance of rolling a 2 of the second die. Again, the probability of this taking place is 1/6 × 1/6 = 1/36. So the probability of rolling a 2 and a 3 would be 1/36 + 1/36 = 2/36 or 1/18.
b. 6 and 6
Solution: 1/36
There is only one way to roll two 6’s on a pair of dice: the first die must be a 6 and the second die must be a 6. The probability is 1/6 × 1/6 = 1/36.
c. at least one 6
Solution: 11/36
There are 3 ways in which to get at least one 6 in the roll of two dice. The first is to roll 6 on both dice, which we already determined has a probability of 1/36. The second way is to roll a 6 on the first die (1/6) and something other than a 6 on the second (5/6). When the multiplication rule is applied to this second possibility, the overall probability is 1/6 x 5/6 = 5/36. The third way would be to roll something other than a 6 in the first die (5/6) and a 6 on the second die (1/6) for an overall probability of 1/6 x 5/6 = 5/36. Using the addition rule to add the probabilities of these three different ways to achieve at least one 6, we arrive at the final answer of 1/36 + 5/36 + 5/36 = 11/36 chance.
d. two of the same number
Solution: 6/36 = 1/6
There are several ways to roll two of the same number. You could roll two 1s, two 2s, two 3s, two 4s, two 5s, or two 6s. Using the multiplication rule, the probability of rolling two 1s is 1/6 × 1/6 = 1/36. The same is true of two 2s, two 3s, two 4s, two 5s, and two 6s. Using the addition rule, the probability of rolling either two 1s, two 2s, two 3s, two 4s, two 5s, and two 6s is 1/36 + 1/36 +1/36 +1/36 +1/36 + 1/36 = 6/36 = 1/6.
e. An even number on each die
Solution: 9/36 = 1/4
Three out of the six sides of a die are even numbers, so there is a 3/6 probability of rolling an even number on each of the dice. The chance of having an even number on both dice is (3/6)(3/6) = 9/36, or ¼.
f. An even number on at least one die
Solution: 27/36 = 3/4
Three out of the six sides of a die are even numbers, so the probability of rolling an even number on the one die is 3/6. The probability of not rolling an even number is 3/6. An even number on at least one die could be obtained by rolling (a) an even on the first but not on the second die (3/6 × 3/6 = 9/36), (b) an even on the second die but not on the first (3/6 × 3/6 = 9/36), or (c) an even on both dice (3/6 × 3/6 = 9/36). Using the addition rule to obtain the probability of either a or b or c, we obtain 9/36 + 9/36 + 9/36 = 27/36 = ¾.
In watermelons, bitter fruit (B) is dominant over sweet fruit (b), and yellow spots (S) are dominant over no spots (s). The genes for these two characteristics assort independently. A homozygous plant that has bitter fruit and yellow spots is crossed with a homozygous plant that has sweet fruit and no spots. The F1 are intercrossed to produce the F2.
a. What will be the phenotypic ratios in the F2?
b. If an F1 plant is backcrossed with the bitter, yellow-spotted parent, what phenotypes and proportions are expected in the offspring?
c. If an F1 plant is backcrossed with the sweet, unspotted parent, what phenotypes and proportions are expected in the offspring?
a. What will be the phenotypic ratios in the F2?
Solution: P: homozygous bitter fruit, yellow spots (BB SS) × homozygous sweet fruit and no spots (bb ss) F1: All progeny have bitter fruit and yellow spots (Bb Ss).
The F1 are intercrossed to produce the F2: Bb Ss × Bb Ss. The F2 phenotypic ratios are as follows: 9/16 bitter fruit and yellow spots 3/16 bitter fruit and no spots 3/16 sweet fruit and yellow spots 1/16 sweet fruit and no spots
b. If an F1 plant is backcrossed with the bitter, yellow-spotted parent, what phenotypes and proportions are expected in the offspring?
Solution: The backcross of an F1 plant (Bb Ss) with the bitter, yellow-spotted parent (BB SS) will produce all bitter, yellow-spotted offspring.
c. If an F1 plant is backcrossed with the sweet, unspotted parent, what phenotypes and proportions are expected in the offspring?
Solution: The backcross of a F1 plant (Bb Ss) with the sweet, unspotted parent (bb ss) will produce the following phenotypic proportions in the offspring: ¼ bitter fruit and yellow spots ¼ bitter fruit and no spots ¼ sweet fruit and yellow spots ¼ sweet fruit and no spots
In cats, curled ears result from an allele (Cu) that is dominant over an allele (cu) for normal ears. Black color results from an independently assorting allele (G) that is dominant over an allele for gray (g). A gray cat homozygous for curled ears is mated with a homozygous black cat with normal ears. All the F1 cats are black and have curled ears.
a. If two of the F1 cats mate, what phenotypes and proportions are expected in the F2?
b. An F1 cat mates with a stray cat that is gray and possesses normal ears. What phenotypes and proportions of progeny are expected from this cross?
a. If two of the F1 cats mate, what phenotypes and proportions are expected in the F2?
Solution: If F1 cats mated, GgCucu × GgCucu, then the following proportions and phenotypes are expected in the F2:
9/16 black with curly ears
3/16 black with normal ears
3/16 gray with curly ears
1/16 gray with normal ears
b. An F1 cat mates with a stray cat that is gray and possesses normal ears. What phenotypes and proportions of progeny are expected from this cross?
Solution: The mating of an F1 cat (GgCucu) with a gray cat with normal ears (ggcucu) is a testcross in which we would expect to produce equal numbers of all the different progeny classes:
¼ black with curly ears
¼ black with normal ears
¼ gray with curly ears
¼ gray with normal ears
The following two genotypes are crossed:
Aa Bb Cc dd Ee x Aa bb Cc Dd Ee
What will the proportion of the following genotypes be among the progeny of the cross?
a. Aa Bb Cc Dd Ee
b. Aa bb Cc dd ee
c. aa bb cc dd ee
d. AA BB CC DD EE
a. Aa Bb Cc Dd Ee Solution: The simplest procedure for determining the proportion of a particular genotype in the offspring is to break the cross down into simple crosses and consider the proportion of the offspring for each cross. AaBbCcddEe × AabbCcDdEe Locus 1: Aa × Aa = ¼ AA, ½ Aa, ¼ aa Locus 2: Bb × bb = ½ Bb, ½ bb Locus 3: Cc × Cc = ¼ CC, ½ Cc, ¼ cc Locus 4: dd × Dd = ½ Dd, ½ dd Locus 5: Ee × Ee = ¼ EE, ½ Ee, ¼ ee ½ (Aa) × ½ (Bb) × ½ (Cc) × ½ (Dd) × ½ (Ee) = 1/32
b. Aa bb Cc dd ee
Solution: ½ (Aa) × ½ (bb) × ½ (Cc) × ½ (dd) × ¼ (ee) = 1/64
c. aa bb cc dd ee
Solution: ¼ (aa) × ½ (bb) × ¼ (cc) × ½ (dd) × ¼ (ee) = 1/256
d. AA BB CC DD EE
Solution: This will not occur. The AaBbCcddEe parent cannot contribute a D allele, and the AabbCcDdEe parent cannot contribute a B allele. Therefore, their offspring cannot be homozygous for the BB and DD gene loci.
