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genetics chapter 4 homework
Terms in this set (33)
he diagram below shows Morgan's reciprocal cross: He first crossed a homozygous red-eyed female with a white-eyed male, and then crossed a homozygous white-eyed female with a red-eyed male. outcome? reciprocal crosses
When Morgan crossed a homozygous red-eyed female with a white-eyed male, all of the offspring had red eyes. However, in the reciprocal cross (homozygous white-eyed female with a red-eyed male), all of the females had red eyes while all of the males had white eyes.
In one of Morgan's experiments, he crossed his newly discovered white-eyed male with a red-eyed female. (Note that all of the females at that time were homozygous for red eyes because the allele for white eyes had not yet propagated through Morgan's flies.) All of the F1 flies produced by this cross (both males and females) had red eyes.
Next, Morgan crossed the red-eyed F1 males with the red-eyed F1 females to produce an F2 generation. The Punnett square below shows Morgan's cross of the F1 males with the F1 females. outcomes for F2 generation ?
When a homozygous red-eyed female was crossed with the white-eyed male (w+w+ × wY), the resulting F1 females were w+w and the F1 males were w+ Y. Crossing the F1 males and F1 females would yield these results:
All the F2 females would have red eyes, although some would be homozygous (w+w+ ) and others would be heterozygous (w+w).
Half the F2 males would have red eyes (w+ Y), and half would have white eyes (wY).
White eye color is an X-linked trait in one line of fruit flies. White eyes is recessive to red eyes. If a red-eyed female and a white-eyed male are crossed,
some of their male progeny may have white eyes
-If the female is heterozygous, approximately half of the male progeny will have white eyes.
You now know that inheritance of eye color in fruit flies is sex-linked: The gene encoding eye color is located on the X chromosome, and there is no corresponding gene on the Y chromosome.
How would the inheritance pattern differ if the gene for eye color were instead located on an autosome (a non-sex chromosome)? Recall that for autosomes, both chromosomes of a homologous pair carry the same genes in the same locations.
Suppose that a geneticist crossed a large number of white-eyed females with red-eyed males.
Consider two separate cases:
Case 1: Eye color exhibits sex-linked inheritance.
Case 2: Eye color exhibits autosomal (non-sex-linked) inheritance. (Note: In this case, assume that the red-eyed males are homozygous.)
Eye color exhibits sex-linked inheritance
case 1 : f there were 100 female offspring, blank would have red eyes and blank would have white eyes.
case 1 2.0: If there were 100 male offspring, blank would have red eyes and blank would have white eyes.
Case 2: Eye color exhibits autosomal (non-sex-linked) inheritance
1.0: If there were 100 female offspring, blank would have red eyes and blank would have white eyes.
2.0: If there were 100 male offspring, blank would have red eyes and blank would have white eyes.
For X-linked traits in Drosophila, the male phenotype is determined by the maternally inherited allele.
Males inherit only one X chromosome. That chromosome is contributed by the female parent.
Eye color in Drosophila is an X-linked trait. White eyes is recessive to red eyes. If a Drosophila male has white eyes, which of the following must also be true?
His mother had at least one white allele.
- Because this male had white eyes, he must have inherited a white allele from his mother.
Studies of large human populations have determined that the penetrance of a particular PRSS1 mutation, Arg122His, is 86%.
What does this means for individuals with the predisposing genotype?
They have an 86% chance of showing the associated phenotype.
-Penetrance is the frequency with which individuals of a given genotype manifest at least some degree of a specific mutant phenotype associated with a trait. Penetrance does not refer to the range of expression of the mutant genotype -- that is expressivity. Therefore, if a mutation is 86% penetrant, then 86% of individuals that possess the predisposing genotype would be expected to express the associated mutant phenotype to some degree.
Imagine that you are part of a research team interested in determining the penetrance of a different PRSS1 mutation, Ala16Val. Your team has identified 120 individuals with the PRSS1 Ala16Val allele and found that 48 of the individuals did not display any evidence of pancreatitis.
From your data, what is the penetrance of the PRSS1 Ala16Val mutation?
Express your answer as a whole number.
Recall that penetrance is the frequency with which individuals of a given genotype manifest at least some degree of a specific mutant phenotype associated with a trait. Of 120 individuals examined, 48 did not show any signs of pancreatitis, indicating that 120-48=72 individuals displayed a phenotype associated with pancreatitis. To determine the penetrance rate, simply divide the number of affected individuals by the total number studied: 72/120= 60%
If a man who is heterozygous for a PRSS1 Ala16Val mutation mates with a woman who is homozygous for normal cationic trypsinogen, what is the probability that their offspring will have hereditary pancreatitis, given the penetrance value obtained in Part B?
30% To solve this problem, you must take into account the probability that the offspring will inherit the dominant allele as well as the probability that the offspring will display the associated phenotype, as the condition has reduced penetrance.
The probability that offspring will inherit the dominant mutation (Aa x aa) is 50%. The probability of expressing the associated phenotype is 60%. Therefore, using the product law, the probability that offspring will have the disorder is 50% x 60% = 30%.
A woman with type A blood (whose father was type O) has children with a man that has type O blood. Both individuals are heterozygous for the MN antigen. Recall that MN blood group antigens are independent of the ABO locus, and that the alleles are codominant.
consider the flowing arrangement :
type A with M antigen:
type A with M and N antigens:
type A with N antigens:
type O with M antigen:
type O with M and N antigens:
type O with N antigen:
Expression of the A and B antigens on red blood cells is influenced by the FUT1 gene. This gene encodes fucosyl transferase, an enzyme that helps produce the H substance. Which of the following statements about the H substance is incorrect?
Individuals that are heterozygous for the FUT1 gene cannot produce the H substance.
-Individuals that are homozygous recessive for the FUT1 gene will appear to be type O because they are unable to produce the H substance, which serves as a substrate for the enzymes produced by the IA and IB genes. These enzymes add the appropriate terminal sugar to the H substance producing the A and B antigens, respectively. If the H substance is not produced, the A and B antigens are not able to be created.
all are true
-Individuals that fail to produce the H substance are said to have the Bombay phenotype.
-Individuals that cannot produce the H substance appear to be type O even if they have functional IA and/or IB alleles.
-The H substance is a substrate for the enzymes produced by the IA and IB genes. These enzymes add the appropriate terminal sugar to the H substance producing the A and B antigens, respectively.
How would the results from Part A change if both parents are also heterozygous for the FUT1 gene controlling the synthesis of the H substance (Hh)?
type A with M antigen:
type A with M and N antigens:
type A with N antigen:
type O with M antigen:
type O with M and N antigens:
type O with N antigen:
Epistasis is the interaction between genes such that one gene influences or interferes with the expression of another gene, leading to a specific phenotype.
Epistatic genes can be dominant or recessive.
In freshwater snails, pigment color is influenced by two genes. If two heterozygous pigmented freshwater snails were crossed and offspring were produced in a ratio of 9 pigmented snails to 7 albino snails, what are the genotypes of the offspring?
Genotype(s) of Pigmented Snails: A_B_
Genotype(s) of Albino Snails:
aabb, aaB_, A_bb
The 9:7 ratio obtained is a modification of the 9:3:3:1 ratio observed in a dihybrid cross of heterozygous individuals. In this case, pigmented snails are produced only when dominant alleles at both loci are present (A_B_) and albino snails result from any of the other possible genotypes (aaB_, A_bb, aabb).
In recessive epistasis,
two recessive alleles mask expression of an allele at a different locus. The 9:7 ratio observed in this example is actually a special case of epistasis called duplicative recessive epistasis, meaning the presence of either aa or bb was sufficient to mask expression of the other gene. In this case, if snails had two copies of either allele a or allele b, pigment production was suppressed, resulting in albino individuals.
In sheep, coat color is influenced by two genes. Gene A influences pigment production, while gene B produces black or brown pigment.
If two heterozygous white sheep were crossed and offspring were produced in a ratio of 12 white sheep to 3 black sheep to 1 brown sheep, what are the genotypes of the offspring?
white sheep: A_bb, A_B_
brown sheep: aabb
The 12:3:1 ratio obtained is a modification of the 9:3:3:1 ratio observed in a dihybrid cross of heterozygous individuals. In this case, white sheep are produced 75% of the time, while sheep with pigmented coats are produced 25% of the time. The 3:1 ratio observed suggests that a dominant allele is responsible for suppressing pigment color. As gene A is responsible for inhibiting pigment production, sheep with an A_ genotype will be white (A_B_ and A_bb).
As for the remaining sheep, 75% are black and 25% are brown, suggesting that black is dominant over brown. Therefore, black sheep can be assigned the aaB_ genotype while brown sheep can be assigned the aabb genotype.
This is an example of dominant epistasis, where one copy of an allele, in this case A, can mask the expression of another gene. Any sheep possessing the A allele (A_B_ and A_bb) had a white coat regardless of the gene present for pigment at the B locus.
In laborador retrievers, pigment color is influenced by two genes. Gene A determines the type of pigment produced and gene B affects whether the pigment gets deposited in the hair shaft.
If two heterozygous black laborator retrievers were crossed and offspring were produced in a ratio of 9 black dogs to 3 brown dogs to 4 yellow dogs, what are the genotypes of the offspring?
black ; A_B_
yellow ; A_bb, aabb
The 9:3:4 ratio obtained is a modification of the 9:3:3:1 ratio observed in a dihybrid cross of heterozygous individuals. In this case, dogs with black or brown coats are produced only when a dominant allele is present at the B loci and yellow dogs result from the presence of recessive alleles, bb.
Of the pigmented dogs, black dogs are produced 75% of the time, while dogs with brown coats are produced 25% of the time. This 3:1 ratio suggests that the black allele is dominant to brown. Therefore, black dogs can be assigned the A_B_ genotype, brown dogs the aaB_ genotype, and yellow dogs the A_bb and aabb genotypes.
This is an example of recessive epistasis, where two recessive alleles mask expression of an allele at a different locus. In this case, the presence of recessive alleles at the B locus (bb) prevented pigment deposition into the hair shaft, causing dogs to be yellow regardless of the genes they carried for pigment color.
Determine the proportion of offspring phenotypes that would result when two merle dogs mate, if one dog is true-breeding for the long-coat trait and the other dog is true-breeding for the short-coat trait.
double merle with short coat:
double merle with long coat
merle with short coat:
merle with long coat:
solid with short coat
solid with long coat
Determine the proportion of offspring phenotypes that would result when two merle dogs mate, if both dogs are heterozygous (Ll) for the gene that regulates coat length.
double merle with short coat:
double merle with long coat:
merle with short coat:
merle with long coat:
solid with short coat:
solid with long coat:
Interestingly, double merles are subject to a variety of health problems, including hearing loss and vision deficiencies. Therefore, it is not recommended that breeders breed for the double merle phenotype.
If you were in the business of breeding dogs, which of the following crosses would you avoid in an effort to not produce double merle puppies?
merle X merle
If breeders selectively mate dogs in an effort to reduce production of double merle puppies, they should avoid mating two merle dogs together. The CMCS x CMCS cross would produce double merle offspring (CMCM ) 25% of the time. Breeders should also avoid mating two double merle dogs together (CMCM x CMCM ) because this combination would produce double merle puppies 100% of the time.
Your mentor asks you to determine if the mutants belong to the same complementation group. What is true about flies that belong to the same complementation group?
They all have some mutation in the same wing-development gene.
Each strain may have a different mutation, but the same gene is mutated in all strains in a complementation group.
All mutations found to be present in a single gene are in the same complementation group, and they will complement mutations in other complementation groups. When large numbers of mutations affecting the same trait are available and studied using complementation analysis, it is possible to predict the total number of genes involved in the determintion of a trait.
What would be the outcome of crossing two strains of wingless flies that belong to the same complementation group?
All offspring will be wingless.
A cross of mutants belonging to the same complementation group will produce only the mutant phenotype in progeny because both parents possess mutations in the same gene. As a result, all progeny will be homozygous for two mutant alleles and display the wingless phenotype because they do not possess a wild-type copy of the gene important for wing development.
The only way that offspring could display the wild-type phenotype and develop normal wings would be if the mutant parental flies belonged to different complementation groups and had mutations in separate genes that affected wing development. In that case, a cross of flies with mutations in two different genes would result in offspring heterozygous for both genes. Because they would have one wild-type allele at each locus, the offspring would develop normal wings.
How many different genes are contributing to the wingless phenotype in theses mutant fly strains?
the number of complication groups = the number of genes since two complication groups are being studied than 2 genes are present
The phenotype of the disease is seen in every generation.If an affected male mates with a normal female, then all daughters and no sons will be affected.If a normal male mates with a heterozygous female, then half of daughters and sons will be affected.The trait appears about equally prevalent in males and females
The phenotype of the disease may skip generations.If an affected male mates with a normal female, then neither daughters nor sons will show the disease phenotype.If a normal male mates with a heterozygous female, then half of sons will be affected and daughters will not show the disease phenotype.There is a greater prevalance of affected males.
compare x linked dominant and recessive
X-linked recessive conditions are expressed in females who are homozygous for the recessive allele and in hemizygous males who carry the recessive allele. Because of male hemizygosity, many more males than females display the phenotype of X-linked recessive traits. If an affected male mates with a normal female, none of the progeny will display the recessive phenotype. This is because all female offspring would be unaffected carriers of the condition, and males would receive the Y chromosome, not the affected X chromosome, from their father.
X-linked dominant conditions are expressed in individuals with a single copy of the dominant allele, and as a result, tend to be equally prevalent in males and females. This kind of inheritance has a few other distinctive characteristics. When heterozygous females are mated to wild-type males, the dominant allele is transmitted to half of their progeny of each sex. When hemizygous males with the dominant allele are mated to wild-type females, the dominant trait is trasmitted to all daughters and no sons, as fathers pass their X chromosome to female offspring and their Y chromosome to sons.
Although the vast majority of women who inherit a single copy of an X-linked recessive disease allele are not affected by the disorder, a small percentage of women do display the disease phenotype. Which of the following could explain this rare phenomenon?
X-inactivation that favors inactivation of the normal X chromosome
Turner syndrome (a sex-chromosome aneuploidy in which women have only one X chromosome)
While not common, it is possible for females who inherit one copy of an X-linked recessive disease allele to be affected by the disease. This can be explained by an imbalance in X-inactivation, the random cessation of transcriptional activity of maternally or paternally derived X chromosomes. If most of the normal X chromosomes are inactivated, and the majority of X chromosomes containing the disease allele are active, a woman who carries an X-linked recessive disease allele can display the disease phenotype.
Another explanation is that the affected female may have Turner syndrome and have only one copy of the X chromosome. If this X chromosome contains recessive disease alleles, then the woman would be affected, as she would not have normal copies of those genes to compensate.
green x green = all green
aurea x aurea = 2/3 aurea, 1/3 green
green x aurea = 1/2 green, 1/2 aurea
Notice that the results from the green x green cross yielded 100% green progeny. This implies that the plants must be true-breeding, and that parents would be homozygous for this trait. The G allele was used to designate green, so green plants would be GG.
The aurea x aurea cross did not yield 100% aurea progeny, implying that plants with the aurea phenotype are not true-breeding, and thus are not GAGA . The aurea x aurea cross produced both green and aurea progeny, implying that heterozygous plants, GGA , will have the aurea phenotype.
One would expect that a heterozygous cross would produce progeny in a 1 GG: 2 GGA :1 GAGA ratio, but instead the ratio obtained was 2 GGA : 1 GG. This suggests that one of the alleles may be a lethal allele. Lethal alleles are often detected as distortions in Mendelian ratios, where one or more classes of expected offspring are missing.
How can the unusual results from the crosses be explained?
The aurea allele behaves as a recessive lethal allele.
Lethal alleles are usually detected by a distortion in segregation ratios. When aurea plants were crossed, GGA x GGA , the expected outcome was 1 GG: 2 GGA :1 GAGA . Instead, a 2 GGA (aurea): 1 GG (green) ratio was obtained. No GAGA offspring were viable, indicating the GA allele is a lethal allele.
Because having one GA allele was not sufficient to cause lethality (the GGA genotype leads to viable aurea plants), and because offspring that were homozygous for the lethal allele were not viable, GA behaves as a recessive lethal allele.
With respect to snapdragon color, what can be said of the G and GA alleles?
The GA is dominant to the G allele.
While the GA allele had a recessive effect with respect to survival, it is actually dominant with respect to snapdragon leaf color. This can be seen in the heterozygotes, where one copy of the GA allele is sufficient to give plant leaves a yellow color. Therefore, it is important to keep in mind that the kind of dominance an allele exhibits depends on the phenotype that is examined.
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