From genes to genomes solutions manual

Remember that 4 phenotypic classes in the progeny means that 2 genes control the phenotypes. Next, determine the phenotypic ratio for each gene separately.

A monohybrid ratio tells you which phenotype is dominant and that both parents were heterozygous for the trait; in contrast, a ratio results from a testcross where the dominant parent was heterozygous. There are 2 genes in this cross 4 phenotypes. The 1 spiny : 1 smooth ratio indicates a test cross for the pod shape gene. Because all progeny were purple, at least one parent plant must have been homozygous for the P allele of the flower color gene.

This is similar to part b , but here all the progeny were spiny so at least one parent must have been homozygous for the S allele. A 1 purple : 1 white monohybrid ratio denotes a test cross. There is a 3 purple : 1 white ratio among the progeny, so the parents were both heterozygous for the P gene. All progeny have smooth pods so the parents were both homozygous recessive ss. All progeny were white so the parents must have been homozygous recessive pp.

Three characters genes are analyzed in this cross. While we can usually tell which alleles are dominant from the phenotype of the heterozygote, we are not told the phenotype of the heterozygote that is, the original pea plant that was selfed. Instead, use the monohybrid phenotypic ratios to determine which allele is dominant and which is recessive for each gene. Consider height first. Finally, consider flower color. Thus, purple is dominant. You thus know that one gene determines the wing trait and one gene determines the eye trait, and you further know the dominance relationship between the alleles of each gene.

In cross 1, all of the parents and offspring show the tiny wing phenotype so there is no variability in the gene controlling this trait, and all flies in this cross are tt. Note that the eye phenotypes in the offspring are seen in a ratio of 3 oval : 1 narrow. This phenotypic monohybrid ratio means that both parents are heterozygous for the gene Nn.

In cross 2 consider the wing trait first. The female parent is tiny tt so this is a test cross for the wings. The offspring show both tiny and normal in a ratio of 82 : 85 or a ratio of 1 tiny : 1 normal. Therefore the normal male parent must be heterozygous for this gene Tt. For eyes the narrow parent is homozygous recessive nn so again this is a test cross for this gene.

Consider the wing phenotype in the offspring of cross 3. Both wing phenotypes are seen in a ratio of 64 normal flies : 21 tiny or a 3 normal : 1 tiny. Thus both parents are Tt heterozygotes. The male parent is narrow nn , so cross 3 is a test cross for eyes.

Both phenotypes are seen in the offspring in a 1 normal : 1 narrow ratio, so the female parent is heterozygous for this gene. When examining cross 4 you notice a monohybrid phenotypic ratio of 3 normal : 1 tiny for the wings in the offspring.

Thus both parents are heterozygous for this gene Tt. Because the male parent has narrow eyes nn , this cross is a test cross for eyes. All of the progeny have oval eyes, so the female parent must be homozygous dominant for this trait. Find the phenotypic monohybrid ratio separately for each gene in the offspring. Then multiply these monohybrid ratios to find the phenotypic dihybrid ratio. When you multiply each fraction by progeny you will see 75 normal oval : 75 normal narrow : 25 tiny oval : 25 tiny narrow.

The protein specified by the pea color gene is an enzyme called Sgr, which is required for the breakdown of the green pigment chlorophyll. See Fig. The y allele could be a null allele because it does not specify the production of any of the Sgr enzyme. The Y allele is dominant because in the heterozygote, the single Y allele will lead to the production of some Sgr enzyme, even if the y allele cannot specify any Sgr.

The amount of the Sgr enzyme made in heterozygotes is sufficient for yellow color. In yy peas, the green chlorophyll cannot be broken down, so this pigment stays in the peas, which remain green in color. If the amount of Sgr protein is proportional to the number of functional copies of the gene, then YY homozygotes should have twice the amount of Sgr protein as do Yy heterozygotes.

Yet both YY and Yy peas are yellow. Just as was seen in part e , for many genes including that for pea color , half the amount of the protein specified by the gene is sufficient for a normal phenotype. Thus, in most cases, even if the gene is essential, heterozygotes for null alleles will survive.

The advantage of having two copies of essential genes is then that even if one normal allele becomes mutated changed so that it becomes a null allele, the organism can survive because half the normal amount of gene product is usually sufficient for survival. Yes, a single pea pod could contain peas with different phenotypes because a pod is an ovary that contains several ovules eggs , and each pea represents a single fertilization event involving one egg and one sperm from one pollen grain.

If the female plant was Yy, or yy, then it is possible that some peas in the same pod would be yellow and others green. For example, fertilization of a y egg with Y pollen would yield a yellow pea, but if the pollen grain was y, the pea would be green. However, a pea pod could not contain peas with different phenotypes if the female plant was YY, because all the peas produced by this plant would be yellow.

Yes, it is possible that a pea pod could be different in color from a pea growing within it. One reason is that, as just seen in part g , a single pod can contain green and yellow peas. But a more fundamental reason is that one gene controls the phenotype of pea color, while a different gene controls the separate phenotype of pod color.

If the alleles of the pea color and pea shape genes inherited from a parent in the P generation always stayed together and never separated, then the gametes produced by the doubly heterozygous F1 individuals in Fig.

Note that only two possibilities would exist, and these would be in equal frequencies. On a Punnett square male gametes shaded in blue, female gametes in red :. These results make sense because if the alleles of the two genes were always inherited as a unit, you would expect the same ratios as in a monohybrid cross.

Similar to what you saw in Fig. The functional enzyme can synthesize the growth hormone gibberellin, so plants with the L allele are tall. Even half the normal amount of this enzyme is sufficient for the tall phenotype, explaining why Ll heterozygotes are tall. Note: Your copy of the text might be missing the key figure referred to in this Problem.

As in Problem 31 above, the dominant allele P most likely specifies a functional product in this case, the protein bHLH , while the recessive p allele cannot specify any functional protein. The fact that the hybrid is purple as shown on Fig. Alleles specifying functional enzymes would yield purple color, while those that could not produce functional enzymes would cause white color. It is likely that the alleles for purple would be dominant.

Section 2. Recessive - two unaffected individuals have an affected child aa. Therefore the parents involved in the consanguineous marriage must both be carriers Aa. Dominant - the trait is seen in each generation and every affected person A — has an affected parent.

Note that III-3 is unaffected aa even though both his parents are affected; this would not be possible for a recessive trait.

Recessive - two unaffected, carrier parents Aa have an affected child aa , as in part a. Cutis laxa must be a recessive trait because affected child II-4 has normal parents.

Because II-4 is affected she must have received a disease allele CL from both parents. The trait is thus recessive. Thus the probability that II-. In Chapter 21 you will find the definition of a term called the allele frequency; if the value of the allele frequency in the population under study is known, you can calculate the very low likelihood that II-2 is a carrier.

In order to have an affected child II-3 must also be a carrier. Diagram the cross! In humans this is usually done as a pedigree. Remember that the affected siblings must be CF CF. Both families have an affected sibling, so both sets of parents that is, all the people in generation I must have been carriers.

Of the remaining possible genotypes, 2 are heterozygous. There are also two cases in which an unrelated individual must have been a carrier II-4 and either I-1 or I-2 , so the disease allele appears to be common in the population. The inheritance pattern seen in Fig. However, the same pattern of inheritance could be seen if the disease were caused by a common recessive mutation. In the case of a common recessive mutation, all the affected individuals would be HD HD.

If the disease is due to a recessive allele, then III-6 and IV-6 must be homozygotes for this recessive allele, and all their children must have the disease.

Alternatively, you could look at the progeny of matings between unaffected individuals in the pedigree such as III-1 and an unaffected spouse.

If the disease is due to a recessive mutation, then many of these individuals would be carriers, and if the trait is common then at least some of the spouses would also be carriers, so such matings could give affected children.

Again assuming this trait is rare in the population, those people marrying into the family II-1 and II-4 are homozygous normal HH. The chance that III-2 will inherit h is exactly the same. This number is slightly higher than the answer to part a , which was 0. Both diseases are known to be rare, so normal people marrying into the pedigree are assumed to be homozygous normal. Nail-patella N syndrome is dominant because all affected children have an affected parent. Alkaptonuria a is recessive because the affected children are the result of a consanguineous mating between 2.

Because alkaptonuria is a rare disease, it makes sense to assume that III-3 and III-4 inherited the same a allele from a common ancestor. Thus, for each of the phenotypes below you must consider both possible genotypes for IV If both the possible IV-2 genotypes can produce the needed gametes, you will need to sum the two probabilities. For the child to have both syndromes N— aa , IV-2 would have to contribute an n a gamete. This could only occur if IV-2 were nn Aa.

There is no need to sum probabilities in this case because IV-2 cannot produce an n a gamete if his genotype is nn AA. This could also occur if IV-2 were nn AA. For the child to have just alkaptonuria nn aa , IV-2 would have to contribute an n a gamete.

You can make this calculation because there are only the four possible outcomes and you have already calculated the probabilities of three of them. This problem illustrates that much care in interpretation is required when the results of many matings in mixed populations are reported as opposed to the results of matings where individuals have defined genotypes. An equally likely possibility exists that any child produced by this couple will be affected A or unaffected U.

From the list just presented in part a , you can see that there are two possibilities in which only one child is affected: AU and UA. From the list just presented in part a , you can see that there is only one possibility in which no child is affected: UU.

We thus use the product rule to determine the chance that each of those 10 independent events will occur in a particular way — a particular birth order.

In a family of ten children, 10 different outcomes birth orders exist that satisfy the criterion that only 1 child has the disease.

Only the first child could have the disease, only the second child, only the third child, etc. One way to determine the probability that four children in a family of ten will have the disease is to write down all possible outcomes for the criterion, as we did above for the second answer in part d.

If you start to do this…… 1. In short — this is not a good way to find the answer! For questions like this, it is far preferable to use a mathematical tool called the binomial theorem in order to determine the number of possible outcomes that satisfy the criterion.

The binomial theorem looks like this:. Remember that! To apply the binomial theorem to the question at hand assuming you can still remember what the question was! This factor of the binomial theorem equation is the probability of each single birth order, as we saw previously in part d above. Use features like bookmarks, note taking and highlighting while reading Genetics, From Genes to Genomes.

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