We could have a three dominant be one recess it be. And it's just that mirror here. We can have three. Dominant be. And when necessity be and three dominant, see one recess, ISI three dominant. See when recessive see same, Same, same over and over again. Three of the dominance c and one recessive.
Okay, So then the question first is probably that all trades are recessive can again, we know that this one is gonna have a 64 possibilities, but so we say, Oh, are recessive. So we say you go start here and there's gonna be one. Multiply that here. How we get one again and we multiply here and we end up with one out of So that checks The next one is that trades are dominant at each low side. Okay, So we're gonna have to start at the dominant a right, and so get three a.
Then we need the dominant. Be there's gonna be times three dominant be and times three dominant. See, that should be 27 for dominance at all. Three low side and then the next one to investigate are dominant at two, but recessive at a third.
Okay, so it's gonna be a little bit more complicated. So if we started dominance at a here, we could get three a. Then we can either dio times three b times one little C or we could also do three a times one little B, times three see, Or if we were to have won a the recessive, then we can hit the, um three each for the dominant B and C, sometimes to be times three c.
And if you like this, you were getting three times three times three, right. This is also gonna be 27 and finally trades dominant at one gene but recessive at the other two. This I can use black again. So if we're dominant at one gene and retested at the other two, we could either have three a then times one times one where you have one excessive A.
And then, um, times like three B. Then go to recess of C or we could be recessive in a and be but then dominant and see. Okay, so this is just three times three is equal to nine. Then that will actually conclude this problem. In chemistry and physics, an element is a substance that cannot be broken down into a simpler substance by chemical means. A pure element is a substance consisting of a single type of atom, with its chemical properties determined by that atom's atomic number, which is the number of protons in its nucleus.
Examples of elements include carbon, oxygen, aluminum, iron, gold, copper, mercury, and lead. In biology, the elements of life are the essential building blocks that make up living things. They are carbon, hydrogen, nitrogen, oxygen, phosphorus, and sulfur.
The first four of these are the most important, as they are used to construct the molecules that are necessary to make up living cells. These elements form the basic building blocks of the major macromolecules of life, including carbohydrates, lipids, nucleic acids and proteins. Carbon is an important element for all living organisms, as it is used to construct the basic building blocks of life, such as carbohydrates, lipids, and nucleic acids.
Even the cell membranes are made of proteins. Hydrogen is used to construct the molecules water and organic compounds with carbon. Nitrogen is used to construct the basic building blocks of life, such as amino acids, nucleic acids, and proteins. Oxygen is used to construct the basic building blocks of life, such as carbohydrates, lipids, and nucleic acids. Phosphorus is used to construct the basic building blocks of life, such as carbohydrates, lipids, and nucleic acids.
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Laporkan Dokumen Ini. Tandai sebagai konten tidak pantas. Unduh sekarang. Untuk Nanti. Judul terkait. Karusel Sebelumnya Karusel Berikutnya. The recombinants should total Since the F1 inherited a chromosome with heart-shaped leaves and many spines hl ns from one parent and a chromosome with normal leaves and few spines Hl Ns from the other parent, these are the nonrecombinant phenotypes, and together they total The two recombinant phenotypes are heart-shaped leaves with few spines hl Ns and normal-shaped leaves with many spines Hl ns.
Heart-shaped, many spines In tomatoes, tall D is dominant over dwarf d , and smooth fruit P is dominant over pubescent p fruit, which is covered with fine hairs.
A farmer has two tall and smooth tomato plants, which we will call plant A and plant B. The genotypes of both plants are DdPp. If so, what is the map distance between them? The two plants have different coupling configurations. In plant B, they are in repulsion; its chromosomes have D p and d P. The four phenotypic classes are not present in ratios no need for chi-square test , so they are linked. How do you know? In tomatoes, dwarf d is recessive to tall D and opaque light green leaves op is recessive to green leaves Op.
For each of the following crosses, determine the phenotypes and proportions of progeny produced. The most straightforward way is to do a Punnett square, including the. In Drosophila melanogaster, ebony body e and rough eyes ro are encoded by autosomal recessive genes found on chromosome 3; they are separated by 20 map units.
The gene that encodes forked bristles f is X-linked recessive and assorts independently of e and ro. Give the phenotypes of progeny and their expected proportions when each of the following genotypes is test-crossed. A series of two-point crosses were carried out among seven loci a, b, c, d, e, f, and g , producing the following recombination frequencies.
Map the seven loci, showing their linkage groups, the order of the loci in each linkage group, and distances between the loci of each linkage group. So one linkage group consists of a, g, and d. We know that gene g is between a and d because the a to d distance is Waxy endosperm wx , shrunken endosperm sh , and yellow seedling v are encoded by three recessive genes in corn that are linked on chromosome 5. A corn plant homozygous for all three recessive alleles is crossed with a plant homozygous for all the dominant alleles.
The resulting F1 are then crossed with a plant homozygous for the recessive genes in a three-point testcross. The nonrecombinants are Wx Sh V and wx sh v. The double crossovers are wx sh V and Wx Sh v. Comparing the two, we see that they differ only at the v locus, so v must be the middle gene. Fine spines s , smooth fruit tu , and uniform fruit color u are three recessive traits in cucumbers whose genes are linked on the same chromosome. A cucumber plant heterozygous for all three traits is used in a testcross and the progeny at the top of the following page are produced from this testcross.
Nonrecombinants are s u Tu and S U tu. Double crossovers are s u tu and S U Tu. Because Tu differs between the nonrecombinants and the double crossovers, Tu is the middle gene. S-Tu distance: recombinants are S Tu and s tu. U-Tu distance: recombinants are u tu and U Tu. In the correct gene order for the heterozygous parent: s Tu u and S tu U For the testcross parent: s tu u and s tu u.
The interference among these genes is 0. A fly with black body, purple eyes, and vestigial wings is crossed with a fly homozygous for gray body, red eyes, and normal wings. The female progeny are then crossed with males that have black body, purple eyes, and vestigial wings. If progeny are produced from this testcross, what will be the phenotypes and proportions of the progeny? Although we know what the recombination frequencies are between the pairs of genes, these recombination frequencies result from both single crossover sco and double crossover dco progenies.
So we must first calculate how many double crossover progeny we should get. We now use the C. These are therefore the nonrecombinant progeny phenotypes.
We calculated above that there will be four double crossover progeny, so we should expect two progeny flies of each double crossover phenotype. These total 56, or 28 each. These total , or 63 each. The two remaining phenotypic classes are the nonrecombinants.
The locations of six deletions have been mapped to the Drosophila chromosome shown on the following page. Recessive mutations a, b, c, d, e, and f are known to be located in the same region as the deletions, but the order of the mutations on the chromosome is not known. On the basis of these data, determine the relative order of the seven mutant genes on the chromosome. The mutations are mapped to the intervals indicated on the figure above the table. The location of f is ambiguous; it could be in either location shown above.
A panel of cell lines was created from mouse-human somatic-cell fusions. Each line was examined for the presence of human chromosomes and for the production of an enzyme.
The following results were obtained:. On the basis of these results, which chromosome has the gene that codes for the enzyme? The enzyme is produced only in cell lines B and E. Of all the chromosomes, only chromosome 8 is present in just these two cell lines and absent in all the other cell lines that do not produce the enzyme.
Therefore, the gene for the enzyme is most likely on chromosome 8. Each line was examined for the presence of human chromosomes and for the production of three enzymes. On the basis of these results, give the chromosome location of enzyme 1, enzyme 2, and enzyme 3. Enzyme 1 is located on chromosome 9. Chromosome 9 is the only chromosome that is present in the cell lines that produce enzyme 1 and absent in the cell lines that do not produce enzyme 1. Enzyme 2 is located on chromosome 4. Enzyme 3 is located on the X chromosome.
The X chromosome is the only chromosome present in the three cell lines that produce enzyme 3 and absent in the cell line that does not produce enzyme 3. In calculating map distances, we did not concern ourselves with whether double crossovers were two-stranded, three-stranded, or four-stranded; yet, these different types of double crossovers produce different types of gametes.
Can you explain why we do not need to determine how many strands take part in double crossovers in diploid organisms? Hint: Draw out the types of gametes produced by the different types of double crossovers and see how they contribute to the determination of map distances.
The three-stranded double crossovers all generate two recombinant chromosomes and two nonrecombinant chromosomes.
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