P generation АА ВВ aa bb Gamete formation Gamete formation Gametes (А В (a b) Fertilization F1 generation Aa Bb Gamete formation Gametes (A B a b A b a B Original combinations of alleles (nonrecombinant gametes) New combinations of alleles (recombinant gametes) Conclusion: Through recombination, gametes contain new combinations of alleles.

P generation АА ВВ aa bb Gamete formation Gamete formation Gametes (А В (a b) Fertilization F1 generation Aa Bb Gamete formation Gametes (A B a b A b a B Original combinations of alleles (nonrecombinant gametes) New combinations of alleles (recombinant gametes) Conclusion: Through recombination, gametes contain new combinations of alleles.

Biology: The Unity and Diversity of Life (MindTap Course List)

14th Edition

ISBN:9781305073951

Author:Cecie Starr, Ralph Taggart, Christine Evers, Lisa Starr

Publisher:Cecie Starr, Ralph Taggart, Christine Evers, Lisa Starr

Chapter14: Chromosomes And Human Inheritance

Section: Chapter Questions

Problem 3GP: Human females have two X chromosomes (XX); males have one X and one Y chromosome (XY). a. With...

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Human females have two X chromosomes (XX); males have one X and one Y chromosome (XY). a. With respect to X-linked alleles, how many different types of gametes can a male produce? b. If a female is homozygous for an X-linked allele, how many types of gametes can she produce with respect to that allele? c. If a female is heterozygous for an X-linked allele, how many types of gametes can she produce with respect to that allele?
In cats, the genotype AA produces tabby fur color; Aa is also a tabby, and aa is black. Another gene at a different locus is epistatic to the gene for fur color. When present in its dominant W form (WW or Ww), this gene blocks the formation of fur color and all the offspring are white; ww individuals develop normal fur color. What fur colors, and in what proportions, would you expect from the cross AaWw Aa Ww?
A woman who sought genetic counseling is found to be heterozygousfor a chromosomal rearrangement between the second andthird chromosomes. Her chromosomes, compared to those in anormal karyotype, are diagrammed on the next page:(a) This woman is phenotypically normal. Does thissurprise you? Why or why not? Under what circumstancesmight you expect a phenotypic effect of such arearrangement?
A tomato geneticist attempts to assign five recessivemutations to specific chromosomes by using trisomics.She crosses each homozygous mutant (2n) with each ofthree trisomics, in which chromosomes 1, 7, and 10 takepart. From these crosses, the geneticist selects trisomicprogeny (which are less vigorous) and backcrosses themto the appropriate homozygous recessive. The diploidprogeny from these crosses are examined. Her results, inwhich the ratios are wild type:mutant, are as follows:Which of the mutations can the geneticist assign towhich chromosomes? (Explain your answer fully.)
Two mothers give birth to sons at the same time at a busy urbanhospital. The son of mother 1 has hemophilia, a disease causedby an X-linked recessive allele. Neither parent has the disease.Mother 2 has a son without hemophilia, despite the fact thatthe father has hemophilia. Several years later, couple 1 suesthe hospital, claiming that these two newborns were swappedin the nursery following their birth. As a genetic counselor, youare called to testify. What information can you provide the juryconcerning the allegation?
Two linked genes, (A) and (B), are separated by 42 m.u. A man with genotype AB/ab marries a woman who is ab/ab. What is the probability that their first child will be Ab/ab? Round your answer to the nearest %.
A man, Penoy, whose sister died in early childhood from a recessive lethal disease marries a woman Esmae, who has the same family history. Because Penoy has survived beyond childhood, he does not have the disease, but he may be a carrier (i.e. heterozygous, as may also be the case with Esmae). What is the probability that their first child will suffer from the disease? [Hint: first calculate the probability that Penoy is heterozygous; then determine the probability that both parents are carriers. Remember that he has survived to adulthood when calculating this probability].
A man is brachydactylous (very short fingers; rare autosomal dominant), and his wife is not. Both can taste thechemical phenylthiocarbamide (autosomal dominant;common allele), but their mothers could not.a. Give the genotypes of the couple.If the genes assort independently and the couple hasfour children, what is the probability ofb. all of them being brachydactylous?c. none being brachydactylous?d. all of them being tasters?e. all of them being nontasters?f. all of them being brachydactylous tasters?g. none being brachydactylous tasters?h. at least one being a brachydactylous taster?
The accompanying pedigree is for a rare, but relativelymild, hereditary disorder of the skin.a. How is the disorder inherited? State reasons for youranswer.b. Give genotypes for as many individuals in thepedigree as possible. (Invent your own defined allelesymbols.)c. Consider the four unaffected children of parentsIII-4 and III-5. In all four-child progenies from parentsof these genotypes, what proportion is expected tocontain all unaffected children?(picture added)
Mules result from a cross between a horse (2 n = 64) and a donkey(2 n = 62), have 63 chromosomes, and are almost always sterile.However, in the summer of 1985, a female mule named Krause who waspastured with a male donkey gave birth to a male foal (O. A. Ryder et al.1985. Journal of Heredity 76:379–381). Blood tests established that thefoal, appropriately named Blue Moon, was the offspring of Krause andthat Krause was indeed a mule. Both Blue Moon and Krause werefathered by the same donkey (see the accompanying pedigree). The foal,like his mother, had 63 chromosomes—half of them horse chromosomesand the other half donkey chromosomes. Analyses of genetic markersshowed that, remarkably, Blue Moon seemed to have inherited acomplete set of horse chromosomes from his mother, instead of therandom mixture of horse and donkey chromosomes that would beexpected with normal meiosis. Thus, Blue Moon and Krause were notonly mother and son, but also brother and sister.a. With the use of a diagram,…
A couple planning their family are aware that through the pastthree generations on the husband’s side a substantial numberof stillbirths have occurred and several malformed babies wereborn who died early in childhood. The wife has studied geneticsand urges her husband to visit a genetic counseling clinic,where a complete karyotype-banding analysis is performed.Although the tests show that he has a normal complement of46 chromosomes, banding analysis reveals that one memberof the chromosome 1 pair (in group A) contains an inversioncovering 70 percent of its length. The homolog of chromosome1 and all other chromosomes show the normal bandingsequence. Question: How would you explain the high incidence of paststillbirths?
A couple planning their family are aware that through the pastthree generations on the husband’s side a substantial numberof stillbirths have occurred and several malformed babies wereborn who died early in childhood. The wife has studied geneticsand urges her husband to visit a genetic counseling clinic,where a complete karyotype-banding analysis is performed.Although the tests show that he has a normal complement of46 chromosomes, banding analysis reveals that one memberof the chromosome 1 pair (in group A) contains an inversioncovering 70 percent of its length. The homolog of chromosome1 and all other chromosomes show the normal bandingsequence. Question: What can you predict about the probability of abnormality/normality of their future children?