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heterozygous punnett square

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Question:If two heterozygous brown-eyed individuals with dimpled chins were have to have children, punnett square will be 4 x 4 squares. Assume both independent assortment and segregation are occuring?

Answers:If you are wanting to know how to set the Punnett square up: I'm making a lot of assumptions here.... #1 - Brown eyes is dominant to whatever color.. #2 - Dimples are dominant to non dimples. Brown = B Not brown = b Dimples = D No dimples = d Across the 4 boxes on top use: BD, Bd, bD, bd Use the same for the boxes going down. I'm not doing the Punnett Square but I think your final ratio should be 9:3:3:1 Fill it out and count them. Good luck

Question:1. Show the Punnett Square cross results between two Dalmatian dogs that are heterozygous for spots? (For Dalmatian dogs, the spotted condition is dominant to non-spotted) 2. In humans, brown eyes are dominant over blue eyes. A woman with blue eyes marries a man with brown eyes. This man's father had blue eyes. (Please answer all 3 questions) a) What would be the genotype of the man that married the blue-eyed woman? b) What would be the genotype of the blue-eyed woman? c) What will be the ratio of children with blue eyes from this marriage? (please show the Punnett Square cross)

Answers:1. S=dominant allele for spots s=recessive allele for no spots Ss x Ss = SS, Ss, Ss, ss 25% offspring would be homozygous dominant with spots (SS) 50% would be heterozygous with spots (Ss) 25% would be homozygous recessive with no spots (ss) 2. B=dominant allele for brown eyes b= recessive allele for blue eyes. Woman is bb since she has blue eyes. If the mans father had blue eyes, the father would be bb too. So since the many has brown eyes, he has to be Bb. a. Bb b. bb. c. Bb x bb= Bb, bb, Bb, bb. 50% are Bb, brown eyes 50% are bb, blue eyes.

Question:QUESTION: how do I know the genotype for the male and female if the problem doesn't state whether or not they are homozygous or heterozygous. Q: In humans, there is a type of blindness due to a dominant gene. Normal vision is the result of a recessive gene. Migraine headaches are due to a dominant gene, and normal (no headaches) is recessive. A male who is blind and does not suffer from headaches marries a woman who has normal vision and suffers from migraines. Could they produce a child with normal vision who does not suffer from heacahes? If yes, can the probability of such child be determined? If B=blindness, b=normal vision, N=migraines and n=no headaches What is the male and females "codes"???? The male would either be: BBnn or Bbnn and the female could be bbNn or bbNN Do I do Punnett Squares for each possibility? Then how do I figure out the probability... or do you assume they are either homozygous or heterozygous............. please help!

Answers:Yes, they could. Whether the genotypes indicate male or female (which they don't) is irrelevant, since the alleles are not sex-linked. You don't need to do Punnett squares for each; they are both heterozygous for the genes in question. If B=blind and M=migraines (I use "M" because migraines is a dominant allele), the male would be Bbmm and the female would be bbMm. Look at the genes separately: Could BB x bb produce normal vision, bb? No, but Bb x bb could (with 50% probability). Could MM x mm produce no migraines, mm? No, but Mm x mm could (with 50% probability) Proof: ______Bm_______bm___ bM__BBMm____BbMm__ bm__Bbmm____bbmm___ child with normal vision, no migraines; 25% which makes sense, since 50% x 50% = 25%

Question:A. is not a carrier of the disorder. B. will not have the disorder. C. cannot have offspring with the disorder. D. will get the disorder late in life. 2. The Punnett square in Figure 7.1 shows a cross between two parents who are heterozygous for an autosomal genetic disorder caused by a recessive allele. People with which genotype will have the disorder? A. Ss parent B. ss offspring C. SS offspring D. Ss offspring 3. An XX female will express a recessive sex-linked trait if she A. has several codominant alleles for the gene. B. fully expresses her Y chromosome. C. carries the allele on an epistatic gene. D. is homozygous for the recessive allele. 4. Suppose a mouse is homozygous for alleles that produce black fur and homozygous for alleles of an epistatic gene that produces albinism. What color fur will the mouse have? A. Black B. White C. Gray D. Spotted 5. The crossing of wild type fruit flies with mutant fruit flies resulted in the conclusion that some A. traits are inherited as a group. B. chromosomes are inherited as a group. C. genes assort independently. D. traits assort independently. 6. A female is born with attached earlobes, which is a recessive phenotype. Which of the following statements about her parents must be true? A. Neither has the codominant allele. B. Both parents have the recessive allele. C. Her father has an inactivated allele. D. Her mother carries the dominant allele. 7. If more males than females in a family have a recessive sex-linked disorder, what can you infer about patterns of inheritance in that family? A. The males would pass on the disorder to sons. B. All females would be carriers of the disorder. C. Only females would be carriers of the disorder. D. Females would not develop the disorder. 8. Down syndrome is characterized by having an extra copy of at least a portion of chromosome twenty-one. Which of the following methods would quickly identify the disorder? A. Linkage map B. Meiosis map C. Karyotype D. Pedigree chart 9. __________ occurs in cells of female mammals. A. X chromosome revitalization B. Y chromosome inactivation C. X chromosome inactivation D. Y chromosome expression 10. A plant that is homozygous for red flowers is crossed with a plant that is homozygous for white flowers. The flowers of the offspring are all pink. This is an example of A. complete recession. B. incomplete recession. C. complete dominance. D. incomplete dominance. 11. Studying genetic disorders has resulted in much of what we know about A. single-expression dominance. B. somatic gene disruptions. C. multiple-gene traits in humans. D. single-gene traits in humans. 12. Traits that result from many genes are called A. monogenic traits. B. multigenic traits. C. polygenic traits. D. monolithic traits. 13. Suppose that the cross-over frequency between two genes is 7%. The distance between the two genes on a linkage map is A. .07 units. B. .7 units. C. 7 units. D. 70 units. 14. Gene expression is influenced by many factors. Which of the following is a factor in gene expression? A. Karyotype B. Pedigree C. Environment D. Phenotype 15. Two parents have the genotype Gg for a genetic disorder caused by a dominant allele. What is the chance that any of their children will inherit the disorder? (Hint: complete a Punnett square cross on a separate piece of paper.) A. 25% B. 50% C. 75% D. 100% 16. Human height occurs in a continuous range because it is affected by the interaction of several genes, making it a(n) A. autosomal trait. B. sex-linked trait. C. polygenic trait. D. codominant trait. 17. The gene linkage map shown in Figure 7.2 shows the order of genes A, B, and C. Which of the following statements about the genes is true? A. The distance between A and B is 14.5 map units. B. A and B cross over 2.5% of the time. C. A and C are linked 8.5% of the time. D. B and C are most likely to be inherited together. 18. Two genes cross over 6% of the time. This percentage means that the genes are A. inactivated in 6 out of 100 offspring. B. incompletely dominant in 6 out of 100 offspring. C. not inherited together in 6 out of 100 offspring. D. on sex chromosomes in 6 out of 100 offspring. 19. Suppose a person is a carrier for a genetic disorder. Which of the following phrases about this person is true? A. Does not have the disorder but can pass it on B. Will develop the disorder only late in life C. Cannot pass the disorder to sons, just daughters D. The allele is not passed on due to Y chromosome inactivation 20. What is the main reason that sex-linked disorders are most often observed in males? A. The X chromosome only has genes for genetic disorders.

Answers:Someone who is heterozygous for a recessive allele that causes a disorder ? B. will not have the disorder. 2. The Punnett square in Figure 7.1 shows a cross between two parents who are heterozygous for an autosomal genetic disorder caused by a recessive allele. People with which genotype will have the disorder? B. ss offspring 3. An XX female will express a recessive sex-linked trait if she D. is homozygous for the recessive allele. 4. Suppose a mouse is homozygous for alleles that produce black fur and homozygous for alleles of an epistatic gene that produces albinism. What color fur will the mouse have? C. Gray 5. The crossing of wild type fruit flies with mutant fruit flies resulted in the conclusion that some C. genes assort independently. 6. A female is born with attached earlobes, which is a recessive phenotype. Which of the following statements about her parents must be true? B. Both parents have the recessive allele. 7. If more males than females in a family have a recessive sex-linked disorder, what can you infer about patterns of inheritance in that family? C. Only females would be carriers of the disorder. 8. Down syndrome is characterized by having an extra copy of at least a portion of chromosome twenty-one. Which of the following methods would quickly identify the disorder? C. Karyotype 9. __________ occurs in cells of female mammals. C. X chromosome inactivation 10. A plant that is homozygous for red flowers is crossed with a plant that is homozygous for white flowers. The flowers of the offspring are all pink. This is an example of D. incomplete dominance. 11. Studying genetic disorders has resulted in much of what we know about D. single-gene traits in humans. 12. Traits that result from many genes are called C. polygenic traits. 13. Suppose that the cross-over frequency between two genes is 7%. The distance between the two genes on a linkage map is C. 7 units. 14. Gene expression is influenced by many factors. Which of the following is a factor in gene expression? C. Environment 15. Two parents have the genotype Gg for a genetic disorder caused by a dominant allele. What is the chance that any of their children will inherit the disorder? (Hint: complete a Punnett square cross on a separate piece of paper.) C. 75% 16. Human height occurs in a continuous range because it is affected by the interaction of several genes, making it a(n) C. polygenic trait. 18. Two genes cross over 6% of the time. This percentage means that the genes are C. not inherited together in 6 out of 100 offspring. 19. Suppose a person is a carrier for a genetic disorder. Which of the following phrases about this person is true? A. Does not have the disorder but can pass it on

From Youtube

Punnett Square Fun :Dihybrid crosses. Independent assortment. Incomplete dominance. Codominance and multiple alleles.

Punnett Square to determine possible blood types :In this example, the Punnett square is used to determine the possible blood types of the father while the mother is confirmed Type O and all children are confirmed Type O.