A DNA strand has the sequence ACCGAGCIT. Which is the complementary strand of RNA? OTGGCTCGAA O UGGCUCGAA O GTTAGTGCC CAATCTAGG

A person suffering from retrograde amnesia will

ololo11 [35]

lose past memories and the ability to acquire new

memories will remain unaffected.

7 0

2 years ago

Many protists can move. what are some structures mentioned that can help protists move.

Ksivusya [100]

Answer:

cilli and flagella

Explanation:

I think there is one more but not sure hope this helped

(also if spelling is weird sorry I tried)

6 0

2 years ago

Different ratios occur in crosses with single gene pairs or two gene pairs. What types of ratios are likely to occur in crosses

likoan [24]

Answer:

Genotype ratio: 1, 1:1, 1:2:1

Phenotype ratio: 1, 3:1

Explanation:

Single gene pair cross is also known as monohybrid cross. This means that only one gene usually with two alleles is observed and it express one trait.

For example, if we name the gene for a certain trait with A, the possible genotypes are AA (dominant homozygous), aa (recessive homozygous) and Aa (heterozygous). Possible crosses are:

P: AA x AA

F1 : all of them are AA

The same is with aa x aa (all of the offspring are with aa genotype)

P: AA x Aa

F1: AA Aa AA Aa (genotype ratio 1:1) (phenotype ratio 3:1)

The same genotype ratio is in aa x Aa (offspring will have aa Aa aa Aa-(genotype ratio 1:1) (phenotype ratio 1:1)

P: Aa x Aa

F1: AA Aa Aa aa (genotype ratio 1:2:1) (phenotype ratio 3:1)

P: AA x aa

F1: Aa Aa Aa Aa (1)

5 0

3 years ago

4) A homozygous groucho fly ( gro, bristles clumped above the eyes) is crossed with a homozygous rough fly (ro, eye abnormality)

docker41 [41]

Answer and Explanation:

A homozygous groucho fly ( gro, bristles clumped above the eyes) is crossed with a homozygous rough fly (ro, eye abnormality).

The F1 females are testcrossed, producing these offspring: groucho 518 rough 471 groucho, rough 6 wild-type 5 1000 a) What is the linkage distance between the two genes? B) Plot the genes on a map c) If the genes were unlinked and the F1 females were mated with the F1 males, what would be the offspring in the F2 generation?

1st cross:

Parental) grogro ro+ro+ x gro+gro+ roro

F1) gro+gro ro+ro

2nd cross:

Parental) gro+gro ro+ro x grogro roro

Gametes) gro+ro+ gro ro

gro+ro gro ro

gro ro+ gro ro

gro ro gro ro

Punnet square)

gro+ro+ gro+ro gro ro+ gro ro

gro ro gro+gro ro+ro gro+gro roro grogro ro+ro grogro roro

F2)

0.518 grogro ro+ro (518 individuals)

0.471 gro+gro roro (471 individuals)

0.006 grogro roro (6 individuals)

0.005 gro+gro ro+ro (5 individuals)

Total number of individuals 1000

Note: These frequencies were calculated dividing the number of individuals belonging to each genotype by the total number of individuals in the F2.

To know if two genes are linked, we must observe the progeny distribution. If individuals, whos genes assort independently, are test crossed, they produce a progeny with equal phenotypic frequencies 1:1:1:1. If we observe a different distribution, that is that phenotypes appear in different proportions, we can assume that genes are linked in the double heterozygote parent.

In the exposed example we might verify which are the recombinant gametes produced by the F1 di-hybrid, and we can recognize them by looking at the phenotypes with lower frequencies in the progeny.

By performing this cross we know that the phenotypes with lower frequencies in the progeny are groucho, rough and wild-type. So the recombinant gametes are gro+ro+ and gro ro, while the parental gametes are gro+ro and gro ro+.

So, the genotype, in linked gene format, of the double heterozygote individual in the F1 is gro+ro/gro ro+.

To calculate the recombination frequency we will make use of the next formula: P = Recombinant number / Total of individuals. The genetic distance will result from multiplying that frequency by 100 and expressing it in map units (MU). One centiMorgan (cM) equals one map unit (MU).

The map unit is the distance between the pair of genes for which one of every 100 meiotic products results in a recombinant product.

The recombination frequency is:

P = Recombinant number / Total of individuals

P = 6 + 5 / 1000

P = 11 / 1000

P = 0.011

The genetic distance between genes, is 0.011 x 100= 1.1 MU.

Genetic Linkage Map:

Parental Phenotypes)

-----gro+------ro---- -----gro------ro+----

----- gro ------ro---- ---- gro------ ro ----

Recombinant phenotypes)

-----gro+------ro+---- -----gro------ro----

----- gro ------ ro---- -----gro------ro----

If the genes were unlinked and the F1 females were mated with the F1 males, the offspring in the F2 generation would have been

4/16 = 1/4 gro+gro ro+ro

4/16 = 1/4 gro+gro roro

4/16 = 1/4 grogro ro+ro

4/16 = 1/4 grogro roro

Their phenotypic frequencies would be 1:1:1:1 related.

7 0

3 years ago

What’s the answers the question?

Sav [38]

Answer:

B- fold incorrectly

Explanation:

Prions are misfolded proteins with the ability to transmit their misfolded shape onto normal variants of the same protein. They characterize several fatal and transmissible neurodegenerative diseases in humans and many other animals.

3 0

3 years ago

A DNA strand has the sequence ACCGAGCIT. Which is the complementary strand of RNA? OTGGCTCGAA O UGGCUCGAA O GTTAGTGCC CAATCTAGG​

A DNA strand has the sequence ACCGAGCIT. Which is the complementary strand of RNA? OTGGCTCGAA O UGGCUCGAA O GTTAGTGCC CAATCTAGG